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Important Derivation Unit 1: Electrostatics
Coulomb’s Law in Vector Form
Let the position vectors of charges q1 and q2 be r1 and r2 respectively Fig. We denote force on q1 due to q2 by F12 and force on q2 due to q1by F21. The two point charges q1 and q2 have been numbered 1 and 2 for convenience and the vector leading from 1 to 2 is denoted by r21:
r21 = r2 – r1. In the same way, the vector leading from 2 to 1 is denoted by r12: r12 = r1 – r2 = – r21 .The magnitude of the vectors r21 andr12 is denoted by r21 and r12, respectively (r12 = r21). The direction of a vector is specified by a unit vector along the vector. To denote the direction from 1 to 2 (or from 2 to 1), we define the unit vectors:
Coulomb’s force law between two point charges q1 and q2 located at r1 and r2 is then expressed as
If q1 and q2 are of the same sign (either both positive and both negative), F21 is along ˆr 21, which denotes repulsion, as it should be for like charges. If q1 and q2 are of opposite signs, F21 is along – ˆr 21(= ˆr 12), which denotes attraction, as expected for unlike charges.
Thus, we do not have to write separate equations for the cases of like and unlike charges. The force F12 on charge q1 due to charge q2, is obtained from force F21, by simply interchanging 1 and 2, i.e.
Thus, Coulomb’s law agrees with the Newton’s third law.
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Newton’s Third Law Newton’s Third Law of Motion states: ‘To every action there is an equal and opposite reaction’. It must be remembered that action and reaction always act on different objects. The Third Law of Motion indicates that when one object exerts a force on another object, the second object instantaneously exerts a force back on the first object. These two forces are always equal in magnitude, but opposite in direction. These forces act on different objects and so they do not cancel each other. Thus, Newton’s Third Law of Motion describes the relationship between the forces of interaction between two objects. Whenever two bodies interact with each other, the force exerted by the first body on the second is called action. The force exerted by the second body on the first body is called reaction. The action and reaction are equal and opposite. For example, when we placed a wooden block on the ground, this block exerts a force equal to its weight, W = mg acting downwards to the ground. This is the action force. The ground exerts an equal and opposite force W’ = mg on the block in the upward direction. This is the reaction force.
Applications of Newton’s Third Law of Motion 1. A gun recoils when a bullet is fired from it: When a bullet is fired from a gun, the gun exerts a force on the bullet in the forward direction. This is the action force. The bullet also exerts an equal force on the gun in the backward direction. This is the reaction force. Due to the large mass of the gun it moves only a little distance backward by giving a jerk at the shoulder of the gunman. The backward movement of the gun is called the recoil of the gun.
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2. A man walking on the ground
3. Rowing a boat
4. A person is moving forward during swimming
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5. Rocket propulsion
Electric Field Intensity on Axial Line & Equatorial Line of an Electric Dipole
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Consider an electric dipole consisting of two point changes −q−q and +q+q separated by a small distance AB=2aAB=2a with center O and dipole moment P=q(2a)P=q(2a) Let OP=rOP=r E1E1= electric intensity at P due to change -q at A. ∴|E1−→|=14π∈0qAP2∴|E1→|=14π∈0qAP2 AP2=OP2+OA2AP2=OP2+OA2 =r2+a2=r2+a2 ∴|E1−→|=14π∈0qr2+a2∴|E1→|=14π∈0qr2+a2 E1−→=Pc−→E1→=Pc→ Let ∠PBA=∠PAB=θ∠PBA=∠PAB=θ. E1E1 has two components E1E1 can θθ along PR∥BAPR∥BA and E1sinθE1sinθ along PE⊥BAPE⊥BA . If E2E2 is the electric intensity at P, due to change +q at B, then E2−→=14π∈0qBp2=14π∈0q(r2+a2)E2→=14π∈0qBp2=14π∈0q(r2+a2) E2−→E2→ is along PD−→−PD→ This has two components, E2E2 is along PR∥BAPR∥BA and E2sinθE2sinθ along PF. E→=2E1cosθE→=2E1cosθ =24π∈0.q(r2+q2)=24π∈0.q(r2+q2)cosθcosθ =24π∈0.q(r2+a2)(OAAD)=24π∈0.q(r2+a2)(OAAD) =24π∈0.q(r2+q2)ar2+a2−−−−−√=24π∈0.q(r2+q2)ar2+a2 =q×2a4π∈0(r2+a2)3/2=q×2a4π∈0(r2+a2)3/2 But q×2a=|P→|q×2a=|P→|, the dipole moment ∴|E→|=|P→|4π∈0(r2+a2)3/2∴|E→|=|P→|4π∈0(r2+a2)3/2 The direction of E→E→ is along . PR−→−||BA−→−PR→||BA→ (ie) opposite to P→P→ or E→=−P→4π∈0(r2+a2)3/2E→=−P→4π∈0(r2+a2)3/2 If the dipole is short 2a<
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Electric Potential due to a point charge Consider two points 'a' and 'b' in an electrostatic field of a single isolated point charge +q.
If a unit positive charge 'q' moves from 'a' to 'b' without acceleration, then the potential difference between 'a' and 'b' is given as
= - Edl But dl = - dr [This is because when we move a distance 'dl' towards the source, we move in the direction of decreasing of 'r']
From equation (1), we have
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If the point 'a' is at infinity, then
From the above, it is evident that for a given charge 'q', potential depends only on 'r'. Therefore, if the charge is positive, potential is positive and if the charge is negative, potential is negative. Electrical Potential on Axial Line & Equatorial Line of Electric Dipole To calculate electric field created by a dipole on the axial line ( on the same line joining the two charges),
All the measurement of distances are to be taken from the centre(O).
Let the distance between O to +q and O to –q be ‘l’. So, total length between +q and –q will be ‘2l’.
Take a point ‘p’ on the axial line at the distance ‘r’ from the centre as shown in figure.
Now, we wish to calculate electric field at point ‘P’.
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By using the formula for electric field due to point charge, Electric field due to +q = +1 4πε0 q (r - l)2 The distance between (P and +q) = (r-l) Electric field due to -q = -1 4πε0 q (r + l)2 The distance between (P and -q) = (r + l) (Electric field due to +q will be positive and electric field due to- q will be negative) Since electric field is a vector quantity so, the net electric field will be the vector addition of the two. So, the net electric field E = E1 + E2 E = 1 4πε0 q (r - l)2 - 1 4πε0 q (r + l)2 E = q 4πε0 [1 (r - l)2 - 1 (r + l)2 ] On solving the equation we get − E = q 4πε0 [(r + l)2 - (r - l)2 (r - l)2(r + l)2 ] E = q 4πε0 4rl (r2 - l2)2 ......(1) We know that the dipole moment or effectiveness of dipole (P) is given by − P = 2ql Therefore, putting this value in eq(1), we get E = 1 4πε0 2Pr (r2 - l2)2 ......(2) Certain assumptions are made based on this equation − Since, the dipole is very small so ‘l’ is also very small as compared to the distance ‘r’. So, on neglecting ‘r’ with respect to ‘l’ we get − E = 1 4πε0 2Pr r4 (from eq(2)) ⇒ E = 1 4πε0 2P r2 E = 1 4πε0 2P r2
Electric Field Intensity at various point due to a uniformly charged spherical shell (thin and thick both)
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To find the magnitude of the electric field outside the charged shall, we evaluate the electric flux by using a spherical G concentric with the shell.
Since the electric field is everywhere perpendicular to the Gaussian surface, the angle between and the elemental a value at all points on the surface, since they are equidistant from the charged shell. Being constant, E can be taken out fr
According to Gauss's law, Q is the net charge in the Gaussian surface and is given as 'q'.
This is a surprising result, for it is the same as that for a point charge. Thus, the electric field outside a uniformly charge charge q were concentrated as a point charge at the centre of the shell. Electric field inside the charged shell :
In this case we select a spherical Gaussian surface that lies inside the shell, concentric with it and with radius r(r < R). Ins must also have spherical symmetry. So the electric flux through the Gaussian surface is
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According to Gauss's law,
But the net charge enclosed by the surface is zero, since all the charge lies on the shell that is outside the Gaussian surfac
Hence the electric field due to a uniformly charged spherical shell is zero at all points inside the shell. Electric field on the surface of shell : In this case, the Gaussian surface is the shell itself i.e., r = R
where is the surface charge density The variation of electric field intensity with the distance from the centre of a uniformly charged spherical shell is represent that the electric field inside the shell is zero, maximum on the surface of the shell. The electric field intensity varies inver the centre, at a point outside the spherical shell.
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Electric Potential due to a thin infinite plan sheet of charge You know that if you have a point charge with charge QQ, then the potential difference VV between spatial infinity and any point a distance rr from the charge is given by Vpoint=kQr.Vpoint=kQr. You also know that the electric field from an infinite sheet of charge with charge density σσ is given by Esheet=2πkσ.Esheet=2πkσ. Because the electric field is uniform, you correctly concluded that there must be an infinite potential difference between any point and spatial infinity. You are surprised because this seems at odds with the first formula for VpointVpoint. However, there is a good explanation. If kQrkQr is originally for a point charge, what values of QQ and rrshould we plug in for the case of a sheet? Well, notice that the sheet has an infinite amount of charge, so that perhaps QQ should be infinite. This explains why we might get an infinite potential difference. However, there is a competing effect occuring with rr. As you go farther out on the infinite sheet, you get farther and farther away from the point where you are trying to compute the potential, so it seems like maybe rr should be very big, maybe infinitely big as well. Let's see how to do the problem correctly. To do the problem correctly, you need to realize that each point on the infinite sheet acts like a little point charge, so each point gives its own kQrkQr contribution. The total potential, by superposition, is the sum of these contributions. We can sum up the contributions by integration. Let's first pick a coordinate system where the plate is on the xx-yy plane, and the point where we want to know the potential is on the zz axis. We can switch to cylindrical coordinates where ρ=x2+y2−−−−−−√ρ=x2+y2. Then the distance rr between the point with coordinate zz on the zz axis and a point with coordinate ρρ is given by r=z2+ρ2−−−−−−√r=z2+ρ2, and so, applying the kQ/rkQ/r formula, the contribution dVdV to the potential from a bit of charge dQdQ a distance ρρ from the origin is given by dV=kdQz2+ρ2−−−−−−√.dV=kdQz2+ρ2. Integrating this over all ρρ we find V=∫∞02πkσρdρz2+ρ2−−−−−−√=πkσ∫∞0duz2+u−−−−−√=2πkσ(∞+z2−−−−−−√−|z|).V=∫0∞2πkσρdρz2+ρ2= πkσ∫0∞duz2+u=2πkσ(∞+z2−|z|).
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Because of the infinity in the square root, the potential above is in fact infinite, even though were started with a finite kQ/rkQ/r law. This infinity was possible because we had infinitely much QQ. Notice the electric field still works out because the infinite part does not have a spatial gradient: E=−dVdz=−2πkσ(z∞+z2−1)z^=2πkσz^. Capacitance of parallel plate capacitor (with or without dielectric medium) Let us take a parallel plate capacitor. Suppose the separation distance between the plates is d. Use air or vacuum as a medium for this experiment. Suppose +Q is the charge on one plate and –Q is charge on the second plate. Bring a rectangular slab made up of conducting material between the plates of the capacitor. The thickness of the slab must be less than the distance between the plates of the capacitor. When the electric field will be applied then polarization of molecules will be started. The polarization will take place in the direction same as that of electric field. Consider a vector that must be polarized, name it as P. The polarization vector must be in the direction of electric field Eo. Then this vector will start its functioning and will produce an electric field Ep in the opposite direction to that of Eo. The net electric field in the circuit is shown by the figure.
E=
Eo
–
Ep
The electric field Eo in the outside region of the dielectric will be null. Now the equation of the potential difference between the plates will be : V=o (d-t) + Et But Eo=Er orK Therefore E= Eo / k So
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V= E o (d-t) + Eot / k V= E o [d-t+t/k] As we know Eo = / =Q/A V= Q / A [d-t+t/k] Capacitance of the capacitor is shown in the equation below: C= Q / V= A / (d-t+t/k) = A /d-t (1-1/K) I.e. C=
A/ d-t (1-1/k) ——- 2.31
So,
C
>
Co
clearly, it is proved that if a dielectric slab is placed in the plates of a capacitor then its capacitance will increase by some amount.
Application of Gauss Law Derivation of Coulumb's Law
Coulumb's law can be derived from Gauss's law. Consider electric field of a single isolated positive charge of magnitude q as shown below in the figure.
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Field of a positive charge is in radially outward direction everywhere and magnitude of electric field intensity is same for all points at a distance r from the charge. We can assume Gaussian surface to be a sphere of radius r enclosing the charge q. From Gauss's law
since E is constant at all points on the surface therefore,
surface area of the sphere is A=4πr2 thus,
Now force acting on point charge q' at distance r from point charge q is
This is nothing but the mathematical statement of Coulomb's law. (B) Electric field due to line charge
Consider a long thin uniformly charged wire and we have to find the electric field intensity due to the wire at any point at perpandicular distance from the wire.
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If the wire is very long and we are at point far away from both its ends then field lines outside the wire are radial and would lie on a plane perpandicular to the wire. Electric field intensity have same magnitude at all points which are at same distance from the line charge. We can assume Gaussian surface to be a right circular cylinder of radius r and length l with its ends perpandicular to the wire as shown below in the figure.
λ is the charge per unit length on the wire. Direction of E is perpendicular to the wire and components of E normal to end faces of cylinder makes no contribution to electric flux. Thus from Gauss's law
Now consider left hand side of Gauss's law
Since at all points on the curved surface E is constant. Surface area of cylinder of radius r and length l is A=2πrl therefore,
Charge enclosed in cylinder is q=linear charge density x length l of cylinder, or, q=λl From Gauss's law
Thus electric field intensity of a long positively charged wire does not depends on length of the wire but on the radial distance r of points from the wire.
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Unit No 2: Current Electricity Expression of drift velocity When conductor is subjected to an electric field E, each electron experience a force. F=-eE and acquires an acceleration a = F/m = - eE/m -----(i) Here m = mass of election, e = charge, E = electric field. The average time difference between two consecutive collisions is known as relaxation time of election.
(ii) According to drift velocity expression, relaxation time is the time interval between successive collisions of an electron on increasing temperature, the electrons move faster and more collisions occur quickly. Hence, relaxation time decreases with increase in temperature which implies that drift velocity also decreases with temperature. Relation between drift velocity and current Let l is the length of the conductor and A uniforms area of cross-section. Therefore, the volume of the conductor = Al If n is the number of free electrons per unit volume of the conductor, then the total number of free electrons in the conductor=A/n. If e is the charge on each electron then total charge on all the free electrons in the conductor Q=A/ne (1) Let a constant potential differences V is applied across the ends of the conductor with the help of a battery
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The electric field set up across the conductor is given by E=V/l Due to this field, the free electrons present in the conductor will begin to move with a drift velocity vd towards the left hand side of the conductor. Thus the time taken by the free electrons to cross the conductor is t=l/vd (2) As current I=q/t
(3)
By substituting equation (1) and (2) in equation (3), We get I=Alne/l/vd Or I=Anevd (4) Since A,n and e are constant I∞vd Hence the current flowing through a conductor is directly proportional to the drift velocity. Explanation and Example. In a conductor there are large number of free electrons. When we close the circuit , the electric field is established in the circuit with the speed of electromagnetic wave which causes electron drift at every portion of the circuit. Due to which the current is set up in the entire circuit instantly .The current thus set up does not wait for the electrons to flow from one end of the conductor to another end. It is due to this reason, the electric bulb glows immediately when switch is on.
Series and parallel connection of resistors Resistors are probably the most commonly occurring components in electronic circuits. Practical circuits often contain very complicated combinations of resistors. It is, therefore, useful to have a set of rules for finding the equivalent resistance of some general arrangement of resistors. It turns out that we can always find the equivalent resistance by repeated application of two simple rules. These rules relate to resistors connected in series and in parallel.
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Figure 18: Two resistors connected in series. Consider two resistors connected in series, as shown in Fig. 18. It is clear that the same current flows through both resistors. For, if this were not the case, charge would build up in one or other of the resistors, which would not correspond to a steady-state situation (thus violating the fundamental assumption of this section). Suppose that the potential drop from point to point is . This drop is the sum of the potential drops Thus,
and
across the two resistors
and
, respectively.
(135)
According to Ohm's law, the equivalent resistance potential drop across these points and the current
between and is the ratio of the which flows between them. Thus, (136)
giving (137)
Here, we have made use of the fact that the current is common to all three resistors. Hence, the rule is The equivalent resistance of two resistors connected in series is the sum of the individual resistances.
For
resistors connected in series, Eq. (137) generalizes to
.
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Figure 19: Two resistors connected in parallel. Consider two resistors connected in parallel, as shown in Fig. 19. It is clear, from the figure, that the potential drop across the two resistors is the same. In general, however, the currents
and
which flow through resistors
and
, respectively, are different. According to
Ohm's law, the equivalent resistance between and is the ratio of the potential drop across these points and the current which flows between them. This current must equal the sum of the currents and flowing through the two resistors, otherwise charge would build up at one or both of the junctions in the circuit. Thus, (138)
It follows that (139)
giving (140)
Here, we have made use of the fact that the potential drop is common to all three resistors. Clearly, the rule is The reciprocal of the equivalent resistance of two resistances connected in parallel is the sum of the reciprocals of the individual resistances.
For
resistors connected in parallel, Eq. (140) generalizes to
Application of Gaus Theorem is calculation of electric field intensity due to a thin infinite plane sheet of charge and uniformly charged spherical shell Let point P lies outside the charged sphere , In order to find the intensity of electric field at P,we draw a concentric Gaussian spherical surface of radius r through P. All points on this surface are at the same distance from the centre. Due to spherically symmetric charge distribution the magnitude of intensity of electric field at all points on the Gaussian sphere is same. As q is positive, the field is radially outward in
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direction.Therefore,the electric flux dφ through any area element dS taken on the surface,where E and area vector dS both are directed radially outward ,that is ,the angle between them is zero ,is given by Dφ=E.dS=Edscos00=EdS Thus, the electric flux through the entire Gaussian sphere as discussed in the gauss law for electrostatics is Φ=∫E.dS=∫EdS Or
φ=E∫dS
But ∫dS=4πr2 2 Φ=E(4πr ) (1) According to the Gauss’s law, Φ=q/ε0 (2) By comparing equations(1) and (2),we get E(4πr2)=q/ε0 Or E=q/4πε0r2 (3) Conclusion .The elementary of electric field due to a uniformly charged sphere at an external point (r>R) is identical with the field intensity due to an equal point charge placed at the centre or charged sphere behaved as if its entire charge is concebtrated as its centre. (2) Electric Field Due To A Uniform Charged Sphere At a Point on the surface of the sphere (r=R):If the point P is situated t the surface of uniformly charged sphere,then the distance of P from the centre of the sphere O is equal to its radius ,that is r=R Thus, the intensity of electric field at the surface of the sphere is obtained by replacing r by R in equation(3) Thus E=1/4πε0 q/r2 (3) Electric Field Due To A Uniform Charged Sphere At an Internal Point(r
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Q1=ρ*volume Or q1=[q/(4/3πR3].4/3πr3 Or q1=qr3/R3 (5) According to Gauss’s law Φ=∫E.dS=q1/ε0 (6) Comparing equations (4) and (6),we get E(4πr2)=q1/ε0 (7) Substituting the value of q1 from equation (5) in equation (7),we get E(4πr2)=1/ε0(qr3/R3) Or E=qr/4πε0R3 Conclusion: The intensity of electric field due to a uniformly charged non-conducting sphere at an internal point is directly proportional to the distance of the point from the centre, A current carrying wire generates a magnetic field. According to Biot-Savart’s law, the magnetic field at a point due to an element of a conductor carrying current is, 1. Directly proportional to the strength of the current, i 2. Directly proportional to the length of the element, dl 3. Directly proportional to the Sine of the angle θ between the element and the line joining the element to the point and 4. Inversely proportional to the square of the distance r between the element and the point.
Thus, the magnetic field at O is dB, such that,
Then,
where,
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is the proportionality constant and is called the permeability of free space. Then,
In vector form,
biot-savert law in the calculation of magnetic field due to current carrying circular loop Consider a circular coil of radius r, carrying a current I. Consider a point P, which is at a distance x from the centre of the coil. We can consider that the loop is made up of a large number of short elements, generating small magnetic fields. So the total field at P will be the sum of the contributions from all these elements. At the centre of the coil, the field will be uniform. As the location of the point increases from the centre of the coil, the field decreases.
By Biot- Savart’s law, the field dB due to a small element dl of the circle, centered at A is given by,
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This can be resolved into two components, one along the axis OP, and other PS, which is perpendicular to OP. PS is exactly cancelled by the perpendicular component PS’ of the field due to a current and centered at A’. So, the total magnetic field at a point which is at a distance x away from the axis of a circular coil of radius r is given by,
If there are n turns in the coil, then
where µ0 is the absolute permeability of free space.
Since this field Bx from the coil is acting perpendicular to the horizontal intensity of earth’s magnetic field, B0, and the compass needle align at an angle θ with the vector sum of these two fields, we have from the figure
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The horizontal component of the earth’s magnetic field varies greatly over the surface of the earth. For the purpose of this simulation, we will assume its magnitude to be B0 = 3.5x10-5 T. The variation of magnetic field along the axis of a circular coil is shown here.
Ampere’s law and its application
Ampere's circuital law in magnetism is analogous to gauss's law in electrostatics This law is also used to calculate the magnetic field due to any given current distribution This law states that " The line integral of resultant magnetic field along a closed plane curve is equal to μ0 time the total current crossing the area bounded by the closed curve provided the electric field inside the loop remains constant" Thus
where μ0 is the permeability of free space and Ienc is the net current enclosed by the loop as shown below in the figure
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The circular sign in equation (21) means that scalar product B.dl is to be integrated around the closed loop known as Amperian loop whose beginning and end point are same Anticlockwise direction of integration as chosen in figure 9 is an arbitrary one we can also use clockwise direction of integration for our calculation depending on our convenience To apply the ampere's law we divide the loop into infinitesimal segments dl and for each segment, we then calculate the scalar product of B and dl B in general varies from point to point so we must use B at each location of dl Amperion Loop is usually an imaginary loop or curve ,which is constructed to permit the application of ampere's law to a specific situation Proof Of Ampere's Law Consider a long straight conductor carrying current I perpendicular to the page in upward direction as shown below in the figure
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From Biot Savart law, the magnetic field at any point P which is at a distance R from the conductor is given by
Direction of magnetic Field at point P is along the tangent to the circle of radius R withTh conductor at the center of the circle For every point on the circle magnetic field has same magnitude as given by
And field is tangent to the circle at each point The line integral of B around the circle is
since ∫dl=2πR ie, circumference of the circle so,
This is the same result as stated by Ampere law This ampere's law is true for any assembly of currents and for any closed curve though we have proved the result using a circular Amperean loop If the wire lies outside the amperion loop, the line integral of the field of that wire will be zero
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but does not necessarily mean that B=0 everywhere along the path ,but only that no current is linked by the path while choosing the path for integration ,we must keep in mind that point at which field is to be determined must lie on the path and the path must have enough symmetry so that the integral can be evaluated
Force on a moving charge in uniform magnetic and electric fields A moving charge experience a force in a magnetic field. F=BqvsinθF=Bqvsinθ, where
F is force acting on a current carrying conductor B is magnetic flux density q is magnitude of charge v is velocity of charge θθ is the angle the velocity makes with the magnetic field
Derivation of above equation: Consider a positive charge q moving at constant speed v at right angles to a magnetic field of flux density B. Assuming that the charge travels a distance l in time t, its speed is v=ltv=lt . The moving charge can be seen as a current of I=qtI=qt . Hence, the force on the charge is given by: F=BIlF=BIl F=B(qt)lF=B(qt)l, substituting l F=BqvF=Bqv, substituting v Effect of magnetic force on charge Work done on the charged particle is always zero as F is ALWAYS perpendicular to the direction of travel(or displacement) and the distance travelled in the direction of the force is zero.
Hence, no energy is gained or lost by the particle moving through the magnetic field and the particle’s speed is always constant.
Since the force is of constant magnitude and it always at right angles to the displacement, the conditions are met for circular motion. Hence, the magnetic force on a moving charge provides a centripetal force to the charge. mv2r=Bqvmv2r=Bqv, where m is mass of moving charge and r is radius of orbit B=mvqrB=mvqr Making r the subject,
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r=mvqBr=mvqB r∝vr∝v,if charge and mass are constant. If the charged particle enters the uniform field at angles other than right angles, it will take a spiral(or helical) path as shown in the figure below.
Reason: Component of velocity parallel to the magnetic field is unaffected by magnetic field and the particle will continue to drift along parallel to the magnetic field in addition to moving in circular motion
Cyclotron
Cyclotron is a device used to accelerate charged particles to high energies. It was devised by Lawrence.
Principle
Cyclotron works on the principle that a charged particle moving normal to a magnetic field experiences magnetic lorentz force due to which the particle moves in a circular path.
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Construction
It consists of a hollow metal cylinder divided into two sections D1 and D2 called Dees, enclosed in an evacuated chamber (Fig 3.21). The Dees are kept separated and a source of ions is placed at the centre in the gap between the Dees. They are placed between the pole pieces of a strong electromagnet. The magnetic field acts perpendicular to the plane of the Dees. The Dees are connected to a high frequency oscillator.
Working: When a positive ion of charge q and mass m is emitted from the source, it is accelerated towards the Dee having a negative potential at that instant of time. Due to the normal magnetic field, the ion experiences magnetic lorentz force and moves in a circular path. By the time the ion arrives at the gap between the Dees, the polarity of the Dees gets reversed. Hence the particle is once again accelerated and moves into the other Dee with a greater velocity along a circle of greater radius. Thus the particle moves in a spiral path of increasing radius and when it comes near the edge, it is taken out with the help of a deflector plate (D.P). The particle with high energy is now allowed to hit the target T.
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When the particle moves along a circle of radius r with a velocity v, the magnetic Lorentz force provides the necessary centripetal force.
Bqv = (vm2 ) / r v /r = Bq / m = constant ...(1) The time taken to describe a semi-circle t=πr/v
(2)
Substituting equation (1) in (2), t = π m/ Bq
.. (3)
It is clear from equation (3) that the time taken by the ion to describe a semi-circle is independent of (i) the radius (r) of the path and (ii) the velocity (v) of the particle Hence, period of rotation T = 2t T = 2 π m / Bq = constant ...(4) So, in a uniform magnetic field, the ion traverses all the circles in exactly the same time. The frequency of rotation of the particle, v = 1 /T = Bq / 2 πm
.. (5)
If the high frequency oscillator is adjusted to produce oscillations of frequency as given in equation (5), resonance occurs. Cyclotron is used to accelerate protons, deutrons and α - particles.
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Limitations
Maintaining a uniform magnetic field over a large area of the Dees is difficult. At high velocities, relativistic variation of mass of the particle upsets the resonance condition. At high frequencies, relativistic variation of mass of the electron is appreciable and hence electrons cannot be accelerated by cyclotron.
Force on a current carrying in a uniform magnetic field Because charges ordinarily cannot escape a conductor, the magnetic force on charges moving in a conductor is transmitted to the conductor itself.
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Figure 1. The magnetic field exerts a force on a current-carrying wire in a direction given by the right hand rule 1 (the same direction as that on the individual moving charges). This force can easily be large enough to move the wire, since typical currents consist of very large numbers of moving charges. We can derive an expression for the magnetic force on a current by taking a sum of the magnetic forces on individual charges. (The forces add because they are in the same direction.) The force on an individual charge moving at the drift velocity vd is given by F = qvdB sin θ. Taking B to be uniform over a length of wire l and zero elsewhere, the total magnetic force on the wire is then F = (qvdB sin θ)(N), where N is the number of charge carriers in the section of wire of length l. Now, N = nV, where n is the number of charge carriers per unit volume and V is the volume of wire in the field. Noting that V = Al, where A is the cross-sectional area of the wire, then the force on the wire is F = (qvdB sin θ) (nAl). Gathering terms, F=(nqAvd)lBsinθF=(nqAvd)lBsinθ. Because nqAvd = I (see Current), F=IlBsinθF=IlBsinθ is the equation for magnetic force on a length l of wire carrying a current I in a uniform magnetic field B, as shown in Figure 2. If we divide both sides of this expression by l, we find that the magnetic force per unit length of wire in a uniform field is Fl=IBsinθFl=IBsinθ. The direction of this force is given by RHR-1, with the thumb in the direction of the current I. Then, with the fingers in the direction of B, a perpendicular to the palm points in the direction of F, as in Figure 2.
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Figure 2. The force on a current-carrying wire in a magnetic field is F = IlB sin θ. Its direction is given by RHR-1. Force between two parallel current carrying conductors
It is experimentally established fact that two current carrying conductors attract each other when the current is in same direction and repel each other when the current are in opposite direction Figure below shows two long parallel wires separated by distance d and carrying currents I1 and I2
Consider fig 5(a) wire A will produce a field B1 at all near by points .The magnitude of B1 due to current I1 at a distance d i.e. on wire b is B1=μ0I1/2πd ----(8) According to the right hand rule the direction of B1 is in downward as shown in figure (5a)
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Consider length l of wire B and the force experienced by it will be (I2lXB) whose magnitude is
Direction of F2 can be determined using vector rule .F2 Lies in the plane of the wires and points to the left From figure (5) we see that direction of force is towards A if I2 is in same direction as I1 fig( 5a) and is away from A if I2 is flowing opposite to I1 (fig 5b) Force per unit length of wire B is
Similarly force per unit length of A due to current in B is
and is directed opposite to the force on B due to A. Thus the force on either conductor is proportional to the product of the current We can now make a conclusion that the conductors attract each other if the currents are in the same direction and repel each other if currents are in opposite direction
Torque experienced by a current loop in uniform magnetic fields Let us consider a rectangular loop PQRS of length l and breadth b (Fig 3.24). It carries a current of I along PQRS. The loop is placed in a uniform magnetic field of induction B. Let θ be the angle between the normal to the plane of the loop and the direction of the magnetic field.
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Magnitude of the force F4 = BIl sin 90o = BIl F4 acts perpendicular to the plane of the paper and inwards. The forces F3 and F4 are equal in magnitude, opposite in direction and have different lines of action. So, they constitute a couple. Hence, Torque = BIl × PN = BIl × PS × sin θ (Fig 3.25) = BIl × b sin θ = BIA sin θ If the coil contains n turns, τ = nBIA sin θ So, the torque is maximum when the coil is parallel to the magnetic field and zero when the coil is perpendicular to the magnetic field. 1. Moving coil galvanometer Moving coil galvanometer is a device used for measuring the current in a circuit. Principle Moving coil galvanometer works on the principle that a current carrying coil placed in a magnetic field experiences a torque. Construction It consists of a rectangular coil of a large number of turns of thin insulated copper wire wound over a light metallic frame (Fig 3.26). The coil is suspended between the pole pieces of a horse-shoe magnet by a fine phosphor – bronze strip from a movable torsion head. The lower end of the coil is connected to a hair spring (HS) of phosphor bronze having only a few turns. The other end of the spring is connected to a binding screw. A soft iron cylinder is placed symmetrically inside the coil. The hemispherical magnetic poles produce a radial magnetic field in which the plane of the coil is parallel to the magnetic field in all its positions (Fig 3.27). A small plane mirror (m) attached to the suspension wire is used along with a lamp and scale arrangement to measure the deflection of the coil.
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Let PQRS be a single turn of the coil (Fig 3.28). A current I flows through the coil. In a radial magnetic field, the plane of the coil is always parallel to the magnetic field. Hence the sides QR and SP are always parallel to the field. So, they do not experience any force. The sides PQ and RS are always perpendicular to the field. PQ = RS = l, length of the coil and PS = QR = b, breadth of the coil Force on PQ, F = BI (PQ) = BIl. According to Fleming’s left hand rule, this force is normal to the plane of the
coil and acts outwards.
Force on RS, F = BI (RS) = BIl.
This force is normal to the plane of the coil and acts inwards. These two equal, oppositely directed parallel forces having different lines of action constitute a couple and deflect the coil. If there are n turns in the coil,
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moment of the deflecting couple = n BIl × b (Fig 3.29) moment of the deflecting couple = nBIA When the coil deflects, the suspension wire is twisted. On account of elasticity, a restoring couple is set up in the wire. This couple is proportional to the twist. If θ is the angular twist, then, moment of the restoring couple = Cθ where C is the restoring couple per unit twist At equilibrium, deflecting couple = restoring couple nBIA = Cθ
i.e I α θ. Since the deflection is directly proportional to the current flowing through the coil, the scale is linear and is calibrated to give directly the value of the current. 2. Pointer type moving coil galvanometer The suspended coil galvanometers are very sensitive. They can measure current of the order of 108 ampere. Hence these galvanometers have to be carefully handled. So, in the laboratory, for experiments like Wheatstone’s bridge, where sensitivity is not required, pointer type galvanometers are used. In this type of galvanometer, the coil is pivoted on ball bearings. A lighter aluminium pointer attached to the coil moves over a scale when current is passed. The restoring couple is provided by a spring. 3. Current sensitivity of a galvanometer.
The current sensitivity of a galvanometer is defined as the deflection produced when unit current passes through the galvanometer. A galvanometer is said to be sensitive if it produces large deflection for a small current.
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The current sensitivity of a galvanometer can be increased by
1. increasing the number of turns 2. increasing the magnetic induction 3.
increasing the area of the coil
4. decreasing the couple per unit twist of the suspension wire. This explains why phosphor-bronze wire is used as the suspension wire which has small couple per unit twist. 4. Voltage sensitivity of a galvanometer The voltage sensitivity of a galvanometer is defined as the deflection per unit voltage.
where G is the galvanometer resistance. An interesting point to note is that, increasing the current sensitivity does not necessarily, increase the voltage sensitivity. When the number of turns (n) is doubled, current sensitivity is also doubled (equation 1). But increasing the number of turns correspondingly increases the resistance (G). Hence voltage sensitivity remains unchanged. 5. Conversion of galvanometer into an ammeter A galvanometer is a device used to detect the flow of current in an electrical circuit. Eventhough the deflection is directly proportional to the current, the galvanometer scale is not marked in ampere. Being a very sensitive instrument, a large current cannot be passed through the galvanometer, as it may damage the coil. However, a galvanometer is converted into an ammeter by connecting a low resistance in parallel with it. As a result, when large current flows in a circuit, only a small fraction of the current passes through the galvanometer and the remaining larger portion of the current passes through the low resistance. The low resistance connected in parallel with the galvanometer is called shunt resistance. The scale is marked in ampere.
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The value of shunt resistance depends on the fraction of the total current required to be passed through the galvanometer. Let Ig be the maximum current that can be passed through the galvanometer. The current Ig will give full scale deflection in the galvanometer. Galvanometer resistance = G Shunt resistance = S Current in the circuit = I ∴ Current through the shunt resistance = Is = (I–Ig)
Since the galvanometer and shunt resistance are parallel, potential is common.
The shunt resistance is very small because Ig is only a fraction of I. The effective resistance of the ammeter Ra is (G in parallel with S)
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Ra is very low and this explains why an ammeter should be connected in series. When connected in series, the ammeter does not appreciably change the resistance and current in the circuit. Hence an ideal ammeter is one which has zero resistance. 6. Conversion of galvanometer into a voltmeter Voltmeter is an instrument used to measure potential difference between the two ends of a current carrying conductor. A galvanometer can be converted into a voltmeter by connecting a high resistance in series with it. The scale is calibrated in volt. The value of the resistance
Connected in series decides the range of the voltmeter. Galvanometer resistance = G The current required to produce full scale deflection in the galvanometer = Ig
Range of voltmeter = V Resistance to be connected in series = R Since R is connected in series with the galvanometer, the current through the galvanometer,
From the equation the resistance to be connected in series with the galvanometer is calculated. The effective resistance of the voltmeter is Rv = G + R
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Rv is very large, and hence a voltmeter is connected in parallel in a circuit as it draws the least current from the circuit. In other words, the resistance of the voltmeter should be very large compared to the resistance across which the voltmeter is connected to measure the potential difference. Otherwise, the voltmeter will draw a large current from the circuit and hence the current through the remaining part of the circuit decreases. In such a case the potential difference measured by the voltmeter is very much less than the actual potential difference. The error is eliminated only when the voltmeter has a high resistance. An ideal voltmeter is one which has infinite resistance.
Current loop as a magnetic dipole and its magnetic dipole moment When an electric current flows in a closed loop of wire, placed in a uniform magnetic field, the magnetic forces produce a torque which tends to rotate the loop so that area of the loop is perpendicular to the direction of the magnetic field.
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Consider a rectangular coil PQRS placed in an external magnetic field as shown in diagram (a). Let 'I' be the current flowing through the coil. Each part of the coil experiences one Lorentz. Each part of the forces is as shown. The forces are equal in magnitude but act in opposite directions along the same straight line. Hence they cancel out.
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These two forces constitute a couple and so rotates the coil in the anticlockwise. The torque
If the coil has N turns then
where pm = NIA is called the magnetic dipole moment of the loop.
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Sub Topics
Magnetic Dipole Moment of a Revolving Electron The Bohr's atom model pictures the electrons to revolve around stationary heavy nucleus of change +ze. The moving electron constitutes a current of revolution.
where e is the electronic charge and T the time period
Also where v is its orbital velocity and the radius of the orbit. The small magnetic moment ml associated with this revolving current is
Multiplying and dividing R.H.S by me we have
Where l = m v r, the angular momentum. The negative sign indicates
is opposite to
.
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The above expression helps to explain magnetic properties of materials. Besides the orbital magnetic moment , electrons also possess spin magnetic moment. The origin of magnetism in iron and other materials can be traced to this magnetic moment. Magnetic dipole moment of a revolving electron
Let us consider an electron that is revolving around in a circle of radius r with a velocity v. The charge of the electron is e and its mass is me both of which are constants. The time period T of the electrons orbit is: T = CircumferenceVelocity = 2πrv The current i due to the motion of the electron is the charge flowing through that time period, i = −e2πrv = −ev2πr Note that the current is in the opposite direction as the electron is negatively charged. The magnetic moment due to a current loop enclosing an area A is given by: μ = iA Magnetic moment of an electron: μ = −ev2πr A = −ev2πr πr2 μ = −erv2 Let us divide and multiply by the mass of the electron, μ = −e2me me vr We know that the angular momentum L is given by: L = mvr Thus we can write, μ = −e2me L Since the angular momentum is given by the right hand rule with respect to the velocity and the current is in the opposite direction, the negative sign shows that the two quantities are on opposing directions as shown in the figure, μ→ = −e2me L→
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This is an important result as the magnetic moment is only dependent on the angular momentum. This is why the orbital angular momentum and orbital magnetic moment terms are used interchangeably. The same is true for the spin angular moment. For an electron revolving in an atom, the angular momentum is quantized as proposed by Niels Bohr. The angular momentum is given by: L = n h2π, n = 0, ±1, ±2 … Where n is the orbit quantum number and h is the Planck’s constant, μ = n −e2me h2π μ = −n eh4πme The quantity that multiplies with n is a constant known as the Bohr Magneton μB, μB = eh4πme = 9.27 × 10−27 J⁄T The Bohr Magneton is used very widely to express magnetic moments at the atomic scale. The expression we obtained is good for only simple atoms like hydrogen and does not predict all the magnetic moment states of an electron. As you would have learnt in chemistry, the electron does not really revolve around the nucleus. Instead the electron’s orbital magnetic moment is obtained by virtue of being trapped in the nucleuses potential well. The spin magnetic moment and orbital magnetic moment combine vectorially and the magnetic moments of atoms in a sample also combine to produce the net magnetic moment of the sample. It is these magnetic moments obtained by the combination of orbital and spin magnetic moments which determine the magnetic properties of materials.
Magnetic field intensity due to magnetic dipole (bar magnetic) along its axis and its perpendicular axis
The strength of magnetic field at any point around a bar magnet can be calculated. However, the measurements at two points are important: at a point on the axis and at a point on the equatorial line of the bar magnet. These are called end-on position and broadside-on position respectively. End-on position Consider a bar magnet of length 2l and pole strength m. Suppose a point P on the axis of the magnet at a distance d from its center. (d –l) is the distance of P form the N-pole of the magnet. The magnetic field intensity at P due to the north-pole of the magnet is B1=μo4πmr2=μo4πm(d–l)2B1=μo4πmr2=μo4πm(d–l)2
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which is directly away from N-pole. Since the south of the magnet is at a distance r = d + l from P, so magnetic field intensity at P due to S-pole is B2=μo4πmr2=μo4πm(d+l)2B2=μo4πmr2=μo4πm(d+l)2 which is direct towards, the S-pole of the magnet. The magnetic field intensity B at P is the resultant of these two fields, B=B1+(−B2)=B1–B2=μom4π[1(d–l)2−1d+l2]=μom4pi[4ld(d2–l2)2]=μo2Md4π(d2–l2)2B=B1+(−B2)=B1– B2=μom4π[1(d–l)2−1d+l2]=μom4pi[4ld(d2–l2)2]=μo2Md4π(d2–l2)2 where M=m×2lM=m×2l, the magnetic moment of the bar magnet. So, the magnetic field at a point on the axis of a bar magnet is B=μoMd2π(d2–l2)2B=μoMd2π(d2–l2)2 If the length of the magnet is very small, d>>I and the magnetic field intensity is B=μoM2πd3B=μoM2πd3 Broadside on Position
bradside of position
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Suppose a point P is on the equatorial line of the bar magnet. The equatorial line of the magnet is the line perpendicular to the axis of the magnet which bisects the magnet. Let d be the distance of the point P from the centre of the magnet and P due to the North Pole is B1=μo4πmr2=μo4πm(d2+l2)B1=μo4πmr2=μo4πm(d2+l2) directed away from N-pole. the magnetic field B2 at P due to S-pole is B2=μo4πmr2=μo4πm(d2+l2)B2=μo4πmr2=μo4πm(d2+l2) directed towards S-pole. These fields have different directions, but the same magnitude as shown in the figure. Let ∠PSO=θ∠PSO=θ and by symmetry, ∠PNO=θ∠PNO=θ. The angle between B1 and B2 is then 2θ2θ. The resultant magnetic field, B at P is given by B2Since,B1=B2in magnitude, soB2=B21+B22+2B1B2cos2θ=B21(2+2cos2θ)=4B21cos2θ=2B21(1+cos2θ)B2=B12+B22+2B1B2cos2θSince ,B1=B2in magnitude, soB2=B12(2+2cos2θ)=2B12(1+cos2θ)=4B12cos2θ we havecosθSo,B=1d2+l2−−−−−−√=2B1cosθ=2μo4πm(d2+l2)1d2+l2−−−−−−√=μo4π2ml(d2+l2)3/2=μo4πMd2 +l2)3/2we havecosθ=1d2+l2So,B=2B1cosθ=2μo4πm(d2+l2)1d2+l2=μo4π2ml(d2+l2)3/2=μo4πMd2+l2)3/2 The direction of B is parallel to the axis of the magnet, from north to south pole. If the magnet is very short, d>>ld>>l, and the magnetic field at P is B=μo4πMd3 Torque on a Magnetic Dipole in a Uniform Magnetic Field
Consider a short magnetic dipole (or bar magnet) of length 2l placed in a uniform a magnetic field of strength that its magnetic dipole moment direction of poles are − qm and + qm respectively. The magnetic dipole moment
in such a way
makes an angle θ with the
The pole strength of magnetic south and north
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M = qm 2l
......(1)
Its direction is from S pole to N-pole. Force : The force on N-pole is
The force on S-pole is
Obviously, the forces
are equal in magnitude and opposite in direction, hence net force on
magnetic dipole in uniform magnetic field is F = qmB - qmB = 0(zero) Torque : The lines of action of both forces
are different, therefore, these two forces form a
couple (or torque) which tends to rotate the magnet along the direction of magnetic field strength. This couple is called the restoring couple (or torque) and is denoted by Τ. Restoring torque τ = magnitude of one force × perpendicular distance between the lines of action of forces = (qmB)(NA) = (qmB)(2l sin θ) = (qm2l)B sin θ Τ = mB sin θ
[using (i)]
....(2)
Clearly, the magnitude of torque depends on orientation (θ) of magnet with respect to magnetic field. Torque (τ) is a vector quantity whose direction is perpendicular to both
Thus, if a magnetic dipole (or a bar magnet) is placed in a uniform magnetic field in oblique orientation, it experiences no force but experiences a torque. This torque tends to align the dipole moment along the direction of magnetic field. Remark : If the magnetic dipole is placed in a non-uniform magnetic field, it experiences a force and torque both.
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EMF Equation of a Transformer When a sinusoidal voltage is applied to the primary winding of a transformer, alternating flux ϕm sets up in the iron core of the transformer. This sinusoidal flux links with both primary and secondary winding. The function of flux is a sine function. The rate of change of flux with respect to time is derived mathematically. The derivation of EMF Equation of the transformer is shown below. Let
ϕm be the maximum value of flux in Weber f be the supply frequency in Hz N1 is the number of turns in the primary winding N2 is the number of turns in the secondary winding Φ is the flux per turn in Weber
As shown in the above figure that the flux changes from + ϕm to – ϕm in half a cycle of 1/2f seconds. By Faraday’s Law Let E1 is the emf induced in the primary winding
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Where Ψ = N1ϕ
Since ϕ is due to AC supply ϕ = ϕm Sinwt
So the induced emf lags flux by 90 degrees. Maximum valve of emf
But w = 2πf
Root mean square RMS value is
Putting the value of E1max in equation (6) we get
Putting the value of π = 3.14 in the equation (7) we will get the value of E1 as
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Similarly
Now, equating the equation (8) and (9) we get
The above equation is called the turn ratio where K is known as transformation ratio. The equation (8) and (9) can also be written as shown below using the relation (ϕm = Bm x Ai) where Ai is the iron area and Bm is the maximum value of flux density.
For a sinusoidal wave
Self and mutual induction When this emf is induced in the same circuit in which the current is changing this effect is called Selfinduction, ( L ). However, when the emf is induced into an adjacent coil situated within the same
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magnetic field, the emf is said to be induced magnetically, inductively or by Mutual induction, symbol ( M ). Then when two or more coils are magnetically linked together by a common magnetic flux they are said to have the property of Mutual Inductance. Mutual Inductance is the basic operating principal of the transformer, motors, generators and any other electrical component that interacts with another magnetic field. Then we can define mutual induction as the current flowing in one coil that induces a voltage in an adjacent coil. But mutual inductance can also be a bad thing as “stray” or “leakage” inductance from a coil can interfere with the operation of another adjacent component by means of electromagnetic induction, so some form of electrical screening to a ground potential may be required. The amount of mutual inductance that links one coil to another depends very much on the relative positioning of the two coils. If one coil is positioned next to the other coil so that their physical distance apart is small, then nearly all of the magnetic flux generated by the first coil will interact with the coil turns of the second coil inducing a relatively large emf and therefore producing a large mutual inductance value. Likewise, if the two coils are farther apart from each other or at different angles, the amount of induced magnetic flux from the first coil into the second will be weaker producing a much smaller induced emf and therefore a much smaller mutual inductance value. So the effect of mutual inductance is very much dependant upon the relative positions or spacing, ( S ) of the two coils and this is demonstrated below. Mutual Inductance between Coils
The mutual inductance that exists between the two coils can be greatly increased by positioning them on a common soft iron core or by increasing the number of turns of either coil as would be found in a transformer.
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If the two coils are tightly wound one on top of the other over a common soft iron core unity coupling is said to exist between them as any losses due to the leakage of flux will be extremely small. Then assuming a perfect flux linkage between the two coils the mutual inductance that exists between them can be given as.
Where:
µo is the permeability of free space (4.π.10-7)
µr is the relative permeability of the soft iron core
N is in the number of coil turns
A is in the cross-sectional area in m2
l is the coils length in meters
Mutual Induction
Here the current flowing in coil one, L1 sets up a magnetic field around itself with some of these magnetic field lines passing through coil two, L2 giving us mutual inductance. Coil one has a current of I1 and N1 turns while, coil two has N2 turns. Therefore, the mutual inductance, M12 of coil two that exists with respect to coil one depends on their position with respect to each other and is given as:
Likewise, the flux linking coil one, L1 when a current flows around coil two, L2 is exactly the same as the flux linking coil two when the same current flows around coil one above, then the mutual inductance of coil one with respect of coil two is defined as M21. This mutual inductance is true irrespective of the size, number of turns, relative position or orientation of the two coils. Because of this, we can write the mutual inductance between the two coils as: M12 = M21 = M. Then we can see that self inductance characterises an inductor as a single circuit element, while mutual inductance signifies some form of magnetic coupling between two inductors or coils, depending on their distance and arrangement, an hopefully we remember from our tutorials on Electromagnets that the self inductance of each individual coil is given as:
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and
By cross-multiplying the two equations above, the mutual inductance, M that exists between the two coils can be expressed in terms of the self inductance of each coil.
giving us a final and more common expression for the mutual inductance between the two coils of: Mutual Inductance Between Coils
However, the above equation assumes zero flux leakage and 100% magnetic coupling between the two coils, L 1 and L 2. In reality there will always be some loss due to leakage and position, so the magnetic coupling between the two coils can never reach or exceed 100%, but can become very close to this value in some special inductive coils. If some of the total magnetic flux links with the two coils, this amount of flux linkage can be defined as a fraction of the total possible flux linkage between the coils. This fractional value is called the coefficient of coupling and is given the letter k. Coupling Coefficient Generally, the amount of inductive coupling that exists between the two coils is expressed as a fractional number between 0 and 1 instead of a percentage (%) value, where 0indicates zero or no inductive coupling, and 1 indicating full or maximum inductive coupling. In other words, if k = 1 the two coils are perfectly coupled, if k > 0.5 the two coils are said to be tightly coupled and if k < 0.5 the two coils are said to be loosely coupled. Then the equation above which assumes a perfect coupling can be modified to take into account this coefficient of coupling, k and is given as: Coupling Factor Between Coils
or
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When the coefficient of coupling, k is equal to 1, (unity) such that all the lines of flux of one coil cuts all of the turns of the second coil, that is the two coils are tightly coupled together, the resulting mutual inductance will be equal to the geometric mean of the two individual inductances of the coils. Also when the inductances of the two coils are the same and equal, L 1 is equal to L 2, the mutual inductance that exists between the two coils will equal the value of one single coil as the square root of two equal values is the same as one single value as shown.
Mutual Inductance Example No1 Two inductors whose self-inductances are given as 75mH and 55mH respectively, are positioned next to each other on a common magnetic core so that 75% of the lines of flux from the first coil are cutting the second coil. Calculate the total mutual inductance that exists between the two coils.
Mutual Inductance Example No2 When two coils having inductances of 5H and 4H respectively were wound uniformly onto a nonmagnetic core, it was found that their mutual inductance was 1.5H. Calculate the coupling coefficient that exists between.
In the next tutorial about Inductors, we look at connecting together Series and the affect this combination has on the circuit’s mutual inductance, total inductance and their induced voltages. Relation between peak and RMS value current As mentioned earlier, for determining rms and average value of such an alternating current summation would be carried over the period for which current actually flows i.e. 0 to π but would be average for the whole cycle i.e. from 0 to 2π. RMS value of current,
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=
=
=
=
= Average value of current,
=
Peak Factor =
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Form factor = In this lecture complex numbers are used to analyse A.C. series circuits, in particular: • Resistance Capacitance (RC) circuits • Resistance (Pure) Inductance (RL) circuits • Resistance (Pure) Inductance and Capacitance (RLC) circuits • Resistance (Real) Inductance and Capacitance (RLC) circuits
RC series A.C. circuits
The e.m.f. that is supplied to the circuit is distributed the resistor and the capacitor. Since the same current in each element, the resistor and capacitor are in seri common current can often be taken to have the refer phase.
In a series circuit, the potential differences are added the circuit. (In a parallel circuit where the emf is the same across elements, the currents are added).
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On a phasor diagram this is:
The physical current and potentials are:
The applied emf is φ rad behind the current in the circuit.
Example A 255V, 500/π Hz supply is connected in series with a 100R resistor and a 2μF capacitor. Taking the phase of the emf as a reference, find the complex and rms values of (a) the current in the circuit, and (b) the potential difference across each element.
First write the complex emf and how it is distributed around the circuit.
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1.37 radians is about 780. The total impedance of the circuit is seen in the relationship between emf and current. The complex and rms currents are now calculated.
The current leads the applied emf phase reference by 1.37 radians or 780.
The potential differences across the resistor and capacitor are now calculated.
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The resistor potential difference is in phase with the current and the capacitor potential difference lags the current phase by π/2 (or 900).
RC high pass filter circuit Since the impedance of the RC series circuit depends on frequency, as indicated above, the circuit can be used to filter out unwanted low frequencies.
The complex e.m.f. supplied is:
The complex potential across the output resistor is:
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The physical potential across the output resistor is:
A graph of output versus frequency gives:
The output potential is zero for a D.C. potential, and Em for very high frequency. Low frequencies are suppressed and high frequencies are not really affected. The cut-off frequency is arbitrarily chosen as the frequency where only half the input power is output.
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The half power angular frequency is the reciprocal of the time constant RC. The phase will be π/4 at the half power frequency.
RC low pass filter circuit As above, the complex e.m.f. supplied is:
The complex potential across the output capacitor is:
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The physical potential across the output capacitor is:
A graph of output potential versus frequency gives:
The output potential is Em for a D.C. potential, and zero for very high frequency. High frequencies are suppressed and low frequencies are not really affected. The cut-off frequency is also chosen as the frequency where only half the input power is output.
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The half power angular frequency is again the reciprocal of the time constant RC. The phase will also be π/4 at the half power frequency.
RL series A.C. circuits
The e.m.f. that is supplied to the circuit is distributed the resistor and the capacitor. Since the resistor and are in series the common current is taken to have the phase.
Adding the potentials around the circuit:
On a phasor diagram this is:
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The physical current and potentials are:
The applied emf is φ rad ahead of the current in the circuit.
Example A 100V, 1000/π Hz supply is connected in series with a 30R resistor and a 20mH inductor. Take the emf as the reference phase and find: (a) the complex impedance of the circuit (b) the complex, real (i.e. physical) and rms currents, and (c) the complex, real (i.e. physical) and rms potential differences across each element.
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The complex impedance for the circuit is 50 Ω, and the phase angle between current and applied emf is 0.93 radians (or about 530).
The emf is the reference phase.
The real (i.e. physical) current is the imaginary part of complex current and lags behind the applied emf with radians (-530). The rms current is an equivalent dc current of 2 A and phase.
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The complex potential difference across the resistor is in phase with the current. The rms potential difference is 60 V.
The complex potential difference across the inductor emf by 0.64 radians (370). The rms potential difference is 80 V.
RL high pass filter circuit
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The complex e.m.f. supplied is:
The complex potential across the output inductor is:
The physical potential across the output inductor is:
The equations have the same physical form as the RC high pass filter, but with time constant L/R instead of RC. The output potential is Em for a very high frequency, and zero for D.C. potential. Low frequencies are suppressed and high frequencies are not really affected. The half power angular frequency is again the reciprocal of the time constant.
RL low pass filter circuit
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The complex e.m.f. supplied is:
The complex potential across the output resistor is:
The physical potential across the output resistor is:
The equations have the same physical form as the RC low pass filter, but with time constant L/R instead of RC. The output potential is Em for a D.C. potential, and zero for very high frequency. High frequencies are suppressed and low frequencies are not really affected. The half power angular frequency is again the reciprocal of the time constant.
RLC series A.C. circuits
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The e.m.f. that is supplied to the circuit is distributed the resistor, the inductor, and the capacitor. Since th are in series the common current is taken to have the phase.
Adding the potentials around the circuit:
The physical current and potentials are:
On a phasor diagram this is:
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Example A 240V, 250/π Hz supply is connected in series with 60R, 180mH and 50μF. Take the emf as the reference phase and find: (a) the complex impedance of the circuit (b) the complex, real (i.e. physical) and rms currents, and (c) the complex, real (i.e. physical) and rms potential differences across each element.
The complex impedance for the circuit is 78.1 Ω, and the phase angle between current and applied emf is 0.69 radians (or 39.80).
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The emf is the reference phase.
The real (i.e. physical) current is the imaginary part of complex current and lags behind the applied emf with radians (-39.80). The rms current is an equivalent dc current of 3 A and phase.
The complex potential difference across the resistor i with the current. The rms potential difference is 180 V.
The complex potential difference across the inductor emf by 0.88 radians (50.20). The rms potential difference is 270 V.
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The complex potential difference across the capacito emf with -2.27 radians (-129.80). (A negative angle is clockwise from the positive "x" axis). The rms potential difference is 120 V.
Impure or Practical Inductors in A.C. series circuits
In general, an inductor will have resistance because it normally resistive wire. The potential difference acro inductor includes both elements because they canno physically separated.
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Adding the potentials around the circuit:
The physical current and potentials are:
Resonance in case of series LCR circuit
On a phasor diagram this is:
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We have also seen in our tutorial about series RLC circuits that two or more sinusoidal signals can be combined using phasors providing that they have the same frequency supply. But what would happen to the characteristics of the circuit if a supply voltage of fixed amplitude but of different frequencies was applied to the circuit. Also what would the circuits “frequency response” behaviour be upon the two reactive components due to this varying frequency. In a series RLC circuit there becomes a frequency point were the inductive reactance of the inductor becomes equal in value to the capacitive reactance of the capacitor. In other words, XL = XC. The point at which this occurs is called the Resonant Frequency point, ( ƒr ) of the circuit, and as we are analysing a series RLC circuit this resonance frequency produces a Series Resonance. Series Resonance circuits are one of the most important circuits used electrical and electronic circuits. They can be found in various forms such as in AC mains filters, noise filters and also in radio and television tuning circuits producing a very selective tuning circuit for the receiving of the different frequency channels. Consider the simple series RLC circuit below. Series RLC Circuit
Firstly, let us define what we already know about series RLC circuits.
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From the above equation for inductive reactance, if either the Frequency or the Inductance is increased the overall inductive reactance value of the inductor would also increase. As the frequency approaches infinity the inductors reactance would also increase towards infinity with the circuit element acting like an open circuit. However, as the frequency approaches zero or DC, the inductors reactance would decrease to zero, causing the opposite effect acting like a short circuit. This means then that inductive reactance is “Proportional” to frequency and is small at low frequencies and high at higher frequencies and this demonstrated in the following curve: Inductive Reactance against Frequency
The graph of inductive reactance against frequency is a straight line linear curve. The inductive reactance value of an inductor increases linearly as the frequency across it increases. Therefore, inductive reactance is positive and is directly proportional to frequency ( XL ∝ ƒ ) The same is also true for the capacitive reactance formula above but in reverse. If either the Frequency or the Capacitance is increased the overall capacitive reactance would decrease. As the frequency approaches infinity the capacitors reactance would reduce to zero causing the circuit element to act like a perfect conductor of 0Ω’s. But as the frequency approaches zero or DC level, the capacitors reactance would rapidly increase up to infinity causing it to act like a very large resistance acting like an open circuit condition. This means then that capacitive reactance is “Inversely proportional” to frequency for any given value of capacitance and this shown below: Capacitive Reactance against Frequency
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The graph of capacitive reactance against frequency is a hyperbolic curve. The Reactance value of a capacitor has a very high value at low frequencies but quickly decreases as the frequency across it increases. Therefore, capacitive reactance is negative and is inversely proportional to frequency ( XC ∝ ƒ 1 ) We can see that the values of these resistances depends upon the frequency of the supply. At a higher frequency XL is high and at a low frequency XC is high. Then there must be a frequency point were the value of XL is the same as the value of XC and there is. If we now place the curve for inductive reactance on top of the curve for capacitive reactance so that both curves are on the same axes, the point of intersection will give us the series resonance frequency point, ( ƒr or ωr ) as shown below. Series Resonance Frequency
where: ƒr is in Hertz, L is in Henries and C is in Farads. Electrical resonance occurs in an AC circuit when the two reactances which are opposite and equal cancel each other out as XL = XC and the point on the graph at which this happens is were the two
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reactance curves cross each other. In a series resonant circuit, the resonant frequency, ƒr point can be calculated as follows.
We can see then that at resonance, the two reactances cancel each other out thereby making a series LC combination act as a short circuit with the only opposition to current flow in a series resonance circuit being the resistance, R. In complex form, the resonant frequency is the frequency at which the total impedance of a series RLC circuit becomes purely “real”, that is no imaginary impedance’s exist. This is because at resonance they are cancelled out. So the total impedance of the series circuit becomes just the value of the resistance and therefore: Z = R. Then at resonance the impedance of the series circuit is at its minimum value and equal only to the resistance, R of the circuit. The circuit impedance at resonance is called the “dynamic impedance” of the circuit and depending upon the frequency, XC (typically at high frequencies) or XL (typically at low frequencies) will dominate either side of resonance as shown below. Impedance in a Series Resonance Circuit
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Note that when the capacitive reactance dominates the circuit the impedance curve has a hyperbolic shape to itself, but when the inductive reactance dominates the circuit the curve is non-symmetrical due to the linear response of XL. You may also note that if the circuits impedance is at its minimum at resonance then consequently, the circuits admittance must be at its maximum and one of the characteristics of a series resonance circuit is that admittance is very high. But this can be a bad thing because a very low value of resistance at resonance means that the resulting current flowing through the circuit may be dangerously high. We recall from the previous tutorial about series RLC circuits that the voltage across a series combination is the phasor sum of VR, VL and VC. Then if at resonance the two reactances are equal and cancelling, the two voltages representing VL and VC must also be opposite and equal in value thereby cancelling each other out because with pure components the phasor voltages are drawn at +90o and 90o respectively. Then in a series resonance circuit as VL = -VC the resulting reactive voltages are zero and all the supply voltage is dropped across the resistor. Therefore, VR = Vsupply and it is for this reason that series resonance circuits are known as voltage resonance circuits, (as opposed to parallel resonance circuits which are current resonance circuits). Series RLC Circuit at Resonance
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Since the current flowing through a series resonance circuit is the product of voltage divided by impedance, at resonance the impedance, Z is at its minimum value, ( =R ). Therefore, the circuit current at this frequency will be at its maximum value of V/R as shown below. Series Circuit Current at Resonance
The frequency response curve of a series resonance circuit shows that the magnitude of the current is a function of frequency and plotting this onto a graph shows us that the response starts at near to zero, reaches maximum value at the resonance frequency when IMAX = IR and then drops again to nearly zero as ƒ becomes infinite. The result of this is that the magnitudes of the voltages across the inductor, L and the capacitor, C can become many times larger than the supply voltage, even at resonance but as they are equal and at opposition they cancel each other out. As a series resonance circuit only functions on resonant frequency, this type of circuit is also known as an Acceptor Circuit because at resonance, the impedance of the circuit is at its minimum so easily accepts the current whose frequency is equal to its resonant frequency.
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You may also notice that as the maximum current through the circuit at resonance is limited only by the value of the resistance (a pure and real value), the source voltage and circuit current must therefore be in phase with each other at this frequency. Then the phase angle between the voltage and current of a series resonance circuit is also a function of frequency for a fixed supply voltage and which is zero at the resonant frequency point when: V, I and VR are all in phase with each other as shown below. Consequently, if the phase angle is zero then the power factor must therefore be unity. Phase Angle of a Series Resonance Circuit
Notice also, that the phase angle is positive for frequencies above ƒr and negative for frequencies below ƒr and this can be proven by,
Bandwidth of a Series Resonance Circuit If the series RLC circuit is driven by a variable frequency at a constant voltage, then the magnitude of the current, I is proportional to the impedance, Z, therefore at resonance the power absorbed by the circuit must be at its maximum value as P = I2Z.
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If we now reduce or increase the frequency until the average power absorbed by the resistor in the series resonance circuit is half that of its maximum value at resonance, we produce two frequency points called the half-power points which are -3dB down from maximum, taking 0dB as the maximum current reference. These -3dB points give us a current value that is 70.7% of its maximum resonant value which is defined as: 0.5( I2 R ) = (0.707 x I)2 R. Then the point corresponding to the lower frequency at half the power is called the “lower cut-off frequency”, labelled ƒL with the point corresponding to the upper frequency at half power being called the “upper cut-off frequency”, labelled ƒH. The distance between these two points, i.e. ( ƒH – ƒL ) is called the Bandwidth, (BW) and is the range of frequencies over which at least half of the maximum power and current is provided as shown. Bandwidth of a Series Resonance Circuit
The frequency response of the circuits current magnitude above, relates to the “sharpness” of the resonance in a series resonance circuit. The sharpness of the peak is measured quantitatively and is called the Quality factor, Q of the circuit. The quality factor relates the maximum or peak energy stored in the circuit (the reactance) to the energy dissipated (the resistance) during each cycle of oscillation meaning that it is a ratio of resonant frequency to bandwidth and the higher the circuit Q, the smaller the bandwidth, Q = ƒr /BW. As the bandwidth is taken between the two -3dB points, the selectivity of the circuit is a measure of its ability to reject any frequencies either side of these points. A more selective circuit will have a narrower bandwidth whereas a less selective circuit will have a wider bandwidth. The selectivity of a series resonance circuit can be controlled by adjusting the value of the resistance only, keeping all the other components the same, since Q = (XL or XC)/R. Bandwidth of a Series RLC Resonance Circuit
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Then the relationship between resonance, bandwidth, selectivity and quality factor for a series resonance circuit being defined as: 1). Resonant Frequency, (ƒr)
2). Current, (I)
3). Lower cut-off frequency, (ƒL)
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4). Upper cut-off frequency, (ƒH)
5). Bandwidth, (BW)
6). Quality Factor, (Q)
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Series Resonance Example No1 A series resonance network consisting of a resistor of 30Ω, a capacitor of 2uF and an inductor of 20mH is connected across a sinusoidal supply voltage which has a constant output of 9 volts at all frequencies. Calculate, the resonant frequency, the current at resonance, the voltage across the inductor and capacitor at resonance, the quality factor and the bandwidth of the circuit. Also sketch the corresponding current waveform for all frequencies.
Resonant Frequency, ƒr
Circuit Current at Resonance, Im
Inductive Reactance at Resonance, XL
Voltages across the inductor and the capacitor, VL, VC
Note: the supply voltage may be only 9 volts, but at resonance, the reactive voltages across the capacitor, VC and the inductor, VL are 30 volts peak! Quality factor, Q
Bandwidth, BW
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The upper and lower -3dB frequency points, ƒH and ƒL
Current Waveform
Series Resonance Example No2 A series circuit consists of a resistance of 4Ω, an inductance of 500mH and a variable capacitance connected across a 100V, 50Hz supply. Calculate the capacitance require to give series resonance and the voltages generated across both the inductor and the capacitor. Resonant Frequency, ƒr
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Voltages across the inductor and the capacitor, VL, VC
Series Resonance Summary You may have noticed that during the analysis of series resonance circuits in this tutorial, we looked at bandwidth, upper and lower frequencies, -3dB points and quality or Q-factor. All these are terms used in designing and building of Bandpass Filters and indeed, resonance is used in 3-element mains filter design to pass all frequencies within the “passband” range while rejecting all others. However, the main aim of this tutorial is to analyse and understand the concept of how Series Resonance occurs in passive RLC series circuits. Their use in RLC filter networks and designs is outside the scope of this particular tutorial, and so will not be looked at here, sorry.
For resonance to occur in any circuit it must have at least one inductor and one capacitor.
Resonance is the result of oscillations in a circuit as stored energy is passed from the inductor to the capacitor.
Resonance occurs when XL = XC and the imaginary part of the transfer function is zero.
At resonance the impedance of the circuit is equal to the resistance value as Z = R.
At low frequencies the series circuit is capacitive as: XC > XL, this gives the circuit a leading power factor.
At high frequencies the series circuit is inductive as: XL > XC, this gives the circuit a lagging power factor.
The high value of current at resonance produces very high values of voltage across the inductor and capacitor.
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Series resonance circuits are useful for constructing highly frequency selective filters. However, its high current and very high component voltage values can cause damage to the circuit.
The most prominent feature of the frequency response of a resonant circuit is a sharp resonant peak in its amplitude characteristics.
Because impedance is minimum and current is maximum, series resonance circuits are also called Acceptor Circuits.
In the next tutorial about Parallel Resonance we will look at how frequency affects the characteristics of a parallel connected RLC circuit and how this time the Q-factor of a parallel resonant circuit determines its current magnification.
Displacement Current The calculation of the magnetic field of a current distribution can, in principle, be carried out using Ampere's law which relates the path integral of the magnetic field around a closed path to the current intercepted by an arbitrary surface that spans this path:
(35.1) Ampere's law is independent of the shape of the surface chosen as long as the current flows along a continuous, unbroken circuit. However, consider the case in which the current wire is broken and connected to a parallel-plate capacitor (see Figure 35.1). A current will flow through the wire during the charging process of the capacitor. This current will generate a magnetic field and if we are far away from the capacitor, this field should be very similar to the magnetic field produced by an infinitely long, continuous, wire. However, the current intercepted by an arbitrary surface now depends on the surface chosen. For example, the surface shown in Figure 35.1 does not intercept any current. Clearly, Ampere's law can not be applied in this case to find the magnetic field generated by the current.
Figure 35.1. Ampere's law in a capacitor circuit. Although the surface shown in Figure 35.1 does not intercept any current, it intercepts electric flux. Suppose the capacitor is an ideal capacitor, with a homogeneous electric field E between the plates and
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no electric field outside the plates. At a certain time t the charge on the capacitor plates is Q. If the plates have a surface area A then the electric field between the plates is equal to
(35.2) The electric field outside the capacitor is equal to zero. The electric flux, [Phi]E, intercepted by the surface shown in Figure 35.1 is equal to
(35.3) If a current I is flowing through the wire, then the charge on the capacitor plates will be time dependent. The electric flux will therefore also be time dependent, and the rate of change of electric flux is equal to
(35.4) The magnetic field around the wire can now be found by modifying Ampere's law
(35.5) where [Phi]E is the electric flux through the surface indicated in Figure 35.1 In the most general case, the surface spanned by the integration path of the magnetic field can intercept current and electric flux. In such a case, the effects of the electric flux and the electric current must be combined, and Ampere's law becomes
(35.6) The current I is the current intercepted by whatever surface is used in the calculation, and is not necessarily the same as the current in the wires. Equation (35.6) is frequently written as
(35.7)
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where Id is called the displacement current and is defined as
(35.8) Example: Problem 35.8 A parallel-plate capacitor has circular plates of area A separated by a distance d. A thin straight wire of length d lies along the axis of the capacitor and connects the two plates. This wire has a resistance R. The exterior terminals of the plates are connected to a source of alternating emf with a voltage V = V0 sin([omega] t). a) What is the current in the thin wire ? b) What is the displacement current through the capacitor ? c) What is the current arriving at the outside terminals of the capacitor ? d) What is the magnetic field between the capacitor plates at a distance r from the axis ? Assume that r is less than the radius of the plates. a) The setup can be regarded as a parallel circuit of a resistor with resistance R and a capacitor with capacitance C (see Figure 35.2). The current in the thin wire can be obtained using Ohm's law
(35.9)
Figure 35.2. Circuit Problem 35.8. b) The voltage across the capacitor is equal to the external emf. The electric field between the capacitor plates is therefore equal to
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(35.10) The electric flux through the capacitor is therefore equal to
(35.11) The displacement current Id can be obtained by substituting eq.(35.11) into eq.(35.8)
(35.12) The current at the outside terminals of the capacitor is the sum of the current used to charge the capacitor and the current through the resistor. The charge on the capacitor is equal to
(35.13) The charging current is thus equal to
(35.14) The total current is therefore equal to
(35.15) d) The magnetic field lines inside the capacitor will form concentric circles, centered around the resistor (see Figure 35.3). The path integral of the magnetic field around a circle of radius r is equal to
(35.16)
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Figure 35.3. Amperian loop used to determine the magnetic field inside a capacitor. The surface to be used to determine the current and electric flux intercepted is the disk of radius r shown Figure 35.3. The electric flux through this disk is equal to
(35.17) The displacement current intercepted by this surface is equal to
(35.18) The current intercepted by the surface is equal to the current through the resistor (eq.(35.9)). Ampere's law thus requires
(35.19) The strength of the magnetic field is thus equal to
(35.20)
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35.2. Maxwells Equations The fundamental equations describing the behavior of electric and magnetic fields are known as the Maxwell equations. They are
(35.21)
(35.22)
(35.23)
(35.24) Maxwell's equations provide a complete description of the interactions among charges, currents, electric fields, and magnetic fields. All the properties of the fields can be obtained by mathematical manipulations of these equations. If the distribution of charges and currents is given, than these equations uniquely determine the corresponding fields. Lens-Makers Formula It is a relation that connects focal length of a lens to radii of curvature of the two surfaces of the lens and refractive index of the material of the lens. The following assumptions are made for the derivation:
The lens is thin, so that distances measured from the poles of its surfaces can be taken as equal to the distances from the optical centre of the lens.
The aperture of the lens is small.
Point object is considered.
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Incident and refracted rays make small angles.
Consider a convex lens (or concave lens) of absolute refractive index m2 to be placed in a rarer medium of absolute refractive index m1. Considering the refraction of a point object on the surface XP1Y, the image is formed at I1 who is at a distance of V1. CI1= P1I1 = V1 (as the lens is thin) CC1 = P1C1 = R1 CO = P1O = u It follows from the refraction due to convex spherical surface XP1Y
The refracted ray from A suffers a second refraction on the surface XP2Y and emerges along BI. Therefore I is the final real image of O. Here the object distance is
(Note P1P2 is very small)
(Final image distance)
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Let R2 be radius of curvature of second surface of the lens. \ It follows from refraction due to concave spherical surface from denser to rarer medium that
Adding (1) & (2)
Refraction at convex surfaces
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Convex Spherical Refracting Surface
Concave Spherical Refracting Surface
XY is the refracting surface. P is the pole of spherical refracting surface. C is centre of curvature of spherical refracting surface. m1, m2 are the absolute refractive indices of the two media. Assumption: In dealing with refraction at spherical refracting surface, we assume.
The object to be a point lying on the principal axis of the spherical refracting surface.
The aperture of the spherical refracting surface is small.
The incident and refracted rays make small angles with the principal axis of the surface so that sini » i and sinr » i
The sign convention used in mirror is applicable for spherical refracting surfaces. Refraction from Rarer to Denser Medium at a Convex Spherical Refracting Surface
Consider a spherical surface XY convex to the incident ray OA. The point O is a point object and I is the image of the point object where the refracted rays actually meet.
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From A draw a perpendicular on the axis so as to meet M
In triangle OAC, i = a + g According to Snell's law
As the aperture is close, M is close to P.
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Using the sign convention, we put PO = -u , PI = +v, PC = R
OR
Note:
For the virtual image, the point lies close to the pole of refracting surface. In this case the refracted rays PC and AB do not meet actually at any point but appear to come from a point I as shown below.
Refraction from denser to rarer medium at a concave spherical refracting surface
Let the point object lie on the principal axis. A ray of light meets the spherical surface concave to the incident ray at A. The refracted ray bends away from the normal C A N and moves along AI.
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Since the two refracted rays AI and BI actually meet, I represent a real image. Now, from Snell's law
(Since refraction occurs from denser to rarer) or m2 sin i = m1 sin ror m2 i = m1 r (as i and r are small angles) In D OAC i=g-a In D AIC r=g-b
From A, draw AM perpendicular to principal axis
For small aperture, M is close to P
Applying the sign convention Following the procedure as in previous case we have
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PO = -u, PI = +v, PC = -R We have
or
A prism is an optical element. It has polished flat surfaces that refract light. The traditional geometric shape of a prism has a triangular base and two rectangular sides. It is called triangular prism. A prism can be made from materials like glass, plastic and fluorite. It can be used to split light into its components. How a Prism Works When light travels from one medium to another medium, it is refracted and enters the new medium at a different angle. The degree of bending of the light's path depends on the angle that the incident beam of light makes with the surface of the prism, and on the ratio between the refractive indices of the two media. This is called Snell's law.
where, n is the i is r is the angle of refraction.
refractive the
index
of angle
the
material of
of
the
prism. incidence.
The refractive index of many materials varies with the wavelength of the light used. This phenomenon is called dispersion. This causes light of different colors to be refracted differently and to leave the prism at different angles, creating an effect similar to a rainbow. This can be used to separate a beam of white light into its constituent spectrum of colors. The relation between Refractive Index (n), Angle of Prism (A) and Angle of Minimum Deviation (D) Consider the following triangular prism.
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The angle A between the two refracting surfaces ABFE and ACDE is called the angle of prism. A ray of light suffers two refractions on passing through a prism. If KL be a monochromatic light falling on the side AB, it is refracted and travels along LM. It once again suffers refraction at M and emerges out along MN. The angle through which the emergent ray deviates from the direction of incident ray is called angle of deviation 'd'.
As the angle of incidence is increased, angle of deviation 'd' decreases and reaches minimum value. If the angle of incidence is further increased, the angle of deviation is increased. A graph is drawn between angle of incidence (i) and angle of deviation (d) by taking angle of incidence (i) along X-axis and angle of deviation (d) along Y-axis. It should be a curved graph.
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The angle of minimum deviation is obtained from the graph. Let D be the angle of minimum deviation, then the refractive index (n) of the material of the prism is calculated using the formula,
Learning outcomes:
Students understand the working of a prism. Students will be better able to do the experiment in a real laboratory by understanding the procedure.
Young's Double Slit Experiment
Let A and B be two fine slits, a small distance 'd' apart. Let them be illuminated by a monochromatic light of wavelength l.
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MN in the screen is at a distance D from the slits AB. The waves from A and B superimpose upon each other and an interference pattern is obtained on the screen. The point C is equidistant from A and B and therefore the path difference between the waves will be zero and so the point C is of maximum intensity. It is called the central maximum. For another point P at a distance 'x' from C, the path difference at P = BP - AP. Now AB = EF = d, AE = BF = D \D BPF [Pythagoras theorem]
Similarly in D APE
(on expanding Binomially)
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For bright fringes (constructive wavelength) the path difference is integral multiple of wavelength i.e., path difference is nl.
(x therefore represents distance of nth bright fringe from C) Now
and so on.
Therefore separation between the centers of two consecutive bright fringe is the width of a dark fringe.
Similarly for dark fringes,
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The separation between the centers of two consecutive dark interference fringes is the width of a bright fringe.
The separation between the centers of two consecutive dark interference fringes is the width of a bright fringe. All bright and dark fringes are of equal width as b1 = b2.
Diffraction pattern from a single slit
A laser illuminates a single slit and the resultant patten is projected on a distant screen.
The sketch shows the view from above a single slit. Let's assume that the slit is constant width and very tall compared with that width, so that we can consider the system as two-dimensional. With light at normal incidence, the pattern is symmetrical about the axis of the slit. On a distant screen, the light arriving on the axis from all points in the slit has travelled an equal distance from the slit, so the centre of the pattern is a maximum. The next question is what determines its width.
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Notice the broad central maximum, and the equally spaced, successively weaker maxima on either side. First order minima This animated sketch shows the angle of the first order minima: the first minimum on either side of the central maximum. We call the slit width a, and we imagine it divided into two equal halves. Using the Huygens' construction, we consider a point at the very top of the slit, and another point a distance a/2 below it, i.e. a point at the very top of the lower half of the slit. Consider parallel rays from both points, at angle θto the axis of symmetry. (Why parallel? Because the screen is distant. Typically in diffraction experiments, the slit is ~ 10 µm wide, while the distance to the screen might be ~ 1 m.) The ray from the distance a/2 below has to travel an extra distance (a sin θ/2). If this distance is half a wavelength, i.e. if a sin θ = λ then they are π/2 out of phase and they interfere destructively. Now, for every point in the top half of the slit, there is one in the bottom half a distance a/2 below and, at the angle that satisfies a sin θ = λ, they all interfere destructively. So the first mimimum has sin θ = λ/a. On the other side of the axis of symmetry, sin θ = –λ/a is also a minimum. These two minima limit the broad central maximum.
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Higher order minima An argument like the one applies if, in our imagination, we divide the slit into any even number of equal slices. THe diagram shows a division into four. Each half is divided into quarters, and light from a source in the first quarter cancels that from one in the second quarter. Similarly, sources in the third quarter are nullified by those in the fourth quarter. So this diagram represents the second order minima, where sin θ = λ/(a/2), or sin θ = 2λ/a. For the nth order minima, we have a sin θ = n λ , where n is an integer, but not zero. Remember that, on the axis where θ = 0, there is a minimum, so the minima are equally spaced in sin θ, except either side of the central maximum. We can note too that, for light diffracting the throught slits, the slit is usually much wider than a wavelength, so the pattern is usually very small, so the approximation that sin θ = θ is usually good.
Geometry for the second order minima
Next we calculate how the intensity varies with sin θ.
Intensity I(θ) To calculate I(θ), we use Huygens' construction and phasor addition.
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The geometry for the phasor sum,
The arc made of very many phasors has the same length as the central amplitude A0.
The slit of width a is divided into N slits, each of width δa. The path difference between rays from successive slices of the slit are equal, and so too are the angles between successive phasors. The slices of the slit all have equal width and length, so the lengths of all the phasors are equal. For a large number of slices, the phasors approximate the arc of a circle, as shown in red. We'll call the angle that this subtends 2α. Now 2α is also the phase difference between the first and the last phasor, which is 2πa sin θ/λ. α = πa sin θ/λ. The phasor sum has magnitude A, which we can write as R.sin α, where R is the radius of the arc formed by the phasor sum. On the axis, where the phasors are all in phase, the phasor sum is the straight line shown in red at right. This is the amplitude of the diffraction patter at θ /= 0, which we call A0. By the definition of angle, 2α = A0/R, which gives us R = A0/2α. So the magnitude of the phasor sum is A = 2R.sin α = A0sin α /α. As we've seen in several previous chapters, the intensity I is proportional to square of the amplitude A. So, using I0 for the intensity at the centre of the pattern, we have
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I = I0(sin α /α)2
where
α = πa sin θ/λ.
Graphing I(θ) with the phasor sum
The animated phasor diagram (left); the plot of I(θ) and, below it on the same scale, the resulting diffraction pattern. The vertical black line scanning across the graph is in step with the phasor diagram.
As we remarked when looking at the intenstiy of Young's experiement in Interference, the eye does not respond linearly to intensity. To my eyes, at least, the difference in brightness seems much less than the difference in the intensity graph.
Varying the slit width
A slit with variable width. As the sit is narrowed, the pattern expands.
DIffraction effects are most noticeable when the slit with a is not very much larger than the wavelength λ. This apparatus allows us to vary the slit width, but the pattern is still projected on a distant screen to make the diffraction effects cl
Young's experiment with finite slit width.
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Top: a Young's experiment with slit separation d and slit width a. Bottom: a single slit diffraction patter with the same slit width a. In the Young's experiement in Interference, we didn't mention the effect of finite slit width. From the equations above, we can see that, if d is an integral number of times the slit width (d = na), then the nth interference fringe is absent: neither slit raidates power at this angle so there are no rays for constructive interference. Further reading.
This page supports the multimedia tutorial Diffraction.
Refraction through a Prism
A prism is a wedge-shaped body made from a refracting medium bounded by two plane faces inclined to each other at some angle. The two plane faces are called are the refracting faces and the angle included between these two faces is called the angle of prism or the refracting angle. In the below figure (1), ABC represents the principal section of a glass-prism having ∠A as its refracting angle.
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A ray KL is incident on the face AB at the point F where N1LO is the normal and ∠i1 is the angle of incidence. Since the refraction takes place from air to glass, therefore, the refracted ray LM bends toward the normal such that ∠r1 is the angle of refraction. If µ be the refractive index of glass with respect to air, then
µ = sin i/sin r (By Snell’s law) The refracted ray LM is incident on the face AC at the point M where N2MO is the normal and ∠r2 is the angle of incidence. Since the refraction now takes place from denser to rarer medium, therefore, the emergent ray MN such that ∠i2 is the angle of emergence. In the absence of the prism, the incident ray KL would have proceeded straight, but due to refraction through the prism, it changes its path along the direction PMN. Thus, ∠QPN gives the angle of deviation ‘δ’, i.e., the angle through which the incident ray gets deviated in passing through the prism. Thus, δ = i1 – r1 + i2 -r2 ….... (1) δ = i1 + i2 – (r1 + r2 ) Again, in quadrilateral ALOM, ∠ALO + ∠AMO = 2rt∠s So, ∠LAM +∠LOM = 2rt∠s Also in ?LOM, ∠r1 +∠r2 + ∠LOM = 2rt∠s
[Since, ∠ALO = ∠AMO = 90º] [Since, Sum of four ∠s of a quadrilateral = 4 rt∠s] ….... (2) …... (3)
Comparing (2) and (3), we get ∠LAM = ∠r1 +∠r2 A = ∠r1 +∠r2 Using this value of ∠A, equation (1) becomes, δ = i1 + i2 - A or i1 + i2 = A + δ …... (4)
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The angle of deviation of a ray of light in passing through a prism not only depends upon its material but also upon the angle of incidence. The above figure (2) shows the nature of variation of the angle of deviation with the angle of incidence. It is clear that an angle of deviation has the minimum value ‘δm’ for only one value of the angle of incidence. The minimum value of the angle of deviation when a ray of light passes through a prism is called the angle of minimum deviation. The figure (3) shows the prism ABC, placed in the minimum deviation position. If a plane mirror M is placed normally in the path of the emergent ray MN the ray will retrace its original path in the opposite direction NMLK so as to suffer the same minimum deviation dm. In the minimum deviation position, ∠i1 = ∠i2 and so ∠r1 = ∠r2 = ∠r (say) Obviously, ∠ALM = ∠LMA = 90º – ∠r Thus, AL = LM and so LM l l BC
Hence, the ray which suffers minimum deviation possess symmetrically through the prism and is parallel to the base BC. Since for a prism, ∠A = ∠r1 + ∠r2
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So, A = 2r (Since, for the prism in minimum deviation position, ∠r1 = ∠r2 = ∠r) or r = A/2 …...(5) Again, i1 + i2 = A + δ or i1 + i1 = A + δm (Since, for the prism in minimum deviation position, i1 = i2 and δ = δm) 2i1 = A + δm or i1 = (A + δm) / 2 …... (6) Now µ = sin i1/sin r1 = sin i1/sin r µ = sin [(A + δm) / 2] / sin (A/2) …... (7)
Equation (7) gives the relation between the refractive index of the material of the prism and the minimum deviation. Grazing Incidence:When i = 90°, the incident ray grazes along the surface of the prism and the angle of refraction inside the prism becomes equal to the critical angle for glass - air. This is known as grazing incidence. Grazing Emergence:When e = 90°, the emergent ray grazes along the prism surface. This happens when the light ray strikes the second face of the prism at the critical angle for glass - air. This is known as grazing emergence. Maximum Deviation:The angle of deviation is same for both the above cases (grazing incidence & grazing emergence) and it is also the maximum possible deviation if the light ray is to emerge out from the other face without any total internal reflection. Refraction through a Prism for Small Angle of incidence
Figure (4) shows a prism having a very small refracting ∠A. a ray KL is incident on the face AB such that the angle of incidence ‘i’ is very small. Accordingly, the angle of refraction ‘r’ will also be very small. Also angles ‘r2’ and ‘i2’ must also be very small. If ‘µ’ be the refractive index of the material of the prism, then µ = sin i1 / sin r1 µ = i1 / r1 (Since sin i1 = i1 and sin r1 = r1 when angles i and r1 are very small) or i1 = µr1
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Similarly,i2 = µr2 Thus, i1 + i2 = µr1 + µr2 = µ (r1 + r2) Or, i1 + i2 = µA (Since, A = r1 + r2) Also, i1 + i2 = A + δ …... (9) Therefore, from (8) and (9), we get, A + δ = µA Or, δ = µA – A Or, δ = A(µ – 1)
…....(8)
This relation shows that the angle of deviation ‘δ’ is independent of the angle of incidence, provided it is small. In other words, if ‘i1’ changes, ‘i2’ will also change accordingly. Dispersion With the help of Newton’s disc experiment, it can be proved that white light is composed of the seven colours. Newton’s disc is a disc divided into seven parts. Different parts are coated with pure pigment colours in the order: violet, indigo, blue, green, yellow, orange and red. On rotating the disc at a high speed, all the colours merge into one another due to persistence of vision, giving a resultant white impression. When a ray of white light passes through a prism, it splits up into its constituent’s colours, producing a band of colours on the screen XY as shown in the figure. This splitting up of light into its constituent colours is called dispersion. Reason for Dispersion Refractive index of a transparent medium depends upon the nature of light (i.e., its wave-length) passing through it. According to Cauchy’s formula, refractive index of a material is given by, µ = A + B/λ2 +....... (10)
Here, ‘A’ and ‘B’ are constants and ‘λ’ is the wave-length of light. Medium has greater refractive index for light and smaller wave-length. Since violet light has smaller wave-length than that for red light, i.e.,λv < λr. Thus, µv > µr Where ‘µv’ and ‘µr’ are the refractive indices of the medium for violet and red light respectively. In case of a prism, µ = sin [(A + δm) / 2] / sin (A/2)
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Since ‘A’; the angle of prism, is same for all the colours and ‘µv’ and ‘‘µr’ are different, above equation can only be satisfied if all the colours have different values of ‘dm’, i.e., they come out along different paths and hence, the phenomenon of dispersion. Watch this Video for more reference Refraction through a Prism As a ray of light is incident on one of the refracting faces of a prism and proceeds through the prism, it undergoes following two changes: (a) Deviation (b) Dispersion (a) Deviation:- A ray of monochromatic light (light possessing one wave-length only), while passing through a prism suffers a change in its path, the phenomenon is known as deviation.
it can be shown that the deviation ‘d’ suffered by a ray, provided it is incident on one of its faces at a very small angle, is given by δ = ( µ – 1) A ….... (11) where ‘A’ is the refracting angle of the prism and ‘µ’ is the refractive index of the material of prism for that particular wave-length of light. (b) Dispersion:- A ray of light (containing more than one wave-lengths), while passing through the prism splits up into a number of rays. This phenomenon is called dispersion.
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From equation (10) it is clear that deviation suffered by a ray depends upon the refractive index ‘µ’ of the material of prism for that wavelength. So, different wave-lengths contained in the incident ray suffer different deviations. hence, these constituent rays will emerge along different paths. Consider a ray of white light incident on one of the faces AB of prism ABC. After passing through the prism, it splits up into its constituent colours with violet and red as its extreme colours. Let ‘µv’ and ‘µr’ be the refractive indices of the material of prism for violet and red colours; then the corresponding deviations are given by, δv = ( µv – 1) A δr = ( µr – 1) A Since, µv > µr Therefore, ‘δv’ is greater than ‘δr’. Other colours possess deviation in between ‘δv’ and ‘δr’. Therefore, the colours, in the dispersed beam, will be spread with in a cone of angle ‘δv – δr’. ‘δv – δr’ is called angular dispersion produced by the prism and is given by δv – δr = [(µv – 1)A] – [(µr – 1) A] Or, δv – δr = (µv – µr) A …... (12) Dividing equation (12) by (11), we get δv – δr /δ = (µv – µr) A / (µ – 1) A δv – δr / δ = (µv – µr) / (µ – 1) = ω Where ‘ω’ is called the dispersive power of the prism. Dispersive power of a prism is defined as the ratio between angular dispersion to mean deviation produced by the prism. If dµ denotes the difference between the refractive indices of material of prism for violet and red light, ω = δµ / µ – 1
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Here ‘µ’ is the refractive index of prism for a mean colour. A mean colour is that colour whose wavelength lies in between that of violet and red. For white light, yellow colour is, generally, taken to be the mean colour. Since µv is always greater than µr, the dispersive power of a prism is always positive. It depends upon the type of glass used. It is different for crown glass and for flint glass.
Einstein’s Photoelectric Equation:
Since wave theory could not explain the photoelectric effect, Einstein proposed a particle theory of light for the first time He said that radiations are made up of specific and discrete packets of energy called as quanta of radiation energy. Each energy quantum has a value equal to hv, where h = Planck’s constant, and v = frequency of incident light These specific packets of quanta of energy are known as photons When a light of frequency(v) (having energyhv ) is incident on a metal surface of work function(Φo), 3 cases could be possible Case-1-When (hv < Φo), i.e., energy of photon is less than the work function of metal , no photoelectric emission occurs Case-2- When (hv = Φo), i.e., energy of photon is exactly same as the work function of metal, then electrons get enough energy to just escape the metal surface. Case-3- When (hv > Φo),e., energy of photon is greater than the work function of metal. Then electron, apart from getting energy to escape the metal surface, the remaining energy is provided to the electron as kinetic energy. Mathematically, it can be expressed as:
hv = Φo + Kmax Here, Kmax is the maximum kinetic energy of a photoelectron
The above equation is known as Einstein’s photoelectric equation
From the Einstein’s photoelectric equation, following points are clear:
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Photoelectric current (i) is directly proportional to the intensity (I) of radiation. As the intensity rises, number of photons received by metal surface in a unit area per unit time rises, so number of electrons emitted rises, and hence, photoelectric current increases i∝I
Since saturation current is just a maximum value of photoelectric current, saturation current gets higher with increasing intensity of incident light For every metal, there exists a certain minimum frequency below which no photoelectric effect occurs. This frequency is called threshold frequency. hvo = Φo
Here, vo = frequency of incident light, Φo = work function of metal
Stopping potential (Vo) and Maximum kinetic energy (Kmax) doesn’t depend upon the intensity. Because intensity is the number of photons in unit area and unit time, and the photoelectric effect take place when one electron takes one photon Stopping potential (Vo) and Maximum kinetic energy (Kmax) is directly proportional to the frequency (ν) Kmax∝ v Vo∝ v
The relation between stopping potential, maximum kinetic energy and the frequency of incident light could be expressed mathematically as follows: using Einstein’s photoelectric equation hv = Φo + Kmax Kmax = hν - Φo Also, Kmax = eVo
∴ eVo = hν - Φo On rearranging the above equation: Vo = (h/e)v + (Φo/e) Plotting the above equation graphically, we get:
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Photoelectric emission is an instantaneous process, meaning that there is no time gap between incident radiation and electron emission.
Numerical Problems: 1)Question: Caesium metal has work function of 2.14eV. Photoelectric emission takes place when a light of frequency 6×1014 Hz is incident on the metal surface. Calculate the following: a) maximum kinetic energy of the electrons emitted, b) stopping potential, and c) maximum speed of the emitted electrons. Solution: Given, Φo = 2.14eV = 2.14×10-19J, and ν = 6×1014Hz a) Using Einstein’s photoelectric equation: hv = Φo+ Kmax Kmax = hv - Φo = (6.6×10-34×6×1014)J – (2.14×10-19)J
∴Kmax = 1.82×10-19J (ans) b) Stopping potential is given by the equation: eVo = Kmax Vo = Kmax/e = 1.82×10-19 = 1.1375V (ans) c) Maximum speed of emitted electrons can be found using maximum kinetic energy equation:
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Kmax = (1/2)mvmax2
2)Question: Light of wavelength 488nm is incident on an emitter plate. The photoelectrons have a stopping (cut-off) potential of 0.38V. Calculate the work function of the emitter plate. Solution: Given, λ = 488×10-9m,Vo = 0.38V Using Einstein’s photoelectric equation: hv = hc/λ = Φo + Kmax and, Kmax = eVo
De broglie relation
Bohr’s model failed to explain many concepts related to the spectrum of different atoms and splitting of spectral lines in magnetic and electric field. To overcome the shortcomings of Bohr’s atomic model, attempts were made to develop a more general model for atoms. One of those developments which contributed significantly in the formulation was the analysis of dual behavior of matter. De Broglie proposed that as light exhibits both wave like and particle like properties, matter too exhibit wave like and particle like properties. This nature was described as dual behavior of matter. On the basis of his observations, de Broglie derived a relationship between wavelength and momentum of matter. This relationship is known as de Broglie relationship. Considering the particle nature, Einstein equation is given as, E= mc2 —- (1) Where, E= energy m= mass
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c = speed of light Considering the wave nature, the Plank’s equation is given as, E = hν
——– (2)
Where, E= energy h = Plank’s constant ν = frequency
From (1) and (2), mc2 = hν ——– (3) Frequency, ν can be expressed in terms of wavelength, λ as, ν = cλ For a general particle, c can be replaced with the velocity of object, v. Hence, equation (3) can be given as, mv2 = hvλ ⇒λ = hmv The above equation is known as de Broglie relationship and the wavelength, λ is known as de Broglie wavelength. Diffraction of electron beams explains the de Broglie relationship as diffraction is the property of waves. Electron microscope is a common instrument illustrating this fact. Thus, every object in motion has a wavelike character. Due to a large mass, the wavelengths associated with ordinary objects are so short that their wave properties cannot be detected. On the other hand, the wavelengths associated with electrons and other subatomic particles can be detected experimentally.
Bohr Model was introduced after the Rutherford’s model. It retained some of the features like the orbiting of nucleus and the electrons under the action of the Coulomb’s Law of electrostatic attraction. Various other concepts like radiation less orbits and stationary states were also introduced in the Bohr Model. We discuss the postulates of Bohr Atomic theory. Bohr introduced the concept of radiation less orbits in which the electrons revolve as usual around the nucleus but without radiating any kind of energy which is contrary to the laws of electromagnetism. This was a hypothesis, but at least a working one. Radiation occurred only when an electron made a transition from one stationary state to another. The difference between the energies of the two states was radiated as a single photon. Absorption occurred when a transition occurred from a lower stationary state to a higher stationary state. He also introduced the correspondence principle which states that the spectrum is continuous and the frequency of light emitted equals the frequency of the electron.
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Neil Bohr explained the line spectrum while developing the model of an atom. We discuss the various postulates of the Bohr Atomic Model. The Bohr model shows that the electrons in atoms are in orbits of differing energy around the nucleus. The electron in an atom has only certain definite stationary states of motion allowed to it, called as energy levels. Each energy level has a definite energy associated with it. In each of these energy levels, electrons move in circular orbit around the positive nucleus. The necessary centripetal force is provided by the electrostatic attraction of the protons in the nucleus. As one moves away from the
nucleus, the energy of the levels increases. The energy level occupied by an electron usually is called the ground state. It may move to a level of higher energy or a less stable shell by absorbing energy. This higher energy but less stable shell or level is termed as the excited state of the electron.
These states of allowed electronic motion are those in which the angular momentum of an electron is an integral multiple of h/2π or one can say that the angular momentum of an electron is quantized.
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Angular Momentum
where m is the mass, v is the velocity, r is the radius of the orbit, h is Planck's constant and n is a positive integer.
When an atom is in one of these states, it does not radiate any energy but whenever there is a transition from one state to other, energy is emitted or absorbed depending upon the nature of transition. When an electron jumps from higher energy state to the lower energy state, it emits radiations in form of photons or quanta. However, when an electron moves from lower energy state to a higher state, energy is absorbed, again in form of photons. After reaching the state of excitation, the electron can return to its original ground state by releasing the energy it has absorbed. Sometimes the energy released by electrons occupies the portion of the electromagnetic spectrum (the range of wavelengths of energy) that humans detect as visible light. Slight variations in the amount of the energy are seen as light of different colors. The energy of a photon emitted or absorbed is given by using Planck's relation (E = h ). If E1be the energy of any lower energy state and E2 be the energy of any higher energy state then the energy of the photon (emitted or absorbed) is given as ΔE:
Where h = Planck's constant and = frequency of radiation emitted or absorbed. Though the Bohr postulates of atomic model worked quite well it has certain limitations. It is applicable only to one electron atoms like helium He+and lithium Li++ apart from hydrogenH- atom.
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Here, we shall discuss the concept of Bohr’s orbits by defining the radius of orbits around the nucleus and the velocity and energy of an electron in various orbits around the nucleus. We take the following assumptions: rn : radius of nth orbit vn : velocity of electron in nth state (orbit) En : energy of nth state m: mass of an electron (9.1 x 10–31 Kg) Z: atomic number (No. of Protons) K = 1/(4πε0) = constant = 9 x 109 N m2 C–2 h: Planck's constant (6.67 x 10–34 J-s) c: velocity of light (3 x 108 m/s) R: Rydberg constant (1.097 x 107 m–1) e: Charge on an electron (1.6 x 10–19 C) : frequency of the radiation emitted or absorbed : wave number of the spectral line in the atomic spectra From Bohr’s first postulate, mv2 / rn = KZe2 / rn2 FromBohr’s second postulate, mvnrn = nh / 2π Solving for rn and vn, we have: Radius,
= 0.53 × 10–10 n2/Z m = 0.53 n2 / Z A Velocity,
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= (2.165 × 106 Z/n) m/s Energy of an electron in ‘n’ th state En :En = KE + PE En = 1/2 mv2 + (– KZe2 / rn) = –1/2 KZe2 / rn => En = –2π2K2e4Z2 / n2h2 (putting value of rn) –18 2 2 = –2.178 × 10 Z /n J / atom = –13.6 Z2/n2 eV/atom = –1312 Z2/n2 kJ/mol When an electron jumps from an outer orbit (higher energy) n2 to an inner orbit (lower energy) n1, then the energy emitted in form of radiation is given by:
= (2π2e4mZ2K2/h2) (1/n12 – 1/n22) As we know that E = h , c = λ and =1/λ So, =?E/hc =1/λ= (2π2e4mZ2K2/ch3) (1/n12 – 1/n22) Now this can be represented as: ==RZ2 (1/n12 – 1/n22) Where,
is the Rydberg’s constant. Refer this vdeo to know more about on,”Bohr Model”:-
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Points to be noted:Bohr’s Theory was quite successful as it introduced various new concepts about atom. The relation given by Bohr resembles the empirical relation given by Balmer& Rydberg on the spectral lines in H-atom spectra. The value of R as obtained above in the Bohr’s theory is the same as obtained in the empirical relation. Bohr’s atomic model postulates introduced various concepts like: Bohr found that closer the electron to the nucleus the less energy it needs, while if it is far away from the nucleus it requires more energy. Because of this Bohr numbered the energy levels of electrons. If the energy level is high, farther is the electron from the nucleus. He also found that there is a fixed number of electrons that every energy level can hold. The level 1 can hold up to 2 electrons, while the level 2 can hold up to 8 electrons and so on. The final transistor amplifier configuration (Figure below) we need to study is the common-base. This configuration is more complex than the other two, and is less common due to its strange operating characteristics.
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Common-base amplifier It is called the common-base configuration because (DC power source aside), the signal source and the load share the base of the transistor as a common connection point shown in Figure below.
Common-base amplifier: Input between emitter and base, output between collector and base. Perhaps the most striking characteristic of this configuration is that the input signal source must carry the full emitter current of the transistor, as indicated by the heavy arrows in the first illustration. As we know, the emitter current is greater than any other current in the transistor, being the sum of base and collector currents. In the last two amplifier configurations, the signal source was connected to the base lead of the transistor, thus handling the least current possible. Because the input current exceeds all other currents in the circuit, including the output current, the current gain of this amplifier is actually less than 1 (notice how Rload is connected to the collector, thus carrying slightly less current than the signal source). In other words, it attenuates current rather than amplifying it. With common-emitter and common-collector amplifier configurations, the transistor parameter most closely associated with gain was β. In the common-base circuit, we follow another basic transistor parameter: the ratio between collector current and emitter current, which is a fraction always less than 1. This fractional value for any transistor is called the alpha ratio, or α ratio. Since it obviously can’t boost signal current, it only seems reasonable to expect it to boost signal voltage. A SPICE simulation of the circuit in Figure below will vindicate that assumption.
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Common-base circuit for DC SPICE analysis.
common-base amplifier vin 0 1 r1 1 2 100 q1 4 0 2 m od1 v1 3 0 dc 15 rload 3 4 5k .model mod1 npn .dc vi n 0.6 1.2 .02 .plot dc v(3,4) .end
Common-base amplifier DC transfer function. Notice in Figure above that the output voltage goes from practically nothing (cutoff) to 15.75 volts (saturation) with the input voltage being swept over a range of 0.6 volts to 1.2 volts. In fact, the output voltage plot doesn’t show a rise until about 0.7 volts at the input, and cuts off (flattens) at about 1.12 volts input. This represents a rather large voltage gain with an output voltage span of 15.75 volts and an input voltage span of only 0.42 volts: a gain ratio of 37.5, or 31.48 dB. Notice also how the output voltage (measured across Rload) actually exceeds the power supply (15 volts) at saturation, due to the series-aiding effect of the input voltage source. A second set of SPICE analyses (circuit in Figure below) with an AC signal source (and DC bias voltage) tells the same story: a high voltage gain
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Common-base circuit for SPICE AC analysis. As you can see, the input and output waveforms in Figure below are in phase with each other. This tells us that the common-base amplifier is non-inverting.
common-base amplifier vin 5 2 sin (0 0.12 2000 0 0) v bias 0 1 dc 0.95 r1 2 1 100 q1 4 0 5 mod1 v1 3 0 dc 15 rload 3 4 5k .model mod1 npn .tran 0.02m 0.78m .plo t tran v(5,2) v(4) .end
The AC SPICE analysis in Table below at a single frequency of 2 kHz provides input and output voltages for gain calculation. Common-base AC analysis at 2 kHz– netlist followed by output. common-base amplifier vin 5 2 ac 0.1 sin vbias 0 1 dc 0.95 r1 2 1 100 q1 4 0 5 mod1 v1 3 0 dc 15 rload 3 4 5k .model mod1 npn .ac dec 1 2000 2000 .print ac vm(5,2) vm(4,3) .end frequency mag(v(5,2)) mag(v(4 ,3)) -------------------------------------------- 0.000000e+00 1.000000e-01 4.273864e+00 Voltage figures from the second analysis (Table above) show a voltage gain of 42.74 (4.274 V / 0.1 V), or 32.617 dB:
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Here’s another view of the circuit in Figure below, summarizing the phase relations and DC offsets of various signals in the circuit just simulated.
Phase relationships and offsets for NPN common base amplifier. . . . and for a PNP transistor: Figure below.
Phase relationships and offsets for PNP common base amplifier.
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We also saw that a family of curves known commonly as the Output Characteristic Curves, relate the transistors Collector current ( Ic ), to the Collector voltage ( Vce ) for different values of the transistors Base current ( Ib ). All types of transistor amplifiers operate using AC signal inputs which alternate between a positive value and a negative value so some way of “presetting” the amplifier circuit to operate between these two maximum or peak values is required. This is achieved using a process known as Biasing. Biasing is very important in amplifier design as it establishes the correct operating point of the transistor amplifier ready to receive signals, thereby reducing any distortion to the output signal. We also saw that a static or DC load line can be drawn onto these output characteristics curves to show all the possible operating points of the transistor from fully “ON” to fully “OFF”, and to which the quiescent operating point or Q-point of the amplifier can be found. The aim of any small signal amplifier is to amplify all of the input signal with the minimum amount of distortion possible to the output signal, in other words, the output signal must be an exact reproduction of the input signal but only bigger (amplified). To obtain low distortion when used as an amplifier the operating quiescent point needs to be correctly selected. This is in fact the DC operating point of the amplifier and its position may be established at any point along the load line by a suitable biasing arrangement. The best possible position for this Q-point is as close to the center position of the load line as reasonably possible, thereby producing a Class A type amplifier operation, ie. Vce = 1/2Vcc. Consider the Common Emitter Amplifier circuit shown below. The Common Emitter Amplifier Circuit
The single stage common emitter amplifier circuit shown above uses what is commonly called “Voltage Divider Biasing”. This type of biasing arrangement uses two resistors as a potential divider network across the supply with their center point supplying the required Base bias voltage to the transistor. Voltage divider biasing is commonly used in the design of bipolar transistor amplifier circuits.
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This method of biasing the transistor greatly reduces the effects of varying Beta, ( β ) by holding the Base bias at a constant steady voltage level allowing for best stability. The quiescent Base voltage (Vb) is determined by the potential divider network formed by the two resistors, R1, R2 and the power supply voltage Vcc as shown with the current flowing through both resistors. Then the total resistance RT will be equal to R1 + R2 giving the current as i = Vcc/RT. The voltage level generated at the junction of resistors R1 and R2 holds the Base voltage (Vb) constant at a value below the supply voltage. Then the potential divider network used in the common emitter amplifier circuit divides the supply voltage in proportion to the resistance. This bias reference voltage can be easily calculated using the simple voltage divider formula below: Bias Voltage
The same supply voltage, (Vcc) also determines the maximum Collector current, Ic when the transistor is switched fully “ON” (saturation), Vce = 0. The Base current Ib for the transistor is found from the Collector current, Ic and the DC current gain Beta, β of the transistor. Beta Value
Beta is sometimes referred to as hFE which is the transistors forward current gain in the common emitter configuration. Beta has no units as it is a fixed ratio of the two currents, Icand Ib so a small change in the Base current will cause a large change in the Collector current.
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One final point about Beta. Transistors of the same type and part number will have large variations in their Beta value for example, the BC107 NPN Bipolar transistor has a DC current gain Beta value of between 110 and 450 (data sheet value) this is because Beta is a characteristic of their construction and not their operation. As the Base/Emitter junction is forward-biased, the Emitter voltage, Ve will be one junction voltage drop different to the Base voltage. If the voltage across the Emitter resistor is known then the Emitter current, Ie can be easily calculated using Ohm’s Law. The Collector current, Ic can be approximated, since it is almost the same value as the Emitter current. Common Emitter Amplifier Example No1 A common emitter amplifier circuit has a load resistance, RL of 1.2kΩs and a supply voltage of 12v. Calculate the maximum Collector current (Ic) flowing through the load resistor when the transistor is switched fully “ON” (saturation), assume Vce = 0. Also find the value of the Emitter resistor, RE if it has a voltage drop of 1v across it. Calculate the values of all the other circuit resistors assuming an NPN silicon transistor.
This then establishes point “A” on the Collector current vertical axis of the characteristics curves and occurs when Vce = 0. When the transistor is switched fully “OFF”, their is no voltage drop across either resistor RE or RL as no current is flowing through them. Then the voltage drop across the transistor, Vce is equal to the supply voltage, Vcc. This establishes point “B” on the horizontal axis of the characteristics curves. Generally, the quiescent Q-point of the amplifier is with zero input signal applied to the Base, so the Collector sits about half-way along the load line between zero volts and the supply voltage, (Vcc/2). Therefore, the Collector current at the Q-point of the amplifier will be given as:
This static DC load line produces a straight line equation whose slope is given as: -1/(RL + RE) and that it crosses the vertical Ic axis at a point equal to Vcc/(RL + RE). The actual position of the Q-point on the DC load line is determined by the mean value of Ib. As the Collector current, Ic of the transistor is also equal to the DC gain of the transistor (Beta), times the Base current (β x Ib), if we assume a Beta (β) value for the transistor of say 100, (one hundred is a reasonable average value for low power signal transistors) the Base current Ib flowing into the transistor will be given as:
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Instead of using a separate Base bias supply, it is usual to provide the Base Bias Voltage from the main supply rail (Vcc) through a dropping resistor, R1. Resistors, R1 and R2 can now be chosen to give a suitable quiescent Base current of 45.8μA or 46μA rounded off. The current flowing through the potential divider circuit has to be large compared to the actual Base current, Ib, so that the voltage divider network is not loaded by the Base current flow. A general rule of thumb is a value of at least 10 times Ib flowing through the resistor R2. Transistor Base/Emitter voltage, Vbe is fixed at 0.7V (silicon transistor) then this gives the value of R2 as:
If the current flowing through resistor R2 is 10 times the value of the Base current, then the current flowing through resistor R1 in the divider network must be 11 times the value of the Base current. The voltage across resistor R1 is equal to Vcc – 1.7v (VRE + 0.7 for silicon transistor) which is equal to 10.3V, therefore R1 can be calculated as:
The value of the Emitter resistor, RE can be easily calculated using Ohm’s Law. The current flowing through RE is a combination of the Base current, Ib and the Collector current Icand is given as:
Resistor, RE is connected between the Emitter and ground and we said previously that it has a voltage of 1 volt across it. Then the value of RE is given as:
So, for our example above, the preferred values of the resistors chosen to give a tolerance of 5% (E24) are:
Then, our original Common Emitter Amplifier circuit above can be rewritten to include the values of the components that we have just calculated above.
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Completed Common Emitter Circuit
Coupling Capacitors In Common Emitter Amplifier circuits, capacitors C1 and C2 are used as Coupling Capacitors to separate the AC signals from the DC biasing voltage. This ensures that the bias condition set up for the circuit to operate correctly is not effected by any additional amplifier stages, as the capacitors will only pass AC signals and block any DC component. The output AC signal is then superimposed on the biasing of the following stages. Also a bypass capacitor, CE is included in the Emitter leg circuit. This capacitor is effectively an open circuit component for DC biasing conditions, which means that the biasing currents and voltages are not affected by the addition of the capacitor maintaining a good Qpoint stability. However, this parallel connected bypass capacitor effectively becomes a short circuit to the Emitter resistor at high frequency signals due to its reactance. Thus only RL plus a very small internal resistance acts as the transistors load increasing voltage gain to its maximum. Generally, the value of the bypass capacitor, CE is chosen to provide a reactance of at most, 1/10th the value of RE at the lowest operating signal frequency. Output Characteristics Curves Ok, so far so good. We can now construct a series of curves that show the Collector current, Ic against the Collector/Emitter voltage, Vce with different values of Base current, Ib for our simple common emitter amplifier circuit. These curves are known as the “Output Characteristic Curves” and are used to show how the transistor will operate over its dynamic range. A static or DC load line is drawn onto the curves for the load resistor RL of 1.2kΩ to show all the transistors possible operating points.
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When the transistor is switched “OFF”, Vce equals the supply voltage Vcc and this is point Bon the line. Likewise when the transistor is fully “ON” and saturated the Collector current is determined by the load resistor, RL and this is point A on the line. We calculated before from the DC gain of the transistor that the Base current required for the mean position of the transistor was 45.8μA and this is marked as point Q on the load line which represents the Quiescent point or Q-point of the amplifier. We could quite easily make life easy for ourselves and round off this value to 50μA exactly, without any effect to the operating point. Output Characteristics Curves
Point Q on the load line gives us the Base current Q-point of Ib = 45.8μA or 46μA. We need to find the maximum and minimum peak swings of Base current that will result in a proportional change to the Collector current, Ic without any distortion to the output signal. As the load line cuts through the different Base current values on the DC characteristics curves we can find the peak swings of Base current that are equally spaced along the load line. These values are
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marked as points N and M on the line, giving a minimum and a maximum Base current of 20μA and 80μA respectively. These points, N and M can be anywhere along the load line that we choose as long as they are equally spaced from Q. This then gives us a theoretical maximum input signal to the Base terminal of 60μA peakto-peak, (30μA peak) without producing any distortion to the output signal. Any input signal giving a Base current greater than this value will drive the transistor to go beyond point N and into its “cut-off” region or beyond point M and into its Saturation region thereby resulting in distortion to the output signal in the form of “clipping”. Using points N and M as an example, the instantaneous values of Collector current and corresponding values of Collector-emitter voltage can be projected from the load line. It can be seen that the Collectoremitter voltage is in anti-phase (-180o) with the collector current. As the Base current Ib changes in a positive direction from 50μA to 80μA, the Collector-emitter voltage, which is also the output voltage decreases from its steady state value of 5.8v to 2.0v. Then a single stage Common Emitter Amplifier is also an “Inverting Amplifier” as an increase in Base voltage causes a decrease in Vout and a decrease in Base voltage produces an increase in Vout. In other words the output signal is 180o out-of-phase with the input signal. Common Emitter Voltage Gain The Voltage Gain of the common emitter amplifier is equal to the ratio of the change in the input voltage to the change in the amplifiers output voltage. Then ΔVL is Vout and ΔVBis Vin. But voltage gain is also equal to the ratio of the signal resistance in the Collector to the signal resistance in the Emitter and is given as:
We mentioned earlier that as the signal frequency increases the bypass capacitor, CEstarts to short out the Emitter resistor due to its reactance. Then at high frequencies RE = 0, making the gain infinite.
However, bipolar transistors have a small internal resistance built into their Emitter region called Re. The transistors semiconductor material offers an internal resistance to the flow of current through it and is generally represented by a small resistor symbol shown inside the main transistor symbol. Transistor data sheets tell us that for a small signal bipolar transistors this internal resistance is the product of 25mV ÷ Ie (25mV being the internal volt drop across the Emitter junction layer), then for our common Emitter amplifier circuit above this resistance value will be equal to:
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This internal Emitter leg resistance will be in series with the external Emitter resistor, RE, then the equation for the transistors actual gain will be modified to include this internal resistance so will be:
At low frequency signals the total resistance in the Emitter leg is equal to RE + Re. At high frequency, the bypass capacitor shorts out the Emitter resistor leaving only the internal resistance Re in the Emitter leg resulting in a high gain. Then for our common emitter amplifier circuit above, the gain of the circuit at both low and high signal frequencies is given as: At Low Frequencies
At High Frequencies
One final point, the voltage gain is dependent only on the values of the Collector resistor, RL and the Emitter resistance, (RE + Re) it is not affected by the current gain Beta, β (hFE) of the transistor. So, for our simple example above we can now summarise all the values we have calculated for our common emitter amplifier circuit and these are:
Minimum
Mean
Maximum
Base Current
20μA
50μA
80μA
Collector Current
2.0mA
4.8mA
7.7mA
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Output Voltage Swing
2.0V
Amplifier Gain
-5.32
5.8V
9.3V
-218
Common Emitter Amplifier Summary Then to summarise. The Common Emitter Amplifier circuit has a resistor in its Collector circuit. The current flowing through this resistor produces the voltage output of the amplifier. The value of this resistor is chosen so that at the amplifiers quiescent operating point, Q-point this output voltage lies half way along the transistors load line. The Base of the transistor used in a common emitter amplifier is biased using two resistors as a potential divider network. This type of biasing arrangement is commonly used in the design of bipolar transistor amplifier circuits and greatly reduces the effects of varying Beta, ( β ) by holding the Base bias at a constant steady voltage. This type of biasing produces the greatest stability. A resistor can be included in the emitter leg in which case the voltage gain becomes -RL/RE. If there is no external Emitter resistance, the voltage gain of the amplifier is not infinite as there is a very small internal resistance, Re in the Emitter leg. The value of this internal resistance is equal to 25mV/IE In the next tutorial about transistor amplifiers we will look at the Junction Field Effect Amplifier commonly called the JFET Amplifier. Like the transistor, the JFET is used in a single stage amplifier circuit making it easier to understand. There are several different kinds of field effect transistor that we could use but the easiest to understand is the junction field effect transistor, or JFET which has a very high input impedance making it ideal for amplifier circuits. AM (or Amplitude Modulation) and FM (or Frequency Modulation) are ways of broadcasting radio signals. Both transmit the information in the form of electromagnetic waves. AM works by modulating (varying) the amplitude of the signal or carrier transmitted according to the information being sent, while the frequency remains constant. This differs from FM technology in which information (sound) is encoded by varying the frequency of the wave and the amplitude is kept constant. Comparison chart AM versus FM comparison chart
Stands for
AM
FM
AM stands for Amplitude Modulation
FM stands for Frequency Modulation
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AM versus FM comparison chart AM Origin AM method of audio transmission was first successfully carried out in the mid 1870s.
FM FM radio was developed in the United states in the 1930s, mainly by Edwin Armstrong.
Modulating In AM, a radio wave known as the differences "carrier" or "carrier wave" is modulated in amplitude by the signal that is to be transmitted. The frequency and phase remain the same.
In FM, a radio wave known as the "carrier" or "carrier wave" is modulated in frequency by the signal that is to be transmitted. The amplitude and phase remain the same.
Pros and cons AM has poorer sound quality compared with FM, but is cheaper and can be transmitted over long distances. It has a lower bandwidth so it can have more stations available in any frequency range.
FM is less prone to interference than AM. However, FM signals are impacted by physical barriers. FM has better sound quality due to higher bandwidth.
Frequency AM radio ranges from 535 to 1705 KHz Range (OR) Up to 1200 bits per second.
Bandwidth Twice the highest modulating Requirements frequency. In AM radio broadcasting, the modulating signal has bandwidth of 15kHz, and hence the bandwidth of an amplitude-modulated signal is 30kHz. Zero crossing in Equidistant modulated signal Complexity Transmitter and receiver are simple but syncronization is needed in case of SSBSC AM carrier.
Noise AM is more susceptible to noise because noise affects amplitude, which
FM radio ranges in a higher spectrum from 88 to 108 MHz. (OR) 1200 to 2400 bits per second. Twice the sum of the modulating signal frequency and the frequency deviation. If the frequency deviation is 75kHz and the modulating signal frequency is 15kHz, the bandwidth required is 180kHz. Not equidistant
Tranmitter and reciver are more complex as variation of modulating signal has to beconverted and detected from corresponding variation in frequencies.(i.e. voltage to frequency and frequency to voltage conversion has to be done). FM is less susceptible to noise because information in an FM signal is transmitted
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AM versus FM comparison chart AM
FM
is where information is "stored" in an AM signal.
through varying the frequency, and not the amplitude.
Contents: AM vs FM
1 History 2 Differences in Spectrum Range 3 Pros and Cons of AM vs. FM 4 Popularity 5 Technical Details 6 References History AM method of audio transmission was first successfully carried out in the mid 1870s to produce quality radio over telephone lines and the original method used for audio radio transmissions. FM radio was developed in the United states mainly by Edwin Armstrong in the 1930s. Differences in Spectrum Range AM radio ranges from 535 to 1705 kilohertz, whereas FM radio ranges in a higher spectrum from 88 to 108 megahertz. For AM radio, stations are possible every 10 kHz and FM stations are possible every 200 kHz. Pros and Cons of AM vs. FM The advantages of AM radio are that it is relatively easy to detect with simple equipment, even if the signal is not very strong. The other advantage is that it has a narrower bandwidth than FM, and wider coverage compared with FM radio. The major disadvantage of AM is that the signal is affected by electrical storms and other radio frequency interference. Also, although the radio transmitters can transmit sound waves of frequency up to 15 kHz, most receivers are able to reproduce frequencies only up to 5kHz or less. Wideband FM was invented to specifically overcome the interference disadvantage of AM radio. A distinct advantage that FM has over AM is that FM radio has better sound quality than AM radio. The disadvantage of FM signal is that it is more local and cannot be transmitted over long distance. Thus, it may take more FM radio stations to cover a large area. Moreover, the presence of tall buildings or land masses may limit the coverage and quality of FM. Thirdly, FM requires a fairly more complicated receiver and transmitter than an AM signal does. Popularity FM radio became popular in the 1970s and early 80s. By the 1990s most music stations switched from AM and adopted FM due to better sound quality. This trend was seen in America and most of
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the countries in Europe, and slowly FM channels exceeded AM channels. Today, speech broadcasting (such as talk and news channels) still prefers to use AM, while music channels are solely FM. Technical Details
A signal may be carried by an AM or FM radio wave. AM was initially developed for telephone communication. For radio communication, a continuous wave radio signal called double sideband amplitude modulation (DSB-AM) was produced. A sideband is a band of frequencies higher (called upper sideband) or lower (called lower sideband) than the carrier frequencies which is a result of modulation. All forms of modulations produce sidebands. In DSB-AM the carrier and both USB and LSB are present. The power usage in this system proved inefficient and led to the double-sideband suppressed-carrier (DSBSC) signal in which the carrier is removed. For greater efficiency, single-sideband modulation was developed and used in which only a single sideband remained. For digital communication, a simple form of AM called continuous wave (CW) operation is used in which the presence or absence of carrier wave represents binary data. The International Telecommunication Union (ITU) designated different types of amplitude modulation in 1982 which include A3E, double sideband full–carrier; R3E, singlesideband reduced-carrier; H3E, single-sideband full-carrier; J3E, single-sideband suppressed-carrier; B8E, independent-sideband emission; C3F, vestigial-sideband and Lincompex, linked compressor and expander. FM radio characteristics and services include pre-emphasis and de-emphasis, stereophonic FM sound, Quadraphonic sound, Dolby FM and other subcarrier services. Pre-emphasis and deemphasis are processes that require boosting and reducing certain frequencies. This is done to reduce noise at high frequencies. Stereophonic FM radio was developed and formally approved in 1961 in the USA. This uses two or more audio channels independently to produce sound heard from various directions. Quadraphonic is four-channel FM broadcasting. Dolby FM is a noise reduction system used with FM radio, which has not been very successful, commercially.
