DC Pandey Electricity And Magnetism

540 — Electricity and Magnetism 11. A uniform but time varying magnetic field exists in a cylindrical region as shown in the figure. The direction of magnetic field is into the plane of the paper and its magnitude is decreasing at a constant rate of 2 × 10−3 T/s. A particle of charge 1 µC is moved slowly along a circle of radius 1m by an 1m external force as shown in figure. The plane of the circle lies in the plane of the paper and it is concentric with the cylindrical region. The work done by the external force in moving this charge along the circle will be (a) zero (b) 2π × 10−9 J (c) π × 10−9 J (d) 4π × 10−6 J 12. Switch S is closed at t = 0, in the circuit shown. The change in flux in the inductor (L = 500 mH) from t = 0 to an instant when it reaches steady state is 5Ω 5Ω 20 V 50 µF 500 mH 10 V S t=0 5Ω (a) 2 Wb (b) 1.5 Wb (c) 0 Wb (d) None of these 13. An L-R circuit is connected to a battery at time t = 0. The energy stored in the inductor reaches half its maximum value at time (a) R ln  2 (b) L ln  2 −1 L   R    2 − 1   2  (c) L ln  2  (d) R ln  2 −1 R   L    2 − 1  2  14. Electric charge q is distributed uniformly over a rod of length l. The rod is placed parallel to a long wire carrying a current i. The separation between the rod and the wire is a. The force needed to move the rod along its length with a uniform velocity v is (a) µ0iqv (b) µ0iqv 2πa 4πa (c) µ0iqvl (d) µ0iqvl 2πa 4πa 15. AB is an infinitely long wire placed in the plane of rectangular coil of dimensions as shown in the figure. Calculate the mutual inductance of wire AB and coil PQRS BP Q c A S R a (b) µ0c ln b 2π a b (a) µ0b ln a (c) µ 0abc (d) None of these 2π b 2π (b − a)2

Chapter 27 Electromagnetic Induction — 541 16. PQ is an infinite current carrying conductor. AB and CD are smooth conducting rods on which a conductor EF moves with constant velocity v as shown. The force needed to maintain constant speed of EF is P A RC EF l v a D b QB (a) 1 µ oIv ln (b) 2 (b) v µ oIv ln (a ) 2 vR  2π R  2π   (a )  (b)  (c) v µ oIv ln (b) 2 (d) None of these R  2π (a )  17. The figure shows a circular region of radius R occupied by a time varying P R magnetic field B(t) such that dB < 0. The magnitude of induced electric field at dt P Q the point P at a distance r < R is R (a) decreasing with r (b) increasing with r E (c) not varying with r (d) varying as r−2 18. Two circular loops P and Q are concentric and coplanar as shown in l1 figure. The loop Q is smaller than P. If the current I1 flowing in loop P is decreasing with time, then the current I2 in the loop Q (a) flows in the same direction as that of P (b) flows in the opposite direction as that of Q (c) is zero (d) None of the above 19. In the circuit shown in figure, the switch S is closed at t = 0. If VL is the L voltage induced across the inductor and i is the instantaneous current, S the correct variation of VL versus i is given by VL VL E (b) (a) i i O O VL VL E E (c) (d) i i O O

542 — Electricity and Magnetism 20. In the figure shown, a uniform magnetic field |B| = 0.5 T is perpendicular to the plane of circuit. The sliding rod of length l = 0.25 m moves uniformly with constant speed v = 4 ms−1. If the resistance of the slides is 2 Ω, then the current flowing through the sliding rod is B 4 Ω l v 12 Ω 2Ω (a) 0.1 A (b) 0.17 A (c) 0.08 A (d) 0.03 A 21. The figure shows a non-conducting ring of radius R carrying a charge q. In a circular region of radius r, a uniform magnetic field B perpendicular to the plane of the ring varies at a constant rate dB = β. The torque acting on the ring is dt q r R (a) 1 qr2β (b) 1 qR2β (c) qr2β (d) zero 2 2 22. A conducting ring of radius 2R rolls on a smooth horizontal conducting surface as shown in figure. A uniform horizontal magnetic field B is perpendicular to the plane of the ring. The potential of A with respect to O is B A 2R v O (a) 2 BvR (b) 1 BvR (c) 8 BvR 2 (d) 4 BvR 23. A uniformly wound long solenoid of inductance L and resistance R is cut into two parts in the ratio η : 1, which are then connected in parallel. The combination is then connected to a cell of emf E. The time constant of the circuit is (a) L (b) L R (η + 1) R (c)  η L (d)  η + 1 L  η + 1 R η R

Chapter 27 Electromagnetic Induction — 543 24. When a choke coil carrying a steady current is short-circuited, the current in it decreases to β (< 1) times its initial value in a time T . The time constant of the choke coil is (a) T (b) T (c) T (d) T ln β β ln  β1 ln β 25. In the steady state condition, the rate of heat produced in a choke coil is P. The time constant of the choke coil is τ. If now the choke coil is short-circuited, then the total heat dissipated in the coil is (a) P τ (b) 1 P τ (c) P τ 2 ln 2 (d) P τ ln 2 26. In the circuit shown in figure initially the switch is in position 1 for a long L R time, then suddenly at t = 0, the switch is shifted to position 2. It is E required that a constant current should flow in the circuit, the value of resistance R in the circuit (a) should be decreased at a constant rate 21 (b) should be increased at a constant rate (c) should be maintained constant (d) Not possible 27. The figure shows an L-R circuit, the time constant for the circuit is R L R E (a) L (b) 2L (c) 2R (d) R 2R R L 2L 28. In figure, the switch is in the position 1 for a long time, then the switch is shifted to position 2 at t = 0. At this instant the value of i1 and i2 are 12 EL R R i1 i2 L (a) E , 0 (b) E , −E R RR (c) E , −E (d) None of these 2R 2R

544 — Electricity and Magnetism 29. In a decaying L-R circuit, the time after which energy stored in the inductor reduces to one-fourth of its initial value is (a) (ln 2) L (b) 0.5 L R R (c) 2 L (d)  2 L R    2 − 1 R 30. Initially, the switch is in position 1 for a long time and then shifted to position 2 at t = 0 as shown in figure. Just after closing the switch, the magnitude of current through the capacitor is 12 ER L RC (a) zero (b) E (c) E 2R R (d) None of these 31. When the switch S is closed at t = 0, identify the correct statement just after closing the switch as shown in figure C LR SE (a) The current in the circuit is maximum (b) Equal and opposite voltages are dropped across inductor and resistor (c) The entire voltage is dropped across inductor (d) All of the above 32. Two metallic rings of radius R are rolling on a metallic rod. A magnetic field of magnitude B is applied in the region. The magnitude of potential difference between points A and C on the two rings (as shown), will be CA ωω (a) 0 (b) 4 BωR2 (c) 8 BωR2 (d) 2 BωR2

Chapter 27 Electromagnetic Induction — 545 33. In the figure, magnetic field points into the plane of paper and the conducting rod of length l is moving in this field such that the lowest point has a velocity v1 and the topmost point has the velocity v2(v2 > v1). The emf induced is given by v2 v1 (a) Bv1l (b) Bv2l (c) 1 B (v2 + v1 ) l (d) 1 B(v2 − v1 ) l 2 2 34. Find the current passing through battery immediately after key (K ) is closed. It is given that initially all the capacitors are uncharged. (Given that R = 6 Ω and C = 4µF) RC KR RC E = 5V C L RC R (a) 1 A (b) 5 A (c) 3 A (d) 2 A 35. In the circuit shown, the key (K ) is closed at t = 0, the current through the key at the instant t = 10−3 ln 2, is 4Ω 5Ω L=10 mH 20 V K 5Ω (a) 2 A 6 Ω C = 0.1 mF (c) 4 A (b) 8 A (d) zero 36. A loop shown in the figure is immersed in the varying magnetic field B = B0t, b directed into the page. If the total resistance of the loop is R, then the direction a and magnitude of induced current in the inner circle is (a) clockwise B0(πa2 − b2) (b) anti-clockwise B0π (a2 + b2) R R (c) clockwise B0 (πa2 + 4b2) (d) clockwise B0 (4b2 − πa2) R R

546 — Electricity and Magnetism 37. A square loop of side a and a straight long wire are placed in the same plane as shown in figure. The loop has a resistance R and inductance L. The frame is turned through 180° about the axis OO′. What is the electric charge that flows through the loop? O′ a b l O (a) µ 0Ia ln  2a + b (b) µ 0Ia ln  b  2πR b 2πR  − a 2  b2 (c) µ 0Ia ln  a +b2b (d) None of these 2πR More than One Correct Options 1. The loop shown moves with a velocity v in a uniform magnetic field L v of magnitude B, directed into the paper. The potential difference B between points P and Q is e. Then, L P (a) e = 1 BLv L/2 2 Q (b) e = BLv (c) P is positive with respect to Q (d) Q is positive with respect to P 2. An infinitely long wire is placed near a square loop as shown in figure. Choose the correct options. a I a a (a) The mutual inductance between the two is µ0a ln (2) 2π (b) The mutual inductance between the two is µ0a2 ln (2) 2π (c) If a constant current is passed in the straight wire in upward direction and loop is brought close to the wire, then induced current in the loop is clockwise (d) In the above condition, induced current in the loop is anti-clockwise 3. Choose the correct options. (a) SI unit of magnetic flux is henry-ampere (b) SI unit of coefficient of self-inductance is J/A (c) SI unit of coefficient of self-inductance is volt -second ampere (d) SI unit of magnetic induction is weber

Chapter 27 Electromagnetic Induction — 547 4. In the circuit shown in figure, circuit is closed at time t = 0. At time a b t = ln (2) second 2H 2Ω (a) rate of energy supplied by the battery is 16 J/s (b) rate of heat dissipated across resistance is 8 J/s (c) rate of heat dissipated across resistance is 16 J/s 8V S (d) Va − Vb = 4 V 5. Two circular coils are placed adjacent to each other. Their planes are parallel A B i2 and currents through them i1 and i2 are in same direction. Choose the correct options. i1 (a) When A is brought near B, current i2 will decrease (b) In the above process, current i2 will increase (c) When current i1 is increased, current i2 will decrease (d) In the above process, current i2 will increase 6. A coil of area 2 m2 and resistance 4 Ω is placed perpendicular to a uniform magnetic field of 4 T. The loop is rotated by 90° in 0.1 second. Choose the correct options. (a) Average induced emf in the coil is 8 V (b) Average induced current in the circuit is 20 A (c) 2 C charge will flow in the coil in above period (d) Heat produced in the coil in the above period can’t be determined from the given data 7. In L-C oscillations, (a) time period of oscillation is 2π LC (b) maximum current in circuit is q0 LC (c) maximum rate of change of current in circuit is q0 LC (d) maximum potential difference across the inductor is q0 . Here, q0 is maximum charge on 2C capacitor 8. Magnetic field in a cylindrical region of radius R in inward direction is as y shown in figure. (a) an electron will experience no force kept at (2R, 0, 0) if magnetic field increases with time O x (b) in the above situation, electron will experience the force in negative y-axis (c) If a proton is kept at 0, R , 0 and magnetic field is decreasing, then it will 2 experience the force in positive x-direction (d) if a proton is kept at (−R, 0, 0) and magnetic field is increasing, then it will experience force in negative y-axis 9. In the figure shown, q is in coulomb and t in second. At time t = 1 s a 1H b 2F c 4Ω d +– q = 2t2 (a) Va − Vb = 4 V (b) Vb − Vc = 1 V (c) Vc − Vd = 16 V (d) Va − Vd = 20 V

548 — Electricity and Magnetism 10. An equilateral triangular conducting frame is rotated with angular velocity ω a ω in a uniform magnetic field B as shown. Side of triangle is l. Choose the correct options. B c (a) Va − Vc = 0 (b) Va − Vc = Bωl2 2 b (c) Va − Vb = Bωl2 (d) Vc − Vb = − Bωl2 2 2 Comprehension Based Questions Passage I (Q. No. 1 to 3 ) R/2 P R A uniform but time varying magnetic field B = (2t3 + 24t) T is present in a cylindrical region of radius R = 2.5 cm as shown in figure. 1. The force on an electron at P at t = 2.0 s is (a) 96 × 10−21 N (b) 48 × 10−21 N (c) 24 × 10−21 N (d) zero 2. The variation of electric field at any instant as a function of distance measured from the centre of cylinder in first problem is EE EE r r r r (a) (b) (c) (d) 3. In the previous problem, the direction of circular electric lines at t = 1 s is (a) clockwise (b) anti-clockwise (c) no current is induced (d) cannot be predicted Passage II (Q. No. 4 to 7 ) A thin non-conducting ring of mass m, radius a carrying a charge q can rotate freely about its own axis which is vertical. At the initial moment, the ring was at rest in horizontal position and no magnetic field was present. At instant t = 0, a uniform magnetic field is switched on which is vertically downward and increases with time according to the law B = B0t. Neglecting magnetism induced due to rotational motion of ring. 4. The magnitude of induced emf on the closed surface of ring will be (a) πa 2B0 (b) 2a 2B0 (c) zero 1 (d) 2 πa 2B0 5. The magnitude of an electric field on the circumference of the ring is (a) aB0 (b) 2aB0 1 (d) zero (c) 2 aB0

Chapter 27 Electromagnetic Induction — 549 6. Angular acceleration of ring is (b) qB0 4m (a) qB0 2m (d) 2qB0 m (c) qB0 m 7. Find instantaneous power developed by electric force acting on the ring at t = 1 s. (a) 2q2B02a 2 (b) q2B02a 2 14m 8m (c) 3q2B02a 2 (d) q2B02a 2 m 4m Passage III (Q. No. 8 to 10 ) Figure shows a conducting rod of negligible resistance that can slide on smooth U -shaped rail made of wire of resistance 1 Ω /m. Position of the conducting rod at t = 0 is shown. A time dependent magnetic field B = 2t tesla is switched on at t = 0. 20 cm Conducting rod 40 cm 8. The current in the loop at t = 0 due to induced emf is (a) 0.16 A, clockwise (b) 0.08 A, clockwise (c) 0.16 A, anti-clockwise (d) zero 9. At t = 0, when the magnetic field is switched on, the conducting rod is moved to the left at constant speed 5 cm/s by some external means. At t = 2 s, net induced emf has magnitude (a) 0.12 V (b) 0.08 V (c) 0.04 V (d) 0.02 V 10. The magnitude of the force required to move the conducting rod at constant speed 5 cm/s at the same instant t = 2s, is equal to (a) 0.096 N (b) 0.12 N (c) 0.08 N (d) 0.064 N Passage IV (Q. No. 11 to 13) Two parallel vertical metallic rails AB and CD are separated by 1 m . They A R1 C D are connected at the two ends by resistances R1 and R2 as shown in the figure. A horizontal metallic bar L of mass 0.2 kg slides without friction, vertically down the rails under the action of gravity. There is a uniform L horizontal magnetic field of 0.6 T perpendicular to the plane of the rails. It 1m is observed that when the terminal velocity is attained, the powers B dissipated in R1 and R2 are 0.76 W and 1.2 W respectively (g = 9.8 m/ s2) 11. The terminal velocity of the bar L will be R2 (a) 2 m/s (b) 3 m/s (c) 1 m/s (d) None of these

550 — Electricity and Magnetism 12. The value of R1 is (b) 0.82 Ω (d) None of these (a) 0.47 Ω (c) 0.12 Ω (b) 0.5 Ω (d) 0.3 Ω 13. The value of R2 is (a) 0.6 Ω (c) 0.4 Ω Match the Columns 1. Match the following two columns. Column I Column II (a) Magnetic induction (p) [MT−2A−1 ] (b) Coefficient of self-induction (q) [L2T−2] (c) LC (r) [ML2T−2A−2] (d) Magnetic flux (s) None of these 2. In the circuit shown in figure, switch is closed at time t = 0. Match the following two columns. 2H 2Ω VL VR 10 V Column I Column II (a) VL at t = 0 (p) zero (b) VR at t = 0 (c) VL at t = 1 s (q) 10 V (r) 10 V (d) VR at t = 1 s e (s) 1 − 1e 10 V 3. In an L-C oscillation circuit, L = 1 H, C = 1 F and maximum charge in the capacitor is 4 C. 4 Match the following two columns. Note that in Column II, all values are in SI units. Column I Column II (a) Maximum current in the circuit (p) 16 (b) Maximum rate of change of current in the (q) 4 circuit (c) Potential difference across inductor when (r) 2 q = 2C (d) Potential difference across capacitor when (s) 8 rate of change of current is half its maximum value

Chapter 27 Electromagnetic Induction — 551 4. In the circuit shown in figure, switch remains closed for long time. It is opened at time t = 0. Match the following two columns at t = (ln 2) second. ab 9H 9V 6Ω 3Ω dc Column I Column II (a) Potential differences across inductor (p) 9 V (b) Potential difference across 3 Ω resistance (q) 4.5 V (c) Potential difference across 6 Ω resistance (r) 6 V (d) Potential difference between points b and c (s) None of these 5. Magnetic flux passing through a coil of resistance 2 Ω is as shown in φ (Wb) figure. Match the following two columns. In Column II all physical quantities are in SI units. 4 Column I Column II (a) Induced emf produced (p) 4 2 t (s) (b) Induced current (q) 1 (c) Charge flow in 2 s (r) 8 (d) Heat generation in 2 s (s) 2 6. A square loop is placed near a long straight current carrying wire as shown. Match the following two columns. i Column I Column II (a) If current is increased (p) induced current in loop is clockwise (b) If current is decreased (q) induced current in loop is anti-clockwise (c) If loop is moved away from the wire (r) wire will attract the loop (d) If loop is moved towards the wire (s) wire will repel the loop

552 — Electricity and Magnetism Subjective Questions 1. In the circuit diagram shown, initially there is no energy in the inductor and the capacitor. The switch is closed at t = 0. Find the current I as a function of time if R = L/C. RL IR C VS 2. A rectangular loop with a sliding connector of length l is located in a uniform magnetic field perpendicular to the loop plane. The magnetic induction is equal to B. The connector has an electric resistance R, the sides ab and cd have resistances R1 and R2. Neglecting the self-inductance of the loop, find the current flowing in the connector during its motion with a constant velocity v. aB d R1 R v R2 bc 3. A rod of length 2a is free to rotate in a vertical plane, about a horizontal axis O passing through its mid-point. A long straight, horizontal wire is in the same plane and is carrying a constant current i as shown in figure. At initial moment of time, the rod is horizontal and starts to rotate with constant angular velocity ω, calculate emf induced in the rod as a function of time. i d Oω 2a 4. In the circuit arrangement shown in figure, the switch S is closed at t = 0. Find the current in the inductance as a function of time? Does the current through 10 Ω resistor vary with time or remains constant. 10 Ω 5 Ω 1 mH S 36 V

Chapter 27 Electromagnetic Induction — 553 5. In the circuit shown, switch S is closed at time t = 0. Find the current through the inductor as a function of time t. 2 V 2Ω 4 V 1Ω S L = 1 mH 6. In the circuit shown in figure, E = 120 V, R1 = 30.0 Ω , R2 = 50.0 Ω , and L = 0.200 H. Switch S is closed at t = 0. Just after the switch is closed. + S E b a R1 R2 c L d (a) What is the potential difference Vab across the inductor R1? (b) Which point, a or b, is at higher potential? (c) What is the potential difference Vcd across the inductor L? (d) Which point, c or d, is at a higher potential? The switch is left closed for a long time and then is opened. Just after the switch is opened (e) What is the potential difference Vab across the resistor R1? (f) Which point a or b, is at a higher potential? (g) What is the potential difference Vcd across the inductor L? (h) Which point, c or d, is at a higher potential? 7. Two capacitors of capacitances 2C and C are connected in series with an inductor of inductance L. Initially, capacitors have charge such that VB − VA = 4V0 and VC − VD = V0. Initial current in the circuit is zero. Find AB CD 2C C L (a) maximum current that will flow in the circuit, (b) potential difference across each capacitor at that instant, (c) equation of current flowing towards left in the inductor. 8. A 1.00 mH inductor and a 1.00 µF capacitor are connected in series. The current in the circuit is described by i = 20 t, where t is in second and i is in ampere. The capacitor initially has no charge. Determine (a) the voltage across the inductor as a function of time, (b) the voltage across the capacitor as a function of time, (c) the time when the energy stored in the capacitor first exceeds that in the inductor.

554 — Electricity and Magnetism 9. In the circuit shown in the figure, E = 50.0 V, R = 250 Ω and C = 0.500 µF. The switch S is closed for a long time, and no voltage is measured across the capacitor. After the switch is opened, the voltage across the capacitor reaches a maximum value of 150 V. What is the inductance L? R E LC S 10. The conducting rod ab shown in figure makes contact with metal rails ca and db. The apparatus is in a uniform magnetic field 0.800 T, perpendicular to the plane of the figure. xx x ax c xx xx xB x v x 50.0 cm xx xx x bx d xx (a) Find the magnitude of the emf induced in the rod when it is moving toward the right with a speed 7.50 m/s. (b) In what direction does the current flow in the rod? (c) If the resistance of the circuit abdc is 1.50 Ω (assumed to be constant), find the force (magnitude and direction) required to keep the rod moving to the right with a constant speed of 7.50 m/s. You can ignore friction. (d) Compare the rate at which mechanical work is done by the force (Fv) with the rate at which thermal energy is developed in the circuit (I2R). 11. A non-conducting ring of mass m and radius R has a charge Q uniformly distributed over its circumference. The ring is placed on a rough horizontal surface such that plane of the ring is parallel to the surface. A vertical magnetic field B = B0t2 tesla is switched on. After 2 s from switching on the magnetic field the ring is just about to rotate about vertical axis through its centre. (a) Find friction coefficient µ between the ring and the surface. (b) If magnetic field is switched off after 4 s, then find the angle rotated by the ring before coming to stop after switching off the magnetic field. 12. Two parallel long smooth conducting rails separated by a x x x x x x x x distance l are connected by a movable conducting x x x x x x x x connector of mass m. Terminals of the rails are connected R lF C by the resistor R and the capacitor C as shown in figure. x x x x x x x x A uniform magnetic field B perpendicular to the plane of the rail is switched on. The connector is dragged by a x x x x x x x x constant force F. Find the speed of the connector as a function of time if the force F is applied at t = 0. Also find the terminal velocity of the connector.

Chapter 27 Electromagnetic Induction — 555 13. A circuit containing capacitors C1 and C2, shown in the figure is in the steady state with key K1 closed and K 2 opened. At the instant t = 0, K1 is opened and K 2 is closed. 20 V R K1 C1 = 2 µF C2 = 2 µF K2 L = 0.2 mH (a) Find the angular frequency of oscillations of L-C circuit. (b) Determine the first instant t, when energy in the inductor becomes one third of that in the capacitor. (c) Calculate the charge on the plates of the capacitor at that instant. 14. Initially, the capacitor is charged to a potential of 5 V and then connected to position 1 with the shown polarity for 1 s. After 1 s it is connected across the inductor at position 2. 100 Ω 1 2 + + L = 25 mH – E = 10 V 10 µF – (a) Find the potential across the capacitor after 1 s of its connection to position 1. (b) Find the maximum current flowing in the L-C circuit when capacitor is connected across the inductor. Also, find the frequency of LC oscillations. 15. A rod of mass m and resistance R slides on frictionless × × × × × a × × × and resistanceless rails a distance l apart that include a × × × × × × × × source of emf E0. (see figure). The rod is initially at rest. × × E×0 × × × × × Find the expression for the (a) velocity of the rod v (t). (b) current in the loop i(t). × × × × ×b × × × 16. Two metal bars are fixed vertically and are connected on the top by a capacitor C. A sliding conductor of length l and mass m slides with its ends in contact with the bars. The arrangement is placed in a uniform horizontal magnetic field directed normal to the plane of the figure. The conductor is released from rest. Find the displacement x(t) of the conductor as a function of time t. ××××××× ××××××× l ××××××× ×××××××

556 — Electricity and Magnetism 17. A conducting light string is wound on the rim of a metal ring of radius r and mass m. The free end of the string is fixed to the ceiling. A vertical infinite smooth conducting plane is always tangent to the ring as shown in the figure. A uniform magnetic field B is applied perpendicular to the plane of the ring. The ring is always inside the magnetic field. The plane and the string are connected by a resistance R. When the ring is released, find ××××× ××××× × × R× × × ××××× ××××× ××××× ××××× ××××× (a) the current in the resistance R as a function of time. (b) the terminal velocity of the ring. 18. A conducting frame abcd is kept in a vertical plane. A conducting rod ef of mass m and length l can slide smoothly on it remaining always horizontal. The resistance of the loop is negligible and inductance is constant having value L. The rod is left from rest and allowed to fall under gravity and inductor has no initial current. A magnetic field of constant magnitude B is present throughout the loop pointing inwards. Determine b Lc B ef ad (a) position of the rod as a function of time assuming initial position of the rod to be x = 0 and vertically downward as the positive x-axis. (b) the maximum current in the circuit. (c) maximum velocity of the rod 19. A rectangular loop with a sliding conductor of length l is located in a uniform magnetic field perpendicular to the plane of loop. The magnetic induction perpendicular to the plane of loop is equal to B. The part ad and bc has electric resistance R1 and R2, respectively. The conductor starts moving with constant acceleration a0 at time t = 0. Neglecting the self-inductance of the loop and resistance of conductor. Find ab ⊗B l a0 R2 R1 dc (a) the current through the conductor during its motion. (b) the polarity of abcd terminal. (c) external force required to move the conductor with the given acceleration.

Chapter 27 Electromagnetic Induction — 557 20. A conducting circular loop of radius a and resistance per unit length R is moving with a constant velocity v0,parallel to an infinite conducting wire carrying current i0.A conducting rod of length 2a is approaching the centre of the loop with a constant velocity v0 along the direction 2 of the current. At the instant t = 0,the rod comes in contact with the loop at A and starts sliding on the loop with the constant velocity. Neglecting the resistance of the rod and the self-inductance of the circuit, find the following when the rod slides on the loop. v0/2 i0 P C a√3 a Q A v0 O B (a) The current through the rod when it is at a distance of  a  from the point A of the loop. 2 (b) Force required to maintain the velocity of the rod at that instant. 21. U-frame ABCD and a sliding rod PQ of resistance R, start moving with velocities v and 2v respectively, parallel to a long wire carrying current i0. When the distance AP = l at t = 0, determine the current through the inductor of inductance L just before connecting rod PQ loses contact with the U-frame. A 2v B i0 P Ql D v a C a

Answers Introductory Exercise 27.1 2. No 4. Clockwise 1. Anti-clockwise 3. [ML2A–1 T –3 ] 6. 6.74 V 8. 9.0 × 10–7 Wb 5. Same direction, opposite direction. 7. 1600 µC Introductory Exercise 27.2 2. 0.00375 N 4. No 1. 4.4 V, north 3. (a) Bωl 2 (b) −3Bωl 2 22 Introductory Exercise 27.3 2. − 80e −4t, − 40 e −4t 4. (a) 4.5 × 10–5 H (b) 4.5 × 10–3 V 1. 3 (t cos t + sin t ) 3. (a) 0.625 mH (b) 0.13 J, 0.21 J/s Introductory Exercise 27.4 2. (a) 2 H (b) 30 V (c) 1 H 1. 3.125 mH, 0.9375 V 3. (a) 0.27 V, Yes (b) 0.27 V Introductory Exercise 27.5 3. No 2. (a) 0.2 s (b) 10 A (c) 9.93 A Introductory Exercise 27.6 2. With KE as v ⇔ i and m ⇔ L. Therefore, 1 mv2 = 1 Li 2 3. (a) 45.9 µC (b) 23.3 V 4. 20.0 V 22 Introductory Exercise 27.7 1. (a) 3.1 × 10–6 V (b) 2.0 × 10–6 V/ m 2. (a) 8.0 × 10–21 N (downward and to the right perpendicular to r2) (b) 0.36 V/ m (upwards and to the left perpendicular to r1 ) Exercises LEVEL 1 Assertion and Reason 1. (d) 2. (b) 3. (d) 4. (a,b) 5. (a) 6. (d) 7. (b) 8. (c) 9. (d) 10. (c) Objective Questions 4. (c) 14. (c) 1. (c) 2. (a) 3. (b) 24. (b) 5. (c) 6. (b) 7. (b) 8. (a) 9. (c) 10. (b) 11. (a) 12. (b) 13. (c) 34. (d) 15. (a) 16. (a) 17. (c) 18. (a) 19. (d) 20. (b) 21. (a) 22. (d) 23. (c) 25. (a) 26. (c) 27. (d) 28. (d) 29. (b) 30. (d) 31. (b) 32. (a) 33. (a) 35. (c) 36. (b) 37. (c) 38. (c) 39. (a)

Chapter 27 Electromagnetic Induction — 559 Subjective Questions 2. 272 m 1. True 4. (a) µ 0 iv ln 1 + l  (b) a (c) zero 3. e = aSB 0e – at 2π d 5. (a) (0.5) (1 – e –10t ) A (b) 1.50 A – (0.25 A) e –10t 6. 0.5 T 7. (a) 0.011 V (b) zero (c) – 0.011 V I F 8. F0 I0 L 3L 9. e = µ 0 2ia2v 22 4 π x(x + a) X X 3L L O L 3L 3L L O 10. (2 × 10−5 ) ln (10) V – 2 – 2 22 – 2 – 2 11. 3.33 × 10–2 A, 6.67 × 10–2 A – I0 12. (a) 0.25 H (b) 4.5 × 10–4 Wb 13. (a) 1.96 H (b) 7.12 × 10–3 Wb F0 = B 2L2V i0 = BLV 14. 0.37 R R 16. (a) 17.3 µs (b) 30.7 µs (a) (b) 15. (a) 518 mW (b) 328 mW (c) 191 mW 17. (a) E e – R + r  t (b) E 2L 18. 8 A 19. (a) 3.6 mH (b) 1.33 kHz (c) 0.188 ms L 3 r 2r (R + r ) 20. (a) 6.28 × 103 rad/s, 10−3 s (b) 10−4 cos (6.28 × 103 t ) (c) 0.0253 H (d) 0.4 A 21. (a) 1.25 × 10–2 V (b) 8.33 × 10–4 A (c) 3.125 × 10–8 J (d) 4.33 × 10–6 C, 7.8 × 10–7 J LEVEL 2 Single Correct Option 1. (b) 2. (c) 3. (a) 4. (b) 5. (c) 6. (a) 7. (a) 8. (c) 9. (c) 10. (a) 11. (b) 12. (b) 13. (c) 14. (a) 15. (b) 16. (a) 17. (b) 18. (a) 19. (d) 20. (a) 21. (a) 22. (d) 23. (a) 24. (b) 25. (b) 26. (d) 27. (b) 28. (b) 29. (a) 30. (c) 31. (c) 32. (b) 33. (c) 34. (a) 35. (a) 36. (d) 37. (d) More than One Correct Options 6. (b,c,d) 7. (b,c) 8. (b,c,d) 9. (a,b,c) 10. (a,c) 1. (a,c) 2. (a,c) 3. (a,c) 4. (a,b,d) 5. (a,c) Comprehension Based Questions 1. (b) 2. (c) 3. (b) 4. (a) 5. (c) 6. (a) 7. (d) 8. (a) 9.(b) 10.(c) 11. (c) 12. (a) 13. (d) Match the Columns 1. (a) → p (b) → r (c) → s (d) → s (c) → r (d) → s 2. (a) → q (b) → p (c) → s (d) → s (c) → p (d) → p 3. (a) → s (b) → p (c) → s (d) → p (c) → p, r (d) → q, s 4. (a) → s (b) → q 5. (a) → s (b) → q 6. (a) → q, s (b) → p, r

560 — Electricity and Magnetism Subjective Questions 1. I = V 2. i = Bvl where R ′ = R1R2 3. e = µ 0iω d ln  d − a sinωt  +  R 2 π sinωt  +  2a R + R′ R1 + R 2  sinωt d  a sinωt   4. 3.6 (1 − e – t/ τL ) A. Here, τL = 300 µs. Current through 10 Ω resistor varies with time. 5. i = 5(1 − e –2000t/3 ) A 6. (a) 120 V (b) a (c) 120 V (d) c (e) −72 V (f) b (g) −192 V (h) d 3 7. (a) q0ω (b) 3V0 : 3V0 (c) i = q0 sinωt. Here, q0 = 2CV0 and ω = 2LC 8. (a) 20 mV (b) 107 t 2 V (c) 63.2 µs 9. 0.28 H 10. (a) 3 V (b) b to a (c) 0.8 N towards right (d) both are 6 W 11. (a) 2B 0QR (b) B 0Q mg m 12. V = FR (1 − e – αt ) Here, α = mR B 2l2 , vT = FR B 2l2 + RB 2l 2C B 2l2 13. (a) 5 × 104 rad/s (b) 1.05 × 10–5 s (c) 10 3 µC 14. (a) 8.16 V (b) 5.16 A, 10 Hz E0 – B2l 2 t (b) i = E0 – Blv mgt 2 16. x = 2 (m + CB 2l 2 ) 15. (a) v = (1 – e mR ) Bl R mg − −2B2r2 t mgR 17. (a) i = (1 − e mR ) (b) vT = 4B 2r 2 2Br 18. (a) x = v0 (1 – cos ωt ), ω = Bl 2mg g mL ω mL (b) imax = Bl (c) vmax = v0 = Bl 19. (a) i = Bla0t (R1 + R2) (b) Polarity of a, b is positive and polarity of c, d is negative R1R 2 (c) Fext = a0  m + B 2l2t (R1 + R2 )  R1R 2   20. (a) i = 9v0µ 0 i0 ln (3) (b) 9 µ 2 i02 v0 (ln 3)2 21. i =  e  [1 − e −lR /Lv ] , where e = µ 0i0v ln (2) 16aR π 2 0 R 2π 32aR π3

Alternating Current Chapter Contents 28.1 Introduction 28.2 Alternating currents and phasors 28.3 Current and potential relations 28.4 Phasor algebra 28.5 Series L-R circuit 28.6 Series C-R circuit 28.7 Series L-C-R circuit 28.8 Power in an AC circuit

562 — Electricity and Magnetism 28.1 Introduction A century ago, one of the great technological debates was whether the electrical distribution system should be AC or DC. Thomas Edison favoured direct current (DC), that is, steady current that does not vary with time. George Westinghouse favoured alternating current (AC), with sinusoidally varying voltages and currents. He argued that transformers can be used to step the voltage up or down with AC but not with DC. Low voltages are safer for consumer use, but high voltages and correspondingly low currents are best for long distances power transmission to minimize i2R losses in the cables. Eventually, Westinghouse prevailed, and most present day household and industrial power distribution systems operate with alternating current. i or V i or V V0, i0 t + + t + Fig. 28.1 – – –i 0 i = i0 cos ωt –V0 i = i0 sin ωt or or V = V0 cos ωt V = V0 sin ωt A time varying current or voltage may be periodic and non-periodic. In case of periodic current or voltage, the current or voltage is said to be alternating if its amplitude is constant and alternate half cycle is positive and half negative. If the current or voltage varies periodically as sine or cosine function of time, the current or voltage is said to be sinusoidal and is what we usually mean by it. 28.2 Alternating Currents and Phasors The basic principle of the AC generator is a direct consequence of Faraday's law of induction. When a conducting loop is rotated in a magnetic field at constant angular frequency ω a sinusoidal voltage (emf) is induced in the loop. This instantaneous voltage is V =V0 sin ωt …(i) The usual circuit diagram symbol for an AC source is shown in Fig. 28.2. Fig. 28.2 In Eq. (i),V0 is the maximum output voltage of the AC generator or the voltage amplitude and ω is the angular frequency equal to 2π times the frequency f. ω = 2 πf The frequency of AC in India is 50 Hz, i.e. f = 50 Hz So, ω = 2πf ≈ 314 rad/s The time of one cycle is known as time period T, the number of cycles per second the frequency f. T=1 or T = 2π f ω

Chapter 28 Alternating Current — 563 A sinusoidal current might be described as i = i0 sin ωt If an alternating current is passed through an ordinary ammeter or voltmeter, it will record the mean value for the complete cycle, as the quantity to be measured varies with time. The average value of current for one cycle is T 2π/ω ∫ ∫i One cycle = (iav )T idt 0 (i0 sin ωt ) dt 0 = T = 2π/ω =0 ∫ ∫0 0 dt dt Thus, i One cycle = 0 Similarly, the average value of the voltage (or emf) for one cycle is zero. V One cycle = 0 Since, these averages for the whole cycle are zero, the DC instrument will indicate zero deflection. In AC, the average value of current is defined as its average taken over half the cycle. Hence, T /2 π/ω ∫ ∫i Half cycle = (iav )T /2 = idt 0 (i0 sin ωt) dt 0 2 T /2 = π/ω = π i0 ∫ ∫0 0 dt dt This is sometimes simply written as iav . Hence, iav = i Half cycle = 2 i0 ≈ 0.637 i0 π Similarly, Vav = 2 ≈ 0.637V0 π V0 A DC meter can be used in an AC circuit if it is connected in the full wave rectifier circuit. The average value of the rectified current is the same as the average current in any half cycle, i.e. 2 times π the maximum current i0. A more useful way to describe a quantity is the root mean square (rms) value. We square the instantaneous current, take the average (mean) value of i2 and finally take the square root of that average. This procedure defines the root-mean-square current denoted as irms. Even when i is negative, i2 is always positive so irms is never zero (unless i is zero at every instant). Hence, T i2dt 2π/ω (i02 sin 2 ωt) ∫ ∫i2 0 0 dt i02 T 2π/ω 2 = = = dt One cycle ∫ ∫0 0dt ∴ irms = i2 One cycle = i0 ≈ 0.707 i0 2 Thus, i rms = i0 ≈ 0.707 i0 2

564 — Electricity and Magnetism Similarly, we get V rms = V0 ≈ 0.707V0 2 The square root of the mean square value is called the virtual value and is the value given by AC instruments. Thus, when we speak of our house hold power supply as 220 V AC, this means that the rms voltage is 220 V and its voltage amplitude is V0 = 2 Vrms = 311 V i Form Factor i0 i rms = 0.707 i 0 i av = 0.637 i 0 The ratio, = rms value = V0/ 2 π = 1.11 t = average value 2V0/π 2 2 is known as form factor. The different values i0, iav and irms are shown in Fig. 28.3. Fig. 28.3 Note (1) The average value of sin ωt, cos ωt, sin 2ωt, cos 2ωt, etc., is zero because it is positive in half of the time and negative in rest half of the time. Thus, sin ωt = cos ωt = sin 2 ωt = cos 2 ωt = 0 If i = i0 sin ωt then i = i0 sin ωt = i0 sin ωt = 0 (2) The average value of sin2 ωt and cos2 ωt is 1 ⋅ 2 or sin2 ωt = cos2 ωt = 1 2 If i 2 = i02 sin2 ωt = i02 then 2 i2 = i02 sin2 ωt = i02 sin2 ωt (3) Like SHM, general expressions of current/voltage in an sinusoidal AC are i = i0 sin (ωt ± φ) V = V0 sin (ωt ± φ) or i = i0 cos (ωt ± φ) and V = V0 cos (ωt ± φ) (4) Average value of current or voltage over a half cycle can be zero also. This depends on the time interval (of course T /2) over which average value is to be found. Think why? Phasors If an AC generator is connected to a series circuit containing resistors, inductors and capacitors and we want to know the amplitude and time characteristics of the alternating current. To simplify our analysis of circuits containing two or more elements, we use graphical constructions called phasor diagrams. In these constructions, alternating (sinusoidal) quantities, such as current and voltage are rotating vectors called phasors. In these diagrams, the instantaneous value of a quantity that varies sinusoidally with time is represented by the projection onto a vertical axis (if it is a sine function) or onto a horizontal axis (if it

Chapter 28 Alternating Current — 565 is a cosine function) of a vector with a length equal to the amplitude (i0 ) of the quantity. The vector rotates counterclockwise with constant angular velocity ω. ωω i0 sin ωt i0 i0 or ωt ωt O O i 0 cos ωt Fig. 28.4 A phasor is not a real physical quantity with a direction in space, such as velocity, momentum or electric field. Rather, it is a geometric entity that helps us to describe and analyze physical quantities that vary sinusoidally with time. V Example 28.1 Show that average heat produced during a cycle of AC is same as produced by DC with i = irms . Solution For an AC, i = i0 sin ωt Therefore, instantaneous value of heat produced in time dt across a resistance R is dH = i 2 Rdt = i02 R sin 2 ωt dt ∴ Average value of heat produced during a cycle, T 2π/ω (i02 2 0 ∫∫ ∫ ∫Hav =dH R sin ωt ) dt 0= 2π /ω T dt dt 0 0 = i02 R 2ωπ = i 2 RT 2 rms i.e. AC produces same heating effects as DC of value i = i rms . V Example 28.2 If the current in an AC circuit is represented by the equation, i = 5 sin (300t – π /4) Here, t is in second and i in ampere. Calculate (a) peak and rms value of current (b) frequency of AC (c) average current Solution (a) As in case of AC, i = i0 sin (ωt ± φ) Ans. ∴ The peak value, i0 = 5 A Ans. and i rms = i0 = 5 = 3.535 A 22 (b) Angular frequency, ω = 300 rad/s Ans. ∴ f = ω = 300 ≈ 47.75 Hz Ans. (c) 2π 2π iav =  π2 i0 =  π2 (5) = 3.18 A

566 — Electricity and Magnetism 28.3 Current and Potential Relations In this section, we will derive voltage current relations for individual circuit elements carrying a sinusoidal current. We will consider resistors, inductors and capacitors. Resistor in an AC Circuit …(i) Consider a resistor with resistance R through which there is a sinusoidal current given by i = i0 sin ωt R a ib Fig. 28.5 Here, i0 is the current amplitude (maximum current). From Ohm's law, the instantaneous PD between points a and b is VR = iR = (i0R ) sin ωt We can write as i0R =V0 , the voltage amplitude ∴ VR =V0 sin ωt …(ii) From Eqs. (i) and (ii), we can see that current and voltage are in phase if only resistance is in the circuit. Fig. 28.6 shows graphs of i and VR as functions of time. i or VR i0 i = i0 sin ωt i0 V0 t VR = V0 sin ωt V0 Fig. 28.6 ωt O Fig. 28.7 The corresponding phasor diagram is shown in Fig. 28.7. Because i and VR are in phase and have the same frequency, the current and voltage phasors rotate together, they are parallel at each instant. Their projection on vertical axis represents the instantaneous current and voltage respectively. Note Direction of an alternating current is not shown in a circuit, as it keeps on changing. In the figure, the direction of instantaneous current is only shown. Capacitor in an AC Circuit If a capacitor of capacitance C is connected across the alternating source, the instantaneous charge on the capacitor is q = CVC = CV0 sin ωt

Chapter 28 Alternating Current — 567 and the instantaneous current i passing through it is given by V = V0 sin ωt i = dq = CV0ω cos ωt q –q dt ab = V0 sin (ωt + π/2) i 1/ωC Fig. 25.8 or i = i0 sin (ωt + π/2) Here, V0 = i0 ωC This relation shows that the quantity 1 is the effective AC resistance or the capacitive reactance of ωC the capacitor and is represented as X C . It has unit as ohm. Thus, XC =1 ωC It is clear that the current leads the voltage by 90° or the potential drop across the capacitor lags the current passing it by 90°. Fig. 28.9 shows V and i as functions of time t. i,V i0 sin ωt i0 i0 i V0 t VC V0 cos ωt ωt 90° V0 Fig. 28.9 Fig. 28.10 The phasor diagram 28.10 shows that voltage phasor is behind the current phasor by a quarter cycle or 90°. Inductor in an AC Circuit Consider a pure inductor of self-inductance L and zero resistance L connected to an alternating source. Again we assume that an ab instantaneous current i = i0 sin ωt flows through the inductor. i Although, there is no resistance, there is a potential difference VL between the inductor terminals a and b because the current varies with Fig. 28.11 time giving rise to a self-induced emf. VL = Vab = – (induced emf ) = –  – L di  dt or VL = L di = Li0ω cos ωt dt

568 — Electricity and Magnetism or VL = V0 sin ωt + π  …(i) 2 Here, V0 = i0 (ωL) …(ii) or i0 = V0 ωL ∴ i = V0 sin ωt …(iii) ωL Eq. (iii) shows that effective AC resistance, i.e. inductive reactance of inductor is X L = ωL and the maximum current, i0 = V0 XL The unit of X L is also ohm. From Eqs. (i) and (iii), we see that the voltage across the inductor leads the current passing through it by 90°. Fig. 28.12 shows VL and i as functions of time. i , VL i0 sin ωt i0 V0 i0 i t V0 V0 cos ωt VL 90° ωt Fig. 28.12 Fig. 28.13 Phasor diagram in Fig. 28.13 shows that VL leads the current i by 90°. Extra Points to Remember ˜ Circuit elements with AC Circuit elements Amplitude relation Circuit quantity Phase of V in phase with i Resistor V0 = i0R R lags i by 90° Capacitor V0 = i0 XC leads i by 90° Inductor V0 = i0 XL XC = 1 ωC XL = ωL ˜ In DC, ω = 0, therefore, XL = 0 and XC = ∞

Chapter 28 Alternating Current — 569 ˜ The potential of point a with respect to point b is given by L VL = + L di , the negative of the induced emf. This expression gives the a i b dt Fig. 28.14 correct sign of VL in all cases. ˜ If an oscillating voltage of a given amplitude V0 is applied across an inductor, the resulting current will have a smaller amplitude i0 for larger value of ω. Since, XL is proportional to frequency, a high frequency voltage applied to the inductor gives only a small current while a lower frequency voltage of the same amplitude gives rise to a larger current. Inductors are used in some circuit applications, such as power supplies and radio interference filters to block high frequencies while permitting lower frequencies to pass through. A circuit device that uses an inductor for this purpose is called a low pass filter. ˜ The capacitive reactance of a capacitor is inversely proportional to the capacitance C and the angular frequency ω. The greater the capacitance and the higher the frequency, the smaller is the capacitive reactance XC . Capacitors tend to pass high frequency current and to block low frequency current, just the opposite of inductors. A device that passes signals of high frequency is called a high pass filter. ˜ Figure shows the graphs of R, XL and XC as functions of angular frequency ω. R, X XL R XC ω Fig. 28.15 ˜ Remember that we can write, VR = iR, (V0 )R = i0R, (V0 )L = i0 XL and (V0 )C = i0 XC but can't write (for instantaneous voltages). VL = i XL or VC = i XC This is because there is a phase difference between the voltage and current in both an inductor and a capacitor. V Example 28.3 A 100 Ω resistance is connected in series with a 4 H inductor. The voltage across the resistor is V R = (2.0 V ) sin (103 rad /s) t : (a) Find the expression of circuit current (b) Find the inductive reactance (c) Derive an expression for the voltage across the inductor. Solution (a) i = VR = (2.0 V)sin (103 rad /s ) t R 100 = (2.0 × 10–2 A ) sin (103 rad /s) t Ans. (b) X L = ωL = (103 rad /s ) (4 H) Ans. = 4.0 × 103 Ω (c) The amplitude of voltage across inductor, V0 = i0 X L = (2.0 × 10–2 A ) (4.0 × 103 Ω ) = 80 V Ans.

570 — Electricity and Magnetism In an AC, voltage across the inductor leads the current by 90° or π/2 rad. Hence, VL = V0 sin (ωt + π/ 2) = (80 V) sin (103 rad /s ) t + π rad 2 Note That the amplitude of voltage across the resistor (= 2.0 V ) is not same as the amplitude of the voltage across the inductor (= 80 V ), even though the amplitude of the current through both devices is the same. 28.4 Phasor Algebra The complex quantities normally employed in AC circuit analysis, can be added and subtracted like coplanar vectors. Such coplanar vectors which represent sinusoidally time varying quantities are known as phasors. y In Cartesian form, a phasor A can be written as A = a + jb bA where, a is the x-component and b is the y-component of phasor A. θ a x The magnitude of A is |A| = a2 + b2 Fig. 28.16 and the angle between the direction of phasor A and the positive x-axis is θ = tan –1  b  a When a given phasor A, the direction of which is along the x-axis is multiplied by the operator j, a new phasor jA is obtained which will be 90° anti-clockwise from A, i.e. along y-axis. If the operator j is multiplied now to the phasor jA, a new phasor j2A is obtained which is along –x-axis and having same magnitude as of A. Thus, j2A = – A ∴ j2 = – 1 or j = –1 Now, using the j operator, let us discuss different circuits of an AC. 28.5 Series L-R Circuit As we know, potential difference across a resistance in AC is in phase with current and it leads in phase by 90° with current across the inductor. VR VL Fig. 28.17 Suppose in phasor diagram current is taken along positive x-direction. Then,VR is also along positive x-direction and VL along positive y-direction, so, we can write

Chapter 28 Alternating Current — 571 V = VR + jVL = iR + j(iX L ) (as X L = ωL) = iR + j(iωL) = iZ y VL V x iφ VR Fig. 28.18 Here, Z = R + j X L = R + j (ωL) is called as impedance of the circuit. Impedance plays the same role in AC circuits as the ohmic resistance does in DC circuits. The modulus of impedance is | Z| = R 2 + (ωL) 2 The potential difference leads the current by an angle, φ = tan –1 | VL | = tan –1  XL  | VR | R or φ = tan –1  ωL  R 28.6 Series C-R Circuit VC VR Fig. 28.19 Potential difference across a capacitor in AC lags in phase by 90° with the current in the circuit. Suppose in phasor diagram current is taken along positive x-direction. Then, VR is also along positive x-direction but VC is along negative y-direction. So, we can write y i VR x φ VC V Fig. 28.20 V = VR – jVC = iR – j(iXC ) = iR – j  i  = iZ ωC Here, impedance is Z = R – j  1  ωC

572 — Electricity and Magnetism The modulus of impedance is  1  2 ωC | Z| = R 2 + and the potential difference lags the current by an angle, φ = tan –1 VC = tan –1 XC = tan –1 1/ωRC  VR R or φ = tan –1  1  ωRC 28.7 Series L-C-R Circuit Potential difference across an inductor leads the current by 90° in VL VC VR phase while that across a capacitor, it lags in phase by 90°. Suppose in a phasor diagram current is taken along positive x-direction. Then, VR is along positive x-direction, VL along positive y-direction and VC along negative y-direction. Fig. 28.21 V = VR2 + (VL – VC)2 VL VL – VC y i ⇒VR φ x VR VC Fig. 28.22 Let us assume that X L > X C or VL >VC So, we can write V =VR + jVL – jVC = iR + j(iX L ) – j(iX C ) = iR + j[i ( X L – X C )] = iZ Here, impedance is Z=R + j (X L – XC ) =R + j ωL – 1  ωC The modulus of impedance is | Z| = R2 + ωL – 1  2 ωC and the potential difference leads the current by an angle, φ = tan –1 VL − VC = tan –1  X L – XC  VR R or φ = tan –1  ωL – ω1C  R

Chapter 28 Alternating Current — 573 Note Let us take the most general case of a series L-C-R circuit in an AC. | Z | = R 2 + (XL ~ XC )2 If XL = XC or ωL = 1 or ω = 1 or f = 1 the modulus of impedance ωC LC 2π LC |Z|= R and the current is in phase with voltage, i.e. if V = V0 sin ωt, then i = i0 sin ωt where, i0 = V0 = V0 |Z| R Such a condition is known as resonance and frequency known as resonance frequency and is given by f= 1 2π LC The current in such a case is maximum. If XL > XC, then the modulus of the impedance | Z | = R 2 + (XL – XC ) 2 and the voltage leads the current by an angle given by φ = tan–1  XL – XC  R i.e. if V = V0 sin ωt, then i = i0 sin (ωt – φ) where, i0 = V0 |Z| If XC > XL, then the modulus of the impedance is | Z | = R 2 + (XC – XL)2 and the current leads the voltage by an angle given by φ = tan–1  XC – XL R i.e. if V = V0 sin ωt, then i = i0 sin (ωt + φ) where, i0 = V0 |Z| Extra Points to Remember ˜ i0 = V0 , irms = Vrms . But in general i ≠ V. |Z| |Z| |Z| ˜ In L-C-R circuit, whenever voltage across various elements is asked, find rms values unless stated in the question for the peak or instantaneous value. The rms values are VR = i rmsR, VL = i rms XL 2. The instantaneous values across and VC = i rms XC The peak values can be obtained by multiplying the rms values by different elements is rarely asked.

574 — Electricity and Magnetism ˜ Voltage magnification in series resonance circuit At resonance  f = 1  , the PD across the 2 π LC inductor and the capacitor are equal and 180° out of phase and therefore, cancel out. Hence, the applied emf is merely to overcome the resistance opposition only. If an inductance or capacitance of very large reactance (XL or XC ) is connected with XL = XC (at resonance) then PD across them increases to a very high value. The ratio is known as voltage magnification and is given by, PD across inductance (or capacitance) = i rms(ω L) = ω L or i rms  1  1 ωC = Applied emf i rms(R) R i rms (R) ωCR This ratio is greater than unity. ˜ Response curves of series circuit The impedance of an XL, XC, R, Z, i L- C - R circuit depends on the frequency. The dependence is i Z shown in figure. The frequency is taken on logarithmic scale because of its wide range. R From the figure, we can see that at resonance, (i) XL = XC or ω = 1 XL XC LC Resonance log ω (ii) Z = Zmin = R and XL – XC (iii) i is maximum. Fig. 28.23 Note Here, by Z we mean the modulus of Z and i means irms. ˜ Acceptor circuit If the frequency of the AC supply can be varied (e.g. in radio or television signal), then in series L-C-R circuit, at a frequency f = 1/2 π LC maximum current flows in the circuit and have a maximum PD across its inductance (or capacitance). This is the method by which a radio or television set is tuned at a particular frequency. The circuit is known as acceptor circuit. V Example 28.4 An alternating emf 200 virtual volts at 50 Hz is connected to a circuit of resistance 1 Ω and inductance 0.01 H. What is the phase difference between the current and the emf in the circuit? Also, find the virtual current in the circuit. Solution In case of an L-R AC circuit, the voltage leads the current in phase by an angle, φ = tan −1  XL  R Here, X L = ωL = (2πfL) = (2π ) (50) (0.01) = π Ω and R = 1Ω ∴ φ = tan –1 (π ) ≈ 72.3° Ans. Further, i rms = V rms = V rms |Z| R2 + X 2 L Substituting the values, we have 200 = 60.67 A Ans. i rms = (1)2 + (π )2

Chapter 28 Alternating Current — 575 V Example 28.5 A resistance and inductance are connected in series across a voltage, V = 283 sin 314t The current is found to be 4 sin (314 t – π /4) . Find the values of the inductance and resistance. Solution In L-R series circuit, current lags the voltage by an angle, φ = tan −1  XL  R Here, φ= π (ω = 314 rad/s) 4 …(i) ∴ ∴ X L = R or ωL = R 314 L = R Further, V0 = i0 | Z | ∴ 283 = 4 R2 + X 2 L or R2 + (ωL)2 =  2483 2 5005.56 = or 2R 2 = 5005.56 (as ωL = R) ∴ R ≈ 50Ω Ans. and from Eq. (i), L = 0.16 H Ans. V Example 28.6 Find the voltage across the various elements, i.e. resistance, capacitance and inductance which are in series and having values 1000 Ω, 1 µF and 2.0 H, respectively. Given emf is V = 100 2 sin 1000 t volt Solution The rms value of voltage across the source, Vrms = 100 2 = 100 V 2 ∴ ω = 1000 rad /s ∴ i rms = V rms = V rms = V rms |Z| R 2 + (X L ~ X C )2 ωL 1  2 ωC R2 + – = 100 2 (1000)2 +  × 2– × 1 × 10–6  1000 1   1000  = 0.0707 A

576 — Electricity and Magnetism The current will be same every where in the circuit, therefore, PD across resistor, VR = i rms R = 0.0707 × 1000 = 70.7 V PD across inductor, VL = i rms X L = 0.0707 × 1000 × 2 = 141.4 V and PD across capacitor, VC = i rms X C = 0.0707 × 1 1 × 1000 × 10–6 = 70.7 V Ans. Note The rms voltages do not add directly as `VR + VL + VC = 282.8 V which is not the source voltage 100 V. The reason is that these voltages are not in phase and can be added by vector or by phasor algebra. V = VR2 + (VL ~ VC )2 INTRODUCTORY EXERCISE 28.1 1. (a) What is the reactance of a 2.00 H inductor at a frequency of 50.0 Hz? (b) What is the inductance of an inductor whose reactance is 2.00 Ω at 50.0 Hz? (c) What is the reactance of a 2.00 µF capacitor at a frequency of 50.0 Hz? (d) What is the capacitance of a capacitor whose reactance is 2.00 Ω at 50.0 Hz? 2. An electric lamp which runs at 100 V DC and consumes 10 A current is connected to AC mains at 150 V, 50 Hz cycles with a choke coil in series. Calculate the inductance and drop of voltage across the choke. Neglect the resistance of choke. 3. A circuit operating at 360 Hz contains a 1µF capacitor and a 20 Ω resistor. How large an inductor 2π must be added in series to make the phase angle for the circuit zero? Calculate the current in the circuit if the applied voltage is 120 V. 28.8 Power in an AC Circuit In case of a steady current, the rate of doing work is given by P =Vi In an alternating circuit, current and voltage both vary with time and also they differ in time. So, we cannot use P =Vi for the power generated. Suppose in an AC, the voltage is leading the current by an angle φ. Then, we can write V =V0 sin ωt and i = i0 sin (ωt – φ) The instantaneous value of power in that case is P =Vi =V0i0 sin ωt sin (ωt – φ) or P = V0i0 sin 2 ωt cos φ – 1 sin 2ωt sin φ …(i) 2 Now, the average rate of doing work (power) in one cycle will be ∫∫P One cycle = T = 2π/ω …(ii) Pdt 0 T = 2π/ω dt 0

Chapter 28 Alternating Current — 577 Substituting the value of P from Eq. (i) in Eq. (ii) and then integrating it with proper limits, we get P One cycle = 1 V0 i0 cos φ = V0 ⋅ i0 cos φ 2 2 2 or P One cycle = V rms irms cos φ Here, the term cos φ is known as power factor. It is said to be leading if current leads voltage, lagging if current lags voltage. Thus, a power factor of 0.5 lagging means current lags the voltage by 60° (as cos –1 0.5 = 60°). The product of Vrms and irms gives the apparent power. While the true power is obtained by multiplying the apparent power by the power factor cos φ. Thus, Apparent power = Vrms × irms and True power = apparent power × power factor For φ = 0°, the current and voltage are in phase. The power is thus, maximum (=Vrms × irms). For φ = 90°, the power is zero. The current is then stated wattless. Such a case will arise when resistance in the circuit is zero. The circuit is purely inductive or capacitive. The case is similar to that of a frictionless pendulum, where the total work done by gravity upon the pendulum in a cycle is zero. Extra Points to Remember Let us consider a choke coil (used in tube lights) of large inductance L and low resistance R. The power factor for such a coil is given by cos φ = R = R ≈ R (as R << ωL) Z R2 + ω2L2 ωL As R << ωL, cos φ is very small. Thus, the power absorbed by the coil Vrms irms cos φ is very small. On account of its large impedance Z = R2 + ω2L2 , the current passing through the coil is very small. Such a coil is used in AC circuits for the purpose of adjusting current to any required value without waste of energy. The only loss of energy is due to hysteresis in the iron core, which is much less than the loss of energy in the resistance that can also reduce the current instead of a choke coil. V Example 28.7 A 750 Hz, 20 V source is connected to a resistance of 100 Ω, an inductance of 0.1803 H and a capacitance of 10 µF all in series. Calculate the time in which the resistance (thermal capacity 2 J /°C ) will get heated by 10° C. Solution The impedance of the circuit, Z= R 2 + (X L – X C )2 = R2 + (2πfL) – 1 2  (2πfC ) = (100) 2 +  × 3.14 × 750 × 0.1803) – (2 × 1 2 (2 3.14 × 750 × 10–5   ) = 834 Ω

578 — Electricity and Magnetism In case of an AC, V   RZ V  2     P = V rms i rms cos φ = (V rms ) rms = rms R Z Z =  82304 2 × 100 = 0.0575 J/s Now, P × t = S∆θ Ans. Here, S = thermal capacity t = S∆θ = 2 × 10 = 348 s ∴ P 0.0575 Note In AC, the whole energy or power is consumed by resistance. V Example 28.8 In an L-C-R series circuit, R = 150 Ω, L = 0.750 H and C = 0.0180 µF. The source has voltage amplitude V = 150 V and a frequency equal to the resonance frequency of the circuit. (a) What is the power factor? (b) What is the average power delivered by the source? (c) The capacitor is replaced by one with C = 0.0360 µF and the source frequency is adjusted to the new resonance value. Then, what is the average power delivered by the source? Solution (a) At resonance frequency, XL = XC ,Z = R and power factor cos φ= R = 1.0 Z (b) P = V 2 = (150/ 2 )2 = 75 W rms R 150 (c) Again, P = V 2 = 75 W rms R INTRODUCTORY EXERCISE 28.2 1. If a 0.03 H inductor, a 10 Ω resistor and a 2 µF capacitor are connected in series. At what frequency will they resonate? What will be the phase angle at resonance? 2. An arc lamp consumes 10 A at 40 V. Calculate the power factor when it is connected with a suitable value of choke coil required to run the arc lamp on AC mains of 200 V (rms) and 50 Hz.

Chapter 28 Alternating Current — 579 Final Touch Points 1. Frequency of AC in India is 50 Hz. 2. The AC is converted into DC with the help of rectifier while DC is converted into AC with the help of inverter. 3. An AC cannot produce electroplating or electrolysis. 4. The AC is measured by hot wire ammeter. 5. An AC can be stepped up or down with the help of a transformer. 6. An AC can be transmitted over long distances without much power loss. 7. An AC can be regulated by using choke coil without any significant waste of energy. 8. In an AC (sinusoidal), current or voltage can have the following four values (i) instantaneous value (ii) peak value (i 0 or V0) (iii) rms value (irms or Vrms) (iv) average value : In full cycle, average value is zero while in half cycle it is non-zero. Note That in sinusoidal AC the average value in half cycle can also be zero. It depends on the time interval over which half average value is desired. 9. In an series L-C-R circuit, (i) Capacitive reactance, XC = 1 ωC (ii) Inductive reactance, XL = ωL (iii) Impedance, Z = R2 + (XC − X L 2 ) (iv) If XC > XL, current leads and if XL > XC, voltage leads by an angle φ given by R φ = XC ~ XL cos φ = Z R and tan (v) Instantaneous power = instantaneous current × instantaneous voltage (vi) Average power =V i cos φ , where rms rms R cos φ = Z = power factor. Note Power is also equal to P = ir2msR But this is not equal to P ≠ Vr2ms R This is because Vrms = irms Z and cos φ = R . If we substitute in P = Vrms irms cos φ, then we get the first relation Z but not the second one. This implies that power is consumed only across resistance. VV i = i = (vii) 0 or rms 0 rms Z Z (viii) (VC )rms = (irms) XC , (VL)rms = (irms) XL and (VR )rms = (irms) R (ix) V = VR2 + (VC ~ VL)2 Here, V is the rms value of applied voltage VR is the rms value of voltage across resistance. VC across capacitor and VL across inductor etc.

580 — Electricity and Magnetism 10. ω = ωr = 1 is called resonance frequency. LC 11. At ω = ωr, (i) XL = XC (ii) Z = minimum value = R (iii) i = maximum value = Vrms = Vrms Zmin R rms VV i = maximum = = (iv) value 0 0 0 Zmin R (v) Power factor cos φ = 1 12. In one complete cycle, power is consumed only by resistance. No power is consumed by a capacitor or an inductor. 13. Z XL R ωr XC ω XC = 1 ⇒ XC ∝ 1 ωC ω XL = ωL ⇒ XL ∝ ω R does not depend on ω. It is a constant. At ω = ωr : XC = XL and Z =Z =R min 14. For ω > ωr, XL > XC. Hence, voltage will lead the current or circuit is inductive. For ω < ωr, XC > XL. Hence, current will lead the voltage function or circuit is capacitive. At ω = ωr, XC = XL. Hence, current function and voltage function are in same phase. 15. Conditions Phase angle Power factor R=0 90° 0 0° 1 XC = XL ≠ 0 R≠0 0° 1 XC = XL = 0 0° 1 R≠0 ω = ωr In all other cases, phase difference between current function and voltage function is 0° < φ < 90° XC > XL, φ = tan−1 X C − XL  −1  R  R Z If or cos XL > XC, φ = tan−1 XL − XC  −1  R  R Z If or cos

Solved Examples TYPED PROBLEMS Type 1. Based on a real inductor Concept An ideal inductor wire has zero resistance (R = 0). So, in an AC, only X L is the impedance. But, a real inductor has some resistance also. In DC, total resistance is only R. In AC, total resistance called impedance is Z = R2 + X 2 L Therefore, I DC = V DC and R I AC = VAC = V AC Z R2 + X 2 L If VAC = VDC , then I AC < I DC V Example 1 A current of 4 A flows in a coil when connected to a 12 V DC source. If the same coil is connected to a 12 V , 50 rad /s AC source, a current of 2.4 A flows in the circuit. Determine the inductance of the coil. Also, find the power developed in the circuit if a 2500 µF capacitor is connected in series with the coil. Solution (i) A coil consists of an inductance (L) and a resistance (R). In DC, only resistance is effective. Hence, R = V = 12 = 3 Ω i4 In AC, irms = Vrms = Vrms Z R2 + ω2L2 1  Vrms  2  ω2 irms  R2 ∴ L2 = –  L=1  Virrmmss 2 ω  ∴ – R2 Substituting the values, we have L= 1  21.24 2 – (3)2 50 = 0.08 H Ans.

582 — Electricity and Magnetism (ii) When capacitor is connected to the circuit, the impedance is Z = R2 + (XL – XC )2 Here, R =3 Ω XL = ωL = (50) (0.08) = 4 Ω and XC = 1 = (50) 1 × 10–6 ) =8 Ω ωC (2500 ∴ Z = (3)2 + (4 – 8)2 = 5 Ω Now, P = Vrms irms cos φ = Vrms × Vrms × R Z Z  Vrms  2 Z = × R Substituting the values, we have P =  152 2 × 3 = 17.28 W Ans. Type 2. Different time functions in AC Concept In an L-C-R series circuit, there are total five VR VC VL functions of time, V , I , VR , VC and VL . Now, the following points are important in these functions. (i) V and I have a phase difference of φ where I 0° ≤ φ ≤ 90° (ii) VR and I are in same phase V (iii) VC lags behind I by 90° (iv) VL leads I by 90° (all the time) (v) The functions, V = VL + VC + VL V Example 2 30 Ω 1 mF 0.5 H V = 200 sin (100 t + 30°) In the diagram shown in figure, V function is given. Find other four functions of time I, VC , VR and VL . Also, find power consumed in the circuit, V is given in volts and ω in rad/s.

Chapter 28 Alternating Current — 583 Solution Given, ω = 100 rad /s XL = ωL = 50 Ω XC =1 = 100 1 = 10 Ω ωC × 10−3 Z = R2 + (XL − XC )2 = (30)2 + (50 − 10)2 = 50 Ω Current function Maximum value of current, I0 = V0 = 200 =4 A Z 50 XL > XC, therefore voltage leads the current by a phase difference φ where, cos φ = R = 30 = 3 Z 50 5 or φ = 53° ∴ I = 4 sin (100 t + 30° − 53° ) or I = 4 sin (100 t − 23° ) Ans. VR,VC and VL functions Maximum value of VR = I0R = 4 × 30 = 120 volt, VR and I are in same phase. Therefore, VR = 120 sin (100 t − 23° ) Ans. Maximum value of VC = I0 XC = 4 × 10 = 40 volt Now, VC function lags the current function by 90°. Therefore, VC = 40 sin (100 t − 23° − 90° ) or VC = 40 sin (100 t − 113° ) Ans. Maximum value of VL = I0XL = 4 × 50 = 200 volt, VL function leads the current function by 90°. Therefore, VL = 200 sin (100 t − 23° + 90° ) or VL = 200 sin (100 t + 67° ) Ans. Note We can check at any time that, V = VR + VL + VC Power Power is consumed in an AC circuit only across a resistance and this power is given by P = Vrms Irms cos φ = Ir2ms R Let us use the first formula, P =  2020  4   53 2 = 240 watt Ans.

584 — Electricity and Magnetism Type 3. Parallel circuits Concept Two or more than two sine or cosine functions of same ω can be added by vector method. Actually, their amplitudes are added by vectors method. V Example 3 In the circuit shown in figure, I2 R2 C I1 L R1 I V = 200 sin (100 t + 30°) R1 = 30 Ω, R2 = 40 Ω, L = 0.4 H and C = 1 mF. 3 Find seven functions of time I , I1 , I 2 , VR1 , VL , VR2 and VC . Also, find total power consumed in the circuit. In the given potential function, V is in volts and ω in rad/s. Solution Circuit 1 (containing L and R1) I1 : XL = ωL = 100 × 0.4 = 40 Ω R1 = 30 Ω ∴ Z1 = R12 + XL2 = (30)2 + (40)2 = 50 Ω Maximum value of current, I1 = V0 = 200 =4 A Z1 50 Since, there is only XL, so voltage function will lead the current function by an angle φ1, where cos φ1 = R1 = 30 = 3 Z1 50 5 ∴ φ1 = 53° Ans. ∴ I1 = 4 sin (100 t + 30° − 53° ) or I1 = 4 sin (100 t − 23° ) VR1 : VR1 function is in phase with I1 function. Maximum value of VR1 = (maximum value of I1 ) (R1) = (4) (30) = 120 volt ∴ VR1 = 120 sin (100 t − 23° ) Ans. VL : VL function is 90° ahead of I1 function. Maximum value of VL = (maximum value of I1 ) (XL ) = (4)(40) = 160 volt

Chapter 28 Alternating Current — 585 ∴ VL = 160 sin (100 t − 23° + 90° ) or Power VL = 160 sin (100 t + 67° ) Ans. In this circuit, power will be consumed only across R1. This power is given by PR1 = ( rms value of I1 )2 R1 2  4  = 2 (30) = 240 watt Circuit 2 (containing C and R2 ) I2 : XC =1 = 1 10−3 = 30 Ω ωC 1 100 × × 3 R2 = 40 Ω ∴ Z 2 = R22 + XC2 = (40)2 + (30)2 = 50 Ω Maximum value of I2 = V0 = 200 = 4 A Z2 50 Since, there is only XC, so I2 function will lead the V function by an angle φ2, where cos φ2 = R2 = 40 = 4 Z2 50 5 ∴ φ2 = 37° Ans. ∴ VR2 : I2 = 4 sin (100 t + 30° + 37° ) = 4 sin (100 t + 67° ) VR2 function is in phase with I2 function. Maximum value of VR2 = ( maximum value of I2) (R2) = 4 × 40 = 160 volt ∴ VR2 = 160 sin (100 t + 67° ) Ans. VC : VC function lags I2 function by 90° Maximum value of VC = (Maximum value of I2)(XC ) = 4 × 30 = 120 volt ∴ VC = 120 sin (100 t + 67° − 90° ) or Power VC = 120 sin (100 t − 23° ) Ans. In this circuit, power will be consumed only across R2 and this power is given by PR2 = (rms value of I2)2 R2 2  4  = 2 (40) = 320 W Ans. ∴ Total power consumed in the circuit, P = PR1 + PR2 = (240 + 320) W = 560 W

586 — Electricity and Magnetism I : I = I1 + I2 I = 4 sin (100 t − 23° ) + 4 sin (100 t + 67° ) Now, the amplitudes can be added by vector method. 4A 67°22° 4√2 A 23° 100 t 4A Resultant of 4 A and 4 A at 90° is 4 2 A at 45° from both currents or at 22° from 100 t line. ∴ I = 4 2 sin (100 t + 22° ) Ans. Miscellaneous Examples V Example 4 An AC circuit consists of a 220 Ω resistance and a 0.7 H choke. Find the power absorbed from 220 V and 50 Hz source connected in this circuit if the resistance and choke are joined (a) in series (b) in parallel. Solution (a) In series, the impedance of the circuit is Z = R2 + ω2L2 = R2 + (2πfL)2 Z = (220)2 + (2 × 3.14 × 50 × 0.7)2 XL = 311 Ω φ R ∴ irms = Vrms = 220 = 0.707 A Z 311 and cos φ = R = 220 = 0.707 Z 311 ∴ The power absorbed in the circuit, P = Vrms irms cos φ = (220) (0.707) (0.707) = 110.08 W Ans. (b) When the resistance and choke are in parallel, the entire power is absorbed in resistance, as the choke (having zero resistance) absorbs no power. ∴ P = Vr2ms = (220)2 = 220 W Ans. R 220 V Example 5 A sinusoidal voltage of frequency 60 Hz and peak value 150 V is applied to a series L-R circuit, where R = 20 Ω and L = 40 mH. (a) Compute T , ω , XL , Z and φ (b) Compute the amplitudes of current, VR and VL

Chapter 28 Alternating Current — 587 Solution (a) T=1= 1 s Ans. f 60 Ans. Ans. ω = 2πf = (2π )(60) = 377 rad /s XL = ωL = (377) (0.040) Ans. = 15.08 Ω Ans. Z = XL2 + R2 Ans. Ans. = (15.08)2 + (20)2 Ans. = 25.05 Ω φ = tan –1  XL  = tan−1  152.008 = tan–1 (0.754) R = 37° (b) Amplitudes (maximum value) are i0 = V0 = 150 ≈6 A Z 25.05 (V0 )R = i0R = (6)(20) = 120 V (V0 )L = i0XL = (6) (15.08) = 90.5 V Note V0 = (V0 )R2 + (V0 )2L V Example 6 For the circuit shown in figure, find the instantaneous current through each element. V = V0 sin ωt R CL Solution The three current equations are and V = iRR, V = L diL dt V=q ⇒ dV =1 iC …(i) C dt C The steady state solutions of Eq. (i) are iR = V0 sin ωt ≡ (i0 )R sin ωt R iL = – V0 cos ωt ≡ – V0 cos ωt ≡ – (i0 )L cos ωt ωL XL and iC = V0ωC cos ωt ≡ V0 cos ωt ≡ (i0 )C cos ωt XC where, the reactances XL and XC are as defined.

588 — Electricity and Magnetism V Example 7 In the above problem find the total instantaneous current through the source, and find expressions for phase angle of this current and the impedance of the circuit. Solution For the total current, we have i = iR + iL + iC = V0  1 sin ωt + 1 – 1   R  XC XL  cos ωt   Using the trigonometric identity, A sin θ + B cos θ = A2 + B2 sin (θ + φ) where, φ = tan–1 (B/A) We can write, i ≡ i0 sin (ωt + φ) Here, where, i0 = V0 Z and 1=  R1  2 + 1 – 1 2 Z  XC XL  1 – 1  XC XL  tan φ = (1 /R) V Example 8 An L-C-R series circuit with 100 Ω resistance is connected to an AC source of 200 V and angular frequency 300 rad /s. When only the capacitance is removed, the current lags behind the voltage by 60°. When only the inductance is removed, the current leads the voltage by 60° . Calculate the current and the power dissipated in the L-C-R circuit Solution When capacitance is removed, then tan φ = XL R or tan 60° = XL R ∴ XL = 3R …(i) When inductance is removed, then tan φ = XC or R tan 60° = XC R ∴ XC = 3 R …(ii) From Eqs. (i) and (ii), we see that XC = XL So, the L-C-R circuit is in resonance. Hence, Z =R

Chapter 28 Alternating Current — 589 ∴ irms = Vrms = 200 =2 A Ans. Z 100 Ans. P = Vrms irms cos φ At resonance current and voltage are in phase, or φ = 0° ∴ P = (200) (2) (1) = 400 W V Example 9 A series L-C-R circuit containing a resistance of 120 Ω has resonance frequency 4 × 105 rad /s. At resonance the voltages across resistance and inductance are 60 V and 40 V , respectively. Find the values of L and C. At what angular frequency the current in the circuit lags the voltage by π /4? Solution At resonance, XL – XC = 0 and Z = R = 120 Ω ∴ irms = (VR )rms = 60 =1 A R 120 2 Also, irms = (VL )rms ωL ∴ L = (VL )rms = 40  12 ωirms 105 ) (4 × = 2.0 × 10–4 H = 0.2 mH Ans. The resonance frequency is given by C = 1 ω = 1 or ω 2L LC Substituting the values, we have C = (4 × 105 )2 1 × 10–4 ) (2.0 = 3.125 × 10–8 F Ans. Current lags the voltage by 45°, when ωL – 1 ωC tan 45° = R Substituting the values of L, C, R and tan 45°, we get Ans. ω = 8 × 105 rad /s V Example 10 A choke coil is needed to operate an arc lamp at 160 V (rms) and 50 Hz. The lamp has an effective resistance of 5 Ω when running at 10 A (rms). Calculate the inductance of the choke coil. If the same arc lamp is to be operated on 160 V (DC), what additional resistance is required? Compare the power loses in both cases.