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www.sakshieducation.com Alternating Current Voltage and Power 2011 1. In an AC circuit, V and I are given by V = 150sin (150t ) volt , π and I = 150sin 150t + amp . The power dissipated in the circuit is 3 a) 106 W b) 150 W c) 5625 W d) Zero 2010 2. An AC ammeter is used to measure current in a circuit. When a given direct current passes through the circuit, the AC ammeter reads 3A. When another alternating current passes through the circuit, the AC ammeter reads 4A. Then the reading of this ammeter, if DC and AC flow through the circuit simultaneously, is a) 3 A 3. b) 4 A c) 7 A An AC source is 120V – 60Hz. The value of voltage after a) 20.2 V b) 42.4 V d) 5 A 1 s from start will be 720 c) 84.8 V d) 106.8 V 2009 4. The oscillating electric and magnetic vectors of an electromagnetic wave are oriented along a) The same direction but has a phase difference of 90° b) The same direction and are in same phase c) Mutually perpendicular directions and are in same phase d) Mutually perpendicular directions but has a phase difference of 90° 5. Alternating current is transmitted to take places a) At high voltage and low current b) At high voltage and high current c) At low voltage and low current d) At low voltage and high current www.sakshieducation.com www.sakshieducation.com 2008 6. In an AC circuit the emf (e) and the current (i) at any instant are given respectively by e = Eo sin ω t i = I o sin ( ωt − φ ) The average power in the circuit over one cycle of AC is a) 7. Eo I o 2 b) Eo I o sin φ 2 c) Eo I o cos φ 2 d) Eo I o Alternating current cannot be measured by DC ammeter because a) AC cannot pass through DC ammeter b) AC changes direction c) Average value of current for complete cycle is zero d) DC ammeter will get damaged 2007 8. If the power factor changes from 1 1 to then what is the increase in impedance in 2 4 AC? a) 20% 9. b) 50% c) 25% d) 100% The instantaneous voltage through a device of impedance 20Ω is e = 80sin100π t . The effective value of the current is a) 3 A e) 10. b) 2.828 A c) 1.732 A d) 4 A c) Zero resistance d) None of the 2A For high frequency, capacitor offers a) More resistance b) Less resistance above 2006 11. If reading of an ammeter is 10A, the peak value of current is a) 10 A 2 b) 5 A 2 c) 20 2 A www.sakshieducation.com d) 10 2 A www.sakshieducation.com 12. If impedance is a) Zero 13. 3 times of resistance, find phase difference b) 30° c) 60° d) Data is incomplete An alternating voltage V = Vo sin ωt is applied across a circuit. As a result the current I = I o sin ( ωt − π 2) flows in it. The power consumed in the circuit per cycle is a) 0.5Vo I o W 14. b) 0.707 Vo I o W c) 1.919Vo I o W d) Zero An AC is represented by e = 220sin (100π ) t volt and is applied over a resistance of110Ω . The heat produced in 7 min is a) 11 × 103 cal 15. b) 22 × 103 cal c) 33 × 103 cal d) 25 × 103 cal If an AC produces same heat as that produced by a steady current of 4A, then peak value of current is a) 4 A 16. b) 1.56 A c) 5.6 A d) 1.41 A The potential difference across an instrument in an AC circuit of frequencies f is V and the current through it is I such that V = 5cos 2π ft and I = 2sin 2π ft amp . The power dissipated in the instrument is a) Zero b) 10 W c) 5 W d) 2.5 W 2005 17. An alternating current is given by I = I1 cos ωt + I 2 sin ωt . The root mean square current is a) 18. ( I1 + I 2 ) 2 2 2 c) I12 + I 22 2 d) I12 − I 22 2 The peak value of AC voltage on 220V mains is a) 240 2 V 19. b) ( I1 + I 2 ) b) 230 2 V c) 220 2 V d) 200 2 V The time taken by an alternating current of 50Hz in reaching from zero to its maximum value will be a) 0.5 s b) 0.005 s c) 0.05 s www.sakshieducation.com d) 5 www.sakshieducation.com Alternating Current Voltage and Power Key 1) c 2) d 3) c 4) c 5) a 6) c 7) c 8) d 9) b b 11) d 12) c 13) d 14) b 15) c 16) a 17) c Solutions 1. V = 150 sin (150t) volt And I = 150sin (150t + π / 3)amp I 0 = 150amp and V0 = 150volt 1 P = Vo I 0 cos φ 2 P = 0.5 × 150 × 150 × cos 600 2. Quantity of heat liberated in the ammeter of resistance R i) Due to direct current of 3 A, 2 Q = ( 3 ) R / J ) 2 ii) Due to alternating current of 4 A, Q = ( 4 ) R / J ) Total heat produced per second ( 3) = 2 J R ( 4) + 2 J R = I 2 R 25R But, = J J I=5A 3. V = V0 sin ωt V = Vrms 2 sin ωt www.sakshieducation.com 25 R J 18) c 19) b 10) www.sakshieducation.com t= 1 s 720 V = 120 2 sin 2π v t = 120 2 sin 2π × 60 × 6. Pav = Pav = W T ( E0 I 0 cos φ ) T / 2 Pav = 8. T E0 I 0 cos φ 2 cos φ α 1 Z Z1 cos φ1 1/ 4 1 = = = Z 2 cos φ2 1/ 2 2 ∴ Z 2 = 2 Z1 ⇒ Percentage change = 9. 1 = 60 2 = 84.8V 720 2 Z1 − Z1 × 100 =100% Z1 e = 80 sin 100π t Here, e0 = 80V Where e0 is the peak value of voltage Impedance (Z) = 20 Ω I0 = 11. e0 e I0 = 0 Z Z I rms = I0 4 = = 2 2 = 2.828 A 2 2 irms = i0 2 ⇒ i0 = 2 × irms = 2 × 10 A = 10 2 A 2011 13. V = V0 sin ωt and I = I 0 sin (ωt − π / 2 ) www.sakshieducation.com www.sakshieducation.com 1 ∴ P = V0 I 0 cos φ 2 But φ = π / 2 = 900 1 P = V0 I 0 cos 900 2 ( = 0 ∴cos900 = 0 14. H= I v2 RT J ) ( 2) cal = 2 × 100 × 7 × 60 42 = 22 × 103 Cal I 0 220 /100 = = 2 ∵ I ms = 2 2 15. I 0 = 2 I rms I rms = 4 A I 0 = 2 × 4 = 5.6 A 16. P = VI cos φ V= 5cos P = VI cos 2π ft π V = 5sin 2π ft + and I = 2sin 2π ft 2 Here, φ = π 2 P = VI cos φ = VI cos 17. π 2 =0 I = I1 cos ωt + I 2 sin ωt I 0 = I12 + I 22 I rms 18. I = 0 = 2 I12 + I 22 2 = I12 + I 22 2 Vrms = 220V But Vrms = peak voltage VP = 2 2 VP = 220 2 V www.sakshieducation.com www.sakshieducation.com 19. T= ∴ 1 1 = = 0.0.2 s f 50 T 0.02 = = 0.005s 4 4 AC Circuits 2011 1. In an AC circuit an alternating voltage e = 200 2 sin100t volt is connected to a capacitor of capacity 1 µ F . The rms value of the current in the circuit is a) 100 mA 2. b) 200 mA c) 20 mA d) 10 mA An AC voltage is applied to a resistance R and an inductor L in series. If R and the inductive reactance are both equal to 3Ω , the phase difference between the applied voltage and the current in the circuit is 3. a) π 4 b) π 2 c) Zero d) π 6 The peak value of an alternating current is 5A and its frequency is 60Hz. Find its rms value and time taken to reach the peak value of current starting from zero 4. a) 3.536 A, 4.167 ms b) 3.536 A, 15 ms c) 6.07 A, 10 ms d) 2.536 A, 4.167 m An L – C – R series current is under resonance. If I m is current amplitude, Vm is voltage amplitude, R is the resistance, z is impedance, X L is the inductive reactance and X C is the capacitive reactance then a) I m = 5. Vm Z Vm XL b) I m = c) I m = Vm XC d) I m = Vm R In the case of an inductor a) Voltage lags the current by π c) Voltage leads the current by 2 π 3 b) Voltage leads the current by d) Voltage leads the current by www.sakshieducation.com π 2 π 4 www.sakshieducation.com 2010 6. The power factor of an R – L circuit is 1 . If the frequency of AC is doubled, 2 what will be the power factor? a) 7. 1 3 b) 1 5 c) 1 7 d) 1 11 An ideal choke draws a current of 8A when connected to an AC supply of 100V, 50Hz. A pure resistor draws a current of 10A when connected to the same source. The ideal choke and the resistor are connected in series and then connected to the AC source of 150V, 40Hz. The current in the circuit becomes a) 8. 15 A 2 b) 8A c) 18A d) 10A In an AC circuit the potential difference V and current i are given respectively by π V = 100sin (100t ) volt and i = 100sin 100t + mA . The power dissipated in the 3 circuit will be a) 104 W 9. b) 10 W c) 2.5 W d) 5 W An inductor L, a capacitor of 20 µ F and a resistor of 10Ω are connected in series with an AC source of frequency 50Hz. If the current in the phase with the voltage, then the inductance of the inductor is a) 2.00 H 10. b) 0.51 H c) 1.5 H d) 0.99 H The impedance of a circuit, when a resistance R and an inductor of inductance L are connected in series in an AC circuit of frequency f, is 11. a) R + 2π 2 f 2 L2 b) R + 4π 2 f 2 L2 c) R 2 + 4π 2 f 2 L2 d) R 2 + 2π 2 f 2 L2 In a series L – C – R circuit, resistance R = 10 Ω and the impedance Z = 10 Ω . The phase difference between the current and the voltage is a) 0° b) 30° c) 45° www.sakshieducation.com d) 60° www.sakshieducation.com 12. The inductive time constant in an electrical circuit is a) LR 13. b) c) L R d) R L In a L – C – R circuit R = 10 Ω . When capacitance C is removed, the current lags behind the voltage by voltage by π 3 a) 50 Ω 14. L R π 3 , when inductance L is removed, the current leads the . The impedance of the circuit is b) 100 Ω c) 200 Ω d) 400 Ω In an L – C – R series Ac circuit at resonance a) The capacitive reactance is more than the inductive reactance b) The capacitive reactance equals the inductive reactance c) The capacitive reactance is less than the inductive reactance d) The power dissipated is minimum 15. An L – C – R series circuit consists of a resistance of 10 Ω a capacitor of reactance 6.0 Ω and an inductor coil. The circuit is found to resonate when put across a 300V, 100Hz supply. The inductance of coil is (taking π = 3 ) a) 0.1 H 16. b) 0.01 H c) 0.2 H d) 0.02 H A capacitor or capacitance 1 µ F is charged to a potential of 1V. It is connected in parallel to an inductor of inductance 10−3 H . The maximum current that will flow in the circuit has the value a) 1000 mA 17. b) 1 A c) 1 mA d) 1000 mA Assertion: For an electric lamp connected in series with a variable capacitor and AC source, its brightness increases with increase in capacitance Reason: Capacitive reactance decreases with increase in 1) Both assertion and reason are true and reason is the correct explanation of assertion 2) Both assertion and reason are true but reason is not the correct explanation of assertion 3) Assertion is true but reason is false 4) Assertion is false but reason is true www.sakshieducation.com www.sakshieducation.com 18. 0.4 In an L – R circuit, the value of L is H and the value of R is 30Ω . If in the π circuit, an alternating emf of 200V at 50cycle s is connected, the impedances of the circuit and current will be a) 11.4 Ω, 17.5 A b) 30.7 Ω, 6.5 A c) 40.4 Ω, 5 A d) 50 Ω, 4 A 2009 19. Power dissipated in an L – C – R series circuit connected to an AC source of emf ε is a) 20. 2 1 Cω ε 2R d) R 2 2 b) 2 2 1 R + Lω − Cω ε 2 R 2 + Lω − c) ε ε 2R 1 R + Lω − Cω R 2 1 R + Lω − Cω 2 2 An AC voltage is applied to a pure inductor L, drives a current in the inductor. The current in the inductor would be 21. a) Ahead of the voltage by π 2 b) Lagging the voltage by π 2 c) Ahead of the voltage by π 4 d) Lagging the voltage by 3π 4 The average power dissipated in a pure capacitance AC circuit is a) CV 22. b) Zero 1 CV 2 d) 1 CV 2 4 In a pure inductive circuit, current a) Lags behind emf by π b) Leads the emf by 2 c) Lags behind by π 23. c) π 2 d) Leads the emf by π An alternating current of rms value 10A is passed through a 12 Ω resistor. The maximum potential difference across the resistor is a) 20 V b) 90 V c) 169.68 V www.sakshieducation.com d) None of these www.sakshieducation.com 24. Same current is flowing in to alternating circuits. The first circuit contains only inductance and the other contains only a capacitor. If the frequency of the emf of AC is increased, the effect on the value of the current will be a) Increases in the first circuit and decreases in the other b) Increases in both the circuits c) Decreases in both the circuits d) Decreases in the first circuit and increases 2008 25. An AC source of angular frequency ω is fed across a resistor R and a capacitor C in series. The current registered is I. If now the frequency of source is changed to ω 3 (but maintaining the same voltage), the current in the circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency ω a) 26. 3 5 b) 2 5 c) 1 5 d) 4 5 What is the value of inductance L for which the current is a maximum in a series L – C – R circuit with C = 10 µ F and ω = 1000 s −1 ? 27. a) 100 mH b) 1 mH c) Cannot be calculated unless R is known d) 10 mH In non – resonant circuit, what will be the nature of the circuit for frequencies higher than the resonant frequency? a) Resistive 28. b) Capacitive c) Inductive d) None of the above A coil has an inductance of 0.7H and is joined in series with a resistance of 20Ω . When an alternating emf of 220V at 50 cps is applied to it, then the wattles component of the current in the circuit is a) 5 A 29. c) 0.7 A d) 7 A In L – C – R circuit what is the phase angle φ ? a) 90° 30. b) 0.5 A b) 180° c) 0° d) 60° In R – C circuit ω = 100 rad s −1 , R = 100 Ω, C = 20 µ F . What is impedance? a) 510Ω b) 200Ω c) 250Ω www.sakshieducation.com d) 300Ω www.sakshieducation.com 31. The power factor in a circuit connected to an AC power supply has a value which is a) Unity when the circuit contains only inductance b) Unity when the circuit contains only resistance c) Zero when the circuit contains an ideal resistance only d) Unity when the circuit contains an ideal capacitance only 32. In an AC circuit, a resistance of R ohm is connected in series with an inductor of self inductance L. If phase angle between voltage and current be 45° , the value of inductive reactance ( X L ) will be equal to a) R 33. b) R 8 c) R 4 d) R 2 A transistor – oscillator using a resonant circuit with an inductor L (or negligible resistance) and a capacitor C in series produce oscillations of frequency f. If L is doubled and C is changed to 4C, the frequency will be a) f 4 34. d) f 2 The natural frequency ( ω o ) of oscillations in L – C circuit is given by a) 35. c) f 2 2 b) 8f 1 2π 1 LC b) 1 2π LC c) 1 LC d) LC In L – C- R circuit the resonance condition in terms of capacitive reactance ( X C ) and inductive reactance ( X L ) is a) X C + X L = 0 b) X C = 0 c) X L = 0 d) X C − X L = 0 2007 36. An inductor of inductance L and resistor of resistance R are joined in series and connected by a source of frequency ω . Power dissipated in the circuit is R 2 + ω 2 L2 a) V b) (R V 2R 2 + ω 2 L2 ) V c) 2 2 2 R + ω L www.sakshieducation.com d) R 2 + ω 2 L2 V2 www.sakshieducation.com 37. An inductive coil has a resistance of 100Ω . When an AC signal of frequency 1000 Hz is applied to the coil, the voltage leads the current by 45° . The inductance of the coil is a) 38. 1 10 π 2) 1 20 π c) 1 40 π d) 1 60 π ( The reactance of a coil when used in the AC power supply 220V , 50 cycles −1 ) is 50 Ω . The inductance of the coil is nearly a) 0.16 H 39. b) 0.22 H c) 2.2 H d) 1.6 H 10−3 200 An inductance of mH , a capacitance of π F and a resistance of 10 Ω π are connected in series with an AC source 220V, 50Hz. The phase angle of the circuit is a) 40. π b) 6 π c) 4 π 2 d) π 3 In a circuit, the current lags behind the voltage by a phase difference of π 2 , the circuit will contain which of the following? a) Only R 41. b) Only C c) R and C d) Only L A coil of inductive reactance 31Ω has a resistance of 8 Ω . It is placed in series with a condenser of capacitive reactance 25 Ω . The combination is connected to an AC source of 110V. The power factor of the circuit is a) 0.56 42. b) 0.64 d) 0.33 In a choke coil, the reactance X L and resistance R are such that a) X L = R 43. c) 0.80 b) X L >> R c) X L << R d) X L = ∞ The power factor of a series L – C – R circuit when at resonance is a) Zero b) 0.5 c) 1.0 d) Depends on values of L, C and R www.sakshieducation.com www.sakshieducation.com 2006 44. In a series L – C – R circuit the frequency of a 10V, AC voltage source is adjusted in such a fashion that the reactance of the inductor measures 15Ω and that of the capacitor 11Ω . If R = 3Ω , the potential difference across the series combination of L and C will be a) 8 V 45. b) 10 V c) 22 V d) 52 V An L – C circuit is in the state of resonance. If C = 0.1µ F and L = 0.25H, neglecting ohmic resistance of circuit, what is the frequency of oscillations? a) 1007 Hz 46. b) 100 Hz c) 109 Hz d) 500 Hz In an AC circuit, the current lags behind the voltage by π 3 . The components of the circuit are 1) R and L 47. 3) R and C 4) Only R What is the ratio of inductive and capacitive reactance in AC circuit? a) ω 2 LC 48. 2) L and C b) 1 c) zero d) ω 2 L If a circuit made up of a resistance 1Ω and inductance 0.01H, and alternating emf 200V at 50Hz is connected, then the phase difference between the current and the emf in the circuit is a) tan −1 ( π ) 49. π b) tan −1 2 π c) tan −1 4 π d) tan −1 3 A circuit draws 330W from a 110V, 60Hz AC line. The power factor is 0.6 and the current lags the voltage. The capacitance of a series capacitor that will result in a power factor of unity is equal to a) 31 µ F b) 54 µ F c) 151 µ F d) 201 µ F 2005 50. In a circuit, L, C and R are connected in series with an alternating voltage source of frequency f. The current leads the voltage by 45° . The value of C is a) c) 1 2π f ( 2π fL + R ) 1 2π f ( 2π fL − R ) b) 1 π f ( 2π fL + R ) d) 1 π f ( 2π fL − R ) www.sakshieducation.com www.sakshieducation.com AC Circuits Key 1) c 2) a 3) a 4) a 5) b 6) b 7) a 8) c 9) b 10) c 11) a 12) b 13) b 14) b 15) a 16) a 17) a 18) d 19) a 20) a 21) b 22) a 23) c 24) d 25) a 26) a 27) c 28) b 29) c 30) a 31) b 32) a 33) c 34) a 35) d 36) b 37) b 38) a 39) b 40) d 41) c 42) b 43) c 44) a 45) a 6) 47) a 48) a 49) b 50) c a Solutions 1. e = 200 2 sin100t and C = 1µ F Erms = 200V 2. 3. XC = 1 1 = = 104 Ω −6 ωC 1× 10 × 100 irms = Erms Xc irms = 200 = 2 × 10−2 A = 20 mA 4 10 tan φ = X L Lω = R R tan φ = 3Ω 3Ω irms = T= 6. Or φ = π 4 rad i0 5 = = 2.536 A 2 2 1 = 4.167 ms 4 × 60 cos φ = R R 2 + 4ω 2 L2 = R R2 + 4R2 = 1 5 www.sakshieducation.com www.sakshieducation.com 7. R= 100 = 10Ω 10 X L = 2π fL 100 = 2π × 50 × L 8 ⇒ L= X i = 2π fL = 2π × 40 × 1 8π X i = 2π fL = 2π × 40 × 1 = 8π i= 8. 1 H 8π (10 ) 2 + (10 ) = 10 2Ω 2 V 150 15 = = A Z 10 2 2 V0 = 100V , i0 = 100 mA = 100 × 10−3 A and φ = π 3 = 600 V0i0 100 × 100 × 10−3 P= × cos 600 = 2.5 W cos φ = 2 2 9. tan φ = ωL − L= ωL − R 1 ωC = 0 1 ωC 1 ω 2C Here ω = 2π f = 2 × 3.14 × 50s −1 = 314s −1 And C = 20µ F = 20 × 10−6 F ∴L = 11. ( 1 314 s ) ( −1 2 × 20 × 10−6 F ) = 0.51H Z = R 2 + X L2 X L = ω L = 2π fL Z = R 2 + 4π 2 f 2 L2 www.sakshieducation.com www.sakshieducation.com (10 ) ∴ 10 = 2 + ( X L − XC ) ⇒ 100 = 100 + ( X L − X C ) 2 2 ⇒ X L − XC = 0 13. tan π 3 = XL R When L is removed, tan π XL R = 3 Z= R = 100Ω ⇒ i0 = 220 × 2 220 × 2 = 100π 100 × 3.14 I 0 = 1A 15. ω0 = 2π n − 2π × 100 ω0 = 2 × 3 × 100 = 600rad / s Also, ω0 = XC = 1 LC 1 = 60ω Cω0 ⇒ C= 1 1 = ω0 × 60 600 × 60 1 600 − 1 L 3 36 × 10 36 × 103 36 × 10 = L 4 36 × 103 1 L= = = 0.1H 36 × 104 10 16. q = q0 sin ωt Or I 0 = ω q0 = maximum current ω= ( ) 1 1 = = 109 LC 10−9 1/ 2 www.sakshieducation.com www.sakshieducation.com ( ) I 0 = 109 18. 1/ 2 ( ) × 1× (10−6 = 1000 mA X L = ω L = 2π fL = 2π × 50 × 0.4 π = −40Ω R = 30Ω ( 30 ) Z = rR 2 + X L2 = irms = 25. 2 + ( 40 ) = 50Ω 2 vrms 200 = = 4A X 50 Vrms I rms = 1 R + ωC 2 2 I rms = 2 Also, ∴ 3R 2 = 26. Vrms 1 R2 + ωc 3 2 Vrms = R2 + 9 ω C2 2 5 ω C2 2 ⇒ 1 3 = ωC 5 R i= V R2 + ( X L = X C ) 2 But, X L = X C Or ω L = 1 ωC Or L = 1 ω 2C ω = 1000 s −1 , C = 10µ F = 10 × 10−6 ∴L = 28. 1 (1000 ) 2 × 10 × 10−6 =0.1 H Wattless component of current = I v sin θ = = 220 R 2 + ω 2 L2 × ωL R 2 + ω 2 L2 = 220 × ω L (R 2 + ω 2 L2 EV sin θ Z ) www.sakshieducation.com www.sakshieducation.com = 34. 220 × ( 2π × 50 × 0.7 ) ( 220 ) + ( 2π × 50 × 0.7 ) 2 X L = X C ⇒ ωo L = 1 = 0.5 A 2 1 ω0C 1 LC ω02 = 1 LC ω0 = 2π f = f = 2 = 1 LC 1 2π LC 35. Resonance condition is X L = X C or X C − X L = 0 37. tan φ = XL R ⇒ tan 450 = ⇒ L= ⇒ L= 38. R ω = Lω R 100 2π (1000 ) 1 20π X L = ω L = 2π fL X L = 50Ω, f = 50cps L= 39. XL ω tan θ = = XL 50 1 = = = 0.16 H 2π f 2π × 50 2 × 3.14 X L − XC R 200 X L = 2π fL = 2π × 50 × × 10−3 = 20Ω π XC = 1 1× π = = 10Ω 2π fC 2π × 50 × 10−3 www.sakshieducation.com www.sakshieducation.com And R = 10Ω tan θ = 20 − 10 =1 10 tan θ = tan π 4 π θ= 4 1 45. f = 47. X L = ωL 2π LC = 1 2π 0.25 × 0.1 × 10−6 and X C = ≈ 1007 Hz 1 ωC XL ωL = = ω 2 LC X C 1/ ωC 48. XL and X L = ω L = 2π fL = 2π × 50 × 0.01 = πΩ R tan φ = Also R = 1Ω ∴φ = tan −1 (π ) 49. V2 110 × 110 110 = = Ω P 330 3 R= Cos R = V2 110 × 110 110 = = Ω = 0.6. P 330 3 R R 2 + X L2 R + 2 X L2 X L2 = XL = = 0.6 R = 0.6 R2 ( 0.6 ) 2 −R 2 2 ⇒ X L2 R 2 × 0.64 = 0.36 0.8 R 4 R = 0.6 3 And X L = KC XC = 4 R 4 100 440 = × = Ω 3 3 3 9 www.sakshieducation.com www.sakshieducation.com 1 = 2π f C= 440 Ω 9 9 2 × 3.14 × 60 × 440 −0.000054 F = 54 µ F 50. ωL − tan φ = 1 ωC R φ = 45° ∴ tan 45° = ⇒ 1= ⇒ ωL − ωL − 1 ωC R 1 ωC R R = ωL − ⇒ ωC = ⇒ C= 1 ωC 1 (ω L − R) 1 1 = 2 ( ω L − R ) 2π f ( 2π fL − R ) www.sakshieducation.com www.sakshieducation.com Growth and Decay of Current 2010 1. A coil of resistance R and inductance L is connected to a battery of emf E volt. The final current in the coil is a) E R b) E L c) E R + L2 2 d) EL R + L2 2 2008 3. A coil of inductance 300mH and resistance 2Ω is connected to a source of voltage 2V. The current reaches half of its steady state value in a) 0.05 s b) 0.1 s c) 0.15 s d) 0.3 s 2007 4. A coil of 40H inductance is connected in series with a resistance of 8Ω and this combination is connected to the terminals of 2V battery. The inductive time constant of the circuit is (in second) a) 40 5. b) 20 c) 5 d) 0.2 In an L – R circuit, time constant is that time in which current grows from zero to the value a) 0.63 I o b) 0.50 I o c) 0.37 I o d) I o 2004 6. In a series resonant L – C – R circuit, the voltage across R is 100V and R = 1k Ω with C = 2µ F . The resonant frequency ω is 200 rad s −1 . At resonance the voltage across L is a) 2.5 × 10 −2 V b) 40 V c) 250 V www.sakshieducation.com d) 4 × 10−3 V www.sakshieducation.com Growth and Decay of Current Key 1) a 2) c 3) b 4) c 5) a 6) c Hints 2. At time t = 0, i.e., when capacitor is charging, current I = I= 2 = 1mA 2000 ( I = I o 1 − e − Rt 3. L ( Io = I o 1 − e − Rt 2 ( 1 = 1 − e − Rt 2 L ) L ) ) e − Rt L = 1 2 Rt = ln 2 L ∴ L 300 × 10 −3 ln 2 = × 0.693 R 2 = 150 × 0.693 × 10 −3 = 0.10395 s = 0.1 s L 40 = = 5s R 8 4. t= 5. I = I o 1 − e −1 τ ( ) L τ = Is called time constant. R At t = τ ( I = I o 1 − e −1 ) www.sakshieducation.com 2 = 2mA 1000 www.sakshieducation.com 1 Or I = I o 1 − e I = 0.63 I o Or 6. ωL = I= 1 ωC VR 100 = = 0.1A R 1000 VL = IX L = I ω L But ∴ VL = ωL = 1 ωC 1 0.1 = = 250V ωC 200 × 2 × 10−6 www.sakshieducation.com