Title 23. Alternating Current Medium Ans.pdf ✅

1. (c): In the pure resistive circuit current and voltage both are are in phase. Hence graph (c) is correct. 2. (c): Here, P = 100 W, Vrms = 220V Resistance of the bulb is = = = 484 100 (220) P V R 2 rms 2 3. (c): Compare V = 150sin (150t) with V V sin t, = 0  we get V0 =150V Compare I = 150 sin        + 3 150t with I I sing( t ), = 0  +  we get o 0 60 3 I 150A, =  =  = The power dissipated in ac circuit is o 0 0 150 150 cos60 2 1 V I cos 2 1 P =  =    5625W 2 1 150 150 2 1 =    = 4. (b): The equation of alternating voltage is V(t) V sin t = m  5. (a): The given circuit is a pure resistive circuit. In this circuit the voltage and current both are in phase. 6. (c): Pure resistive circuit 7. (c): Here, R 100 ,V 220V, 50Hz =  rms =  = 2.2A 100 220 R V I rms  rms = = = 8. (b): Here, , Vm = 440V 311.1V 2 440 2 V V m  rms = = = 9. (b): Here, , I rms = 25A I 2 I 2 25 35.36A.  m = rms =  = 10. (a) Here, , R = 50,V0 =140V R 0.707V R V I rsm 0  rms = = 50 0.707140 = =1.98A. 11. (d): As the line has some resistance (R  0), voltage and current differ in phase  such that 2    12. (c): V 2 200 Vrms = 200V 2 200 V0 = 2 Vrms = 2  =       =  =   600 1 V V0 sin 2 t 200sin 2 50 100V 2 1 200 6 200sin =  =  = 13. (a): The instantaneous value of voltage is V =100sin(100t)V Compare it with V = V0 sin(t)V we get 1 0 V 100V, 100rads − =  = The rms value of voltage is, V 70.7V 2 100 2 V V 0 rms = = = The instantaneous value of current is mA 3 I 100sin 100t        = + Compare it with I I sin( t ) = 0  +  we get 1 0 I 100mA, 100rads − =  = The rms value of current is mA 70.7mA 2 100 2 I I 0 rms = = = 14. (d): The average value of the voltage is       = 2 / 0 2 / 0 av( ) dt Vdt V vkSlr    + −  =         2 V sin tdt ( V sin t)dt / 0 2 / / 0 0           +  −    =       2 / / 0 0 / 0 V cos t V cos t 2  −  + +  −  =  = 0 2V0 [ cos cos0 cos2 cos ] 2 V 15. (c):
V = V0 for , 2 T 0  t  V = 0 for , t T 2 T   1/ 2 T 0 T / 2 0 T T / 2 2 0 1/ 2 T 0 T 0 2 rms dt V dt (0)dt dt V dt V               + =               =      1/ 2 2 0 1/ 2 2 0 1/ 2 T / 2 0 2 0 2 V 2 T T V [t] T V        =             =      = 2 V V 0 rms = 16. (d): V =120sin(100t)cos(100t)V = 60sin(200t) (sin2 = 2sincos ) Compare it with standard equation V V sin t = 0  We get V0 = 60V and ,  = 200 or 2 = 200 or  =100Hz 17. (b): in an inductor voltage leads the current by 2  or current las the voltage by 2  . 18. (c): Here, , L 30mH 30 10 H −3 = =  V 220V, 100Hz rms =  = Inductive reactance, XL = 2L =     =  − 2 3.14 100 30 10 18.85 3 19. (b): As the rod is inserted, inductance increases and hance the voltage across inductor increases. This causes a drop in the voltage across the bulb and hence it gets dimmer. 20. (b): Inductive reactance, XL = L = 2L  XL   Hence, inductive reactance increases linearly with frequency. 21. (c): Here, , L 44mH 44 10 H −3 = =  V 220V, 50Hz rms =  = The inductive reactance is , 3 XL L 2 L 2 3.14 50 44 10− =  =  =     =13.82 15.9A 13.82 220 X V I L rms  rms = = = 22. (a): Here, , L = 0.01H,R =1 V 220V, 50Hz rms =  = As , R 2 L R L R X tan L  =   = = 3.14 1 2 3.14 50 0.01 tan =     tan (3.14) −1  = rad 180 72 72 o o  = =  The time lag between maximum alternating voltage and current is. s 250 1 2 50 180 72 t =    =    = 23. (b): The current in the inductor coil is given by 2 L V X V I L  = = Since frequency  in the two cases is different, hence the current in two cases will be different. 24. (a):        = R X tan L XL = L = (2L) = (2)(50)(0.01) =  Also, R = 1 tan ( ) 1  =  − 25. (d): The growth of current in LR circuit is given by I I [1 e ] Rt / L 0 − = − Or 2 1 e Rt / L = − or R Lln 2 In 2 t L Rt =  = or 0.1s 2 300 10 0.693 t 3 =   = − 26. (c): Here, , C 5 F 5 10 F,V 200V, 100Hz rms 6 =  =  =  = − The capacitive reactance is , 2 C 1 XC  = C 6 2 3.14 100 5 10 1 X −      = =   2 3.18 10 = 318
27. (c): Capacitive reactance, 2 C 1 XC  = 2 2 X 1 X X 1 2 C C C 2 1 =   =    =    2 X X 1 2 C  C = 28. (b): Here, , Vrms = 220V,I rms = 65mA = 0.065A C 8 F 8 10 F −6 =  =   =100Hz Capacitive reactance, , 2 C 1 XC  = 6 2 3.14 100 8 10 1 −     = =199 Then rms voltage across the capacitor is VCrms = I rmsXC = 0.065199 =12.94V 29. (b): In a capacitive ac circuits, the voltage lags behind the current in phase by / 2 radian. 30. (b): Here, , C 30 F 30 10 F −6 =  =  V 150V, 60Hz rms =  = Capacitive reactance , 3 4 C 1.1 10 9 10 2 C 1 C 1 X = =   =  = rad 1 s − 1.7A 88.46 150 X V I C rms rms = = = 31. (c): Here, C 60 F 60 10 F −6 =  =  V 110V, 60Hz rms =  = 1/ 2 C V X V I rms C rms rms   = = = 2CVrms 2 3.14 60 60 10 110 2.49A 6 =       = − 32. (c): Reactance of a capacitor, 2 C 1 C 1 XC  =  = As frequency increases, XC decreases and therefore current increases. As R does not vary with frequency, therefore, likely elements constituting the circuit may be capacitor and resistor. 33. (d): When an ac voltage of 220 V is applied to a capacitor C, the charge on the plates is in phase with the applied voltage. As the circuit is pure capacitive so, the current developed leads the applied voltage by a phase angle of o 90 Hence, power delivered to the capacitor per cycle is P V I cos90 0 o = rms rms = 34. (d): Electrical energy is transmitted over large distances at high alternating voltage As P = VI, therefore for a given power level, current (I) is lower and hence power loss ( I R) 2 = becomes low. At the receiving end, high voltage can be easily reduced using step down transformers. 35. (a): On comparing V = 200 2 sin(100t) with V V sin t, = 0  we get 1 0 V 200 2V, 100rad s − =  = 200V 2 200 2V 2 V V 0  rms = = = The capacitive reactance is , =    =  = − 4 C 6 10 100 1 10 1 C 1 X ac ammeter reads the rms value of current, therefore, the reading of the ammeter is, 20 10 A 20mA 10 200V X V I 3 4 C rms rms =  =  = = − The average power consumed in the circuit, P = I rmsVrms cos In an pure capacitive circuit, the phase difference between current and voltage is . 2  cos = 0 P = 0 36. (d): Impedance of the circuit, 2 L C 2 Z = R + (X − X ) At resonance, XL = XC Z = R . R V Z V I m m  m = = 37. (b): At resonance frequency, the inductive capacitive reactance are equal. i.e., , XL = XC  Impedance, Z R (X X ) R 0 R 2 2 2 L C 2 = + − = + = 38. (a): Maximum 39. (d): Here, ,R = 0.2K = 200 C 15 F 15 10 F −6 =  =  V 220V, 50Hz rms =  = Capacitive reactance, 2 C 1 XC  = 6 2 3.14 50 15 10 1 −     = = 212
The impedance of the RC circuit is, 2 C 2 Z = R + X = (200) + (212) = 291.5 2 2 40. (a): Here, R 6 ,L 25mH 25 10 H, −3 =  = =  C 750 F 750 10 F, 50Hz 6 =  =   = − =  =     =  − X 2 L 2 3.14 50 25 10 7.85 3 L 2 C 1 XC  = =      = − 4.25 2 3.14 50 750 10 1 6 XL = XC = 7.85− 4.25 = 3.6 Impedence of the series LCR circuit is 2 L C 2 Z = R + (X − X ) 2 2 Z = (6) + (3.6) = 36+12.96 = 7.0 41. (d): Here, , L = 30 mH 30 10 H,R 8 , 50Hz r 3 =  =   = − As , r r  = 2 = 23.1450 = 314Hz  Quality factor, 1.18 8 314 30 10 R L Q 3 r =   =  = − 42. (a): C 6 2 3.14 50 0.1 10 1 2 C 1 X −     =  = 2 8 C 2 Z = R + X = 100+10.2810 =   4 3.2 10 3.14 10 A 3.2 10 100 Z I 3 4 rms rms − =   =  = 3.14mA 43. (b): If the frequency of the ac source equals the natural frequency of the circuit, the impedance Z = R = 20  The average power dissipated per cycle, 20 (200) R V Z V P 2 rms 2 rms 2 av(vkSlr) = = = = 2000W 44. (a): Here, , L 2H,C 32 F 32 10 F −6 = =  =  R =10 6 r 2 32 10 1 LC 1 −    = = 1 3 3 125rads 8 10 64 10 − = = = 20Hz 2 3.14 125 2 r r =  =    = 45. (b): Here, R L 2   = If R resistance increases, quality factor  decreases finitely. 46. (a): R X X tan L − C  = 47. (d): Bandwidth is the frequency range at which current amplitude m max max I 0.7 I 2 1 I = = From figure, Band width, 1 1.2 0.8 0.4rads − = − = 48. (b): Here, P = 2 kW= 2× 3 10 W 4 3 Vrms = 223V,tan  = − As, Z V P rms 2 =  = = = = 24.86 2000 49729 2000 (223) P V Z 2 rms 2 or, Z ≈ 25 4 3 R X X tan C L = − −  = R. 4 3 XC − XL = − As, 2 C L 2 2 Z = R + (X − X ) 2 2 2 R 4 3 (25) R        = + − 16 25R R 16 9 625 R 2 2 2 = + = 400 25 625 16 R 2 =  = R = 20 49. (c): Impedance of an LCR circuit is given by 2 L C 2 Z = R + (X − X ) 2 2 2 C 1 R 2 L        = +  − The variation of Z with is shown in figure.