Tiêu đề 23. Alternating Current Easy.pdf ✅

1. (b) Power loss 2 (Voltage) 1  2. (a)       = = + 2 5 cos 5 sin  V  t  t and i = 2 sin t Power = Vr.m.s. i r.m.s. cos = 0 (Since 2   = , therefore 0 2 cos = cos =   ) 3. (c) P = Vr.m.s. i r.m.s. cos 3 cos 2 100 10 2 100 3    =  − 2.5 watt 4 10 2 1 2 10 10 4 3  = =  = − 4. (b) In dc ammeter, a coil is free to rotate in the magnetic field of a fixed magnet. If an alternating current is passed through such a coil, the torque will reverse it’s direction each time the current changes direction and the average value of the torque will be zero. 5. (b) The coil having inductance L besides the resistance R. Hence for ac it’s effective resistance 2 2 R + XL will be larger than it’s resistance R for dc. 6. (b) ampere i i o r m s 2 2 2 4 2 . . . = = = 7. (c) Effective voltage V V V o r m s 300 2 423 2 . . . = = = 8. (d) The current takes 4 T sec to reach the peak value. In the given question T sec T 100 1 200 2 =   =   Time to reach the peak value sec 400 1 = 9. (c) i r m s 3 2 A 2 6 . . = = 10. (c) 19 Hz 2 22 120 7 2 =   = =    Vr m s 120 2 170 V 2 240 . . . = =  11. (d) 12. (c) Peak value = 220 2 = 311 V 13. (b) Power 2 2 2 2 2 I R R I I R p p =         = = 14. (c) A R V i r m s r m s 5 40 . . . 200 . . . = = =  i0 = i r.m.s. 2 = 7.07 A 15. (b) 16. (d) Time taken by the current to reach the maximum value sec T t 3 5 10 4 50 1 4 1 4 − =   = = =  and io = i rms 2 = 10 2 = 14.14 amp 17. (c) 18. (b) T t E E t E   2 cos cos = 0 = 0 6 10 cos 600 2 50 1 10 cos   =   = = 5 3 volt. 19. (d) Phase angle o  = 90 , so power P = Vi cos  = 0 20. (c) 2 1 , 2 200 Vrms = i rms = P V i watt rms rms 50 3 cos 2 1 2 200  = cos = =   21. (c) 2 = 377  = 60.03 Hz 22. (a) 23. (c) 2 1 / 2 2 2 1 2 2 2 1 ( ) 2 1 2 i i i i i rms = + + = 24. (d) P = Vi cos  Phase difference =  P = zero 2   25. (c) V0 = Vrms  2 = 220  2 = 310 26. (c) Hot wire ammeter reads rms value of current. Hence its peak value = i rms  2 = 14.14 amp 27. (d) 28. (b) Peak voltage = 2  220 = 311 V 29. (c)
30. (d)  P = Vi cos ,  P  cos  31. (d) cos ; sin 90 . o rms rms P = V I  ce  = So P = 0 32. (d) Brightness  R Pconsumed 1  for Bulb, , Rac = Rdc so brightness will be equal in both the cases. 33. (b) W R V P rms 90 10 (30) 2 2 = = = 34. (b) V V Vrms 84.8 1.414 120 2 0 = = = 35. (d) Peak value to r.m.s. value means, current becomes 2 1 times. So from i i t i i sin100 t 2 1 sin100 = 0   0 = 0 t t sec 400 1 sin100 4  sin =   =  2.5 10 . 3 sec − =  36. (c) Phase difference 6 6 3 2 1        =      −  = − = − 37. (a) Va v V Vrms Vrms . 2 2 ( 2) 2 2 0    = =   = 220 198 V 2 2 =  =  38. (b) 39. (a) 40. (c) 41. (a) . 7 5 280 200 i rms = = A So 2 1 . 7 5 i0 = i rms  2 =   A 42. (d) Required time t T sec 3 5 10 4 50 1 / 4 − =   = = 43. (b) V0 = 2 Vrms = 10 2 44. (b) cos .cos 2 1 o o P PPeak P = V i  = 2 3 1 ( ) cos cos 2 1   Ppeak = Ppeak    =  = 45. (c) E = 141 sin(628 t), V E Erms 100 1.41 141 2 0 = = = and 2f = 628  f = 100 Hz 46. (c) V E Erms 500 1.41 707 2 0 = = = 47. (b) 48. (a) 49. (a) The choke coil can be used only in ac circuits, not in dc circuits, because for dc ( = 0) the inductive reactance XL = L of the coil is zero, only the resistance of the coil remains effective which too is almost zero. 50. (b) Because power , 2 = i R if R = 0, then P = 0. 51. (a) 52. (a) A choke coil contains high inductance but negligible resistance, due to which power loss becomes appreciably small. 53. (b) For purely capacitive circuit e e sint = 0       = + 2 sin  i i t o i.e. current is ahead of emf by 2  54. (c) 55. (d) 56. (b) Z = R + X L , X L = L 2 2 and  = 2f 2 2 2 2  Z = R + 4 f L 57. (c) Hz LC     2 10 2 10 10 1 2 1 5 6 4 =  = = − − 58. (b) 59. (b) The applied voltage is given by 2 2 V = VR + VL 2 2 V = (200 ) +(150 ) = 250 volt 60. (b) 2 2 2 2 2 2 100 4 60 20 120 +   = + = R  L  V i =0.016 A 61. (d) For the first circuit 2 2 2 R L V Z V i + = =  Increase in  will cause a decrease in i.
For the second circuit 2 2 2 1 C R V i  + =  Increase in  will cause an increase in i. 62. (b) C C XC  2 1 1 = = ; For dc  = 0,  XC =  63. (a) In a pure inductor (zero resistance), voltage leads the current by 90 i.e.  / 2. o 64. (b) 65. (a) The voltage across a L–R combination is given by 2 2 2 V = VR + VL 400 144 256 16 . 2 2 V V V volt L = − R = − = = 66. (a) Phase angle 3 1 4 300 2 200 tan  =  = =     R L 3 4 tan −1   = 67. (b) At resonance, LCR circuit behaves as purely resistive circuit, for purely resistive circuit power factor = 1 68. (d) Given =  C L   1 LC 2 1  = or 4 3 6 8 10 10 1 10 10 10 1 = =   = − − −  = =  =  − 10 10 10 4 3 XL L 69. (b) 70. (b) Reading of ammeter 2 V0 C X V i C rms rms  = = = 2 10 A 20 mA 2 200 2 100 (1 10 ) 2 6 =  =    = − − 71. (a) Current will be maximum at the condition of resonance. So resonant frequency 6 0 0.5 8 10 1 1 −   = = LC  = 500 rad/s 72. (d) Average power in ac circuits is given by P = Vrmsi rms cos For pure capacitive circuit o  = 90 so P = 0 73. (c) Amplitude of R NB r R NBA R V ac i (2 ) ( ) 2 0 0    = = = = i 6 mA 1 10 (0.3) 60 200 2 2 2 2 0 =       = −    74. (b) 2 2 Z = R + X L 10 (2 60 2) 753 .7 2 2 = +    = i 0.159 A 753 .7 120  = = 75. (c) Resonance frequency in radian/second is rad sec LC 500 / 8 0.5 10 1 1 6 =   = = −  76. (c) 4 1 1 1 2 1 1 2 2 L L L C L C  = =  = 77. (c) Z = XL = 2  60  0.7 ampere Z i 0.455 2 60 0.7 120 120 =    = =  78. (b) 4 3 5 2 2 2 2 Z = R + X = + = 0.6 5 3  cos = = = Z R  79. (d) 80. (a) Z R cos = . In choke coil   90  so cos   0 81. (c) 2 2 ( ) Z = R + X L − X C 2 6 2 10 10 100 1 100 0.5 100         = +  − −   = 189 .72 82. (c) V 46 volts, L = V volts C = 40 , V volts R = 8 E.M.F. of source V 8 (46 40) 10 volts 2 2 = + − = 83. (c) Resonant frequency 2 LC 1 = does not depend on resistance. 84. (b) Frequency 2 LC 1 = So the combination which represents dimension of frequency is 1 / 2 ( ) 1 − = LC LC 85. (c) For series R-L-C circuit, 2 2 ( ) Z = R + X L − X C =           = +  − 500 1000 2 10 (300 ) 1000 0.9 2 6 2 86. (d) 2 2 2 2 Z = R + X = R + (2fL)
=  = + =       +   900 1600 50 0.4 (30) 2 50 2 2   ampere Z V i 4 50 200 = = = 87. (b) Reactance = L  =   50  L 7 22 2 100 2  L = 0.32 henry 88. (b) 89. (a) =     = = −    5 2 4000 25 10 1 2 1 6 C X C 90. (a) C XC 2 1 = 6 2 5 10 1 1000 1 −     =   MHz   100  = 91. (d) A Z V i 0.8 4 (1000 3 10 ) 4 2 3 2 = +   = = − 92. (c) X R R X R X L L o L tan = tan 45 = = 1 = 93. (a) For purely L-circuit P = 0 94. (a) At resonance LCR series circuit behaves as pure resistive circuit. For resistive circuit o  = 0 95. (c) V V V V Volt R ( L C ) (5) (10 10) 5 2 2 2 2 = + − = + − = 96. (b) When dc is supplied = = = 100  1 100 i V R When ac is supplied = = = 200  0.5 100 i V Z 97. (b) 2 2 2 cos R L R Z R   + = = 2 2 2 2 (12) 4 (60) (0.1) 12 +    =  98. (c) 99. (b) 100.(c) Impedance = + = (8) + (6) = 10 2 2 2 2 Z R X 101.(c) o R L 5.5 80 12 2 50 0.21 tan =  =   = =     102.(a) If the current is wattless then power is zero. Hence phase difference o  = 90 103.(a) LC f 2 1 =  C f 1  104.(b) In non resonant circuits impedance 2 2 1 1 1       + − = L C R Z   , with rise in frequency Z decreases i.e. current increases so circuit behaves as capacitive circuit. 105.(b) V VR VL (20) (16) 656 25.6 V 2 2 2 2 = + = + = = 106. (d) 2 2 (20) (2 50 0.2) 220 +    =  i 3.33 A 66 220 = = 107.(a) Impedance of LCR circuit will be minimum at resonant frequency so  LC  2 1 0 = 3 6 2 1 10 0.1 10 1 − −    =  Hz 2 10 5 = 108. (c) o Z R 60 2 1 20 10 cos = = =  = 109.(a) Current in LC circuit becomes maximum when resonance occurs. So 200 / sec 5 1000 1 25 10 1 1 6 rad LC = =   = = −  110. (b, d) 111.(a) R = 6 + 4 = 10  = =   =  − 2000 5 10 10 3 XL L =    = = − 10 2000 50 10 1 1 6 C XC   = + ( − ) = 10  2 2 Z R X L X C Amplitude of current A Z V i 2 10 0 20 = 0 = = = 112.(a) A X L V i L 0.637 2 50 1 200 200 =   = = =   113. (a)