Alternating Current -07-ANSWERSHEET

1. (d) 2. (b) 3. (d) 4. (d) 5. (d) 6. (a) 7. (b) 8. (c) 9. (b) 10. (b) 1. (b) 2. (b) 3. (d) 4. (c) 5. (c) 6. (a) 7. (b) 8. (c) 9. (d) 10. (a) L rms C rm  0   0 . 4. (a) 1.0 J yes, (b)   103 rad/sec (c) q  q0 cost t  T 3T (i) energy stoed completely electrical at 0, , T, 2 ,...... 2 (ii) energy stored completely magnetic (i.e. electrical energy zero) at t  T , 3T , 5T ,.... where T  1  6.3 ms 4 4 4 v (d) At T  T , 3T , 5T ,..... because q  q cos T  q cos     q0 8 8 8 0 8 0 4 2 (e) R damps out the LC oscillations eventually. The whole of the intial energy ( = 1.0 J) is eventually dissipated as heat 5. (a) 1.82 A (b) 3.2 ms 6. a = 3.23 A, b = 1.55 ms 7. 132.3 volt, 0.042 henry 8. (a) for v = v0 sin t I  v0 sin t   if R = 0 L  1   2   C  I0  11.6A , Irms  8.24A (b) vLrms  207V , v C rms  437V  (c) Whatever be the current I is L, actual voltage loads current by 2 consumed by 2 is zero. therefore averave power  (d) For C, Voltage lags by 2 . Again average power consumed by C is zero. (e) Total average power absorbed is zero. 9. (a) Imax  14.1A (b) Pmax  2300W (c)648 and 678; 10A (d) Q = 21.7 10. Power current = 0.197 amp and wattles current = 0.865 amp. 1. 1.0 103 A, 0.01A, 0.05A, 0.1A 2. (a) 0.10A (b) 50V, 30V, 10V 3. (a) 27 Hz (b) 2 mA 4. zero 5. (a) 2.49 A (b) 37.4, 153, 218, 75.0 V. (c) PC  PL  0 , PR  92W 6. 22sin(100t) 7. 4 J 8. 0.386 amp 9. 0.29 amp, cos  0.008 10. 5F , in series condenser 1. 2. 1.5 A 3. 8.5 mV, 1.6 mV, 0.16 mV, 16 V 5. 100 V 7. 155, 710.4 watt 8. (i) 70.20 ohm (ii) vC  100V and vL  27 V 9. Imax  20A 10. 80 mH, 17.2

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