Alternating Current -02- OBJECTIVE SOLVED

SOLVED OBJECTIVE PROBLEMS Problem 1. A coil of resistance 200 ohms and self inductance 1.0 henry has been connected to an a.c. source of frequency 200 /  Hz. The phase difference between voltage and current is : (a) 30º (b) 63º (c) 45º (d) 75º. Ans. (b) Solution: tan   XL  2fL  2(200 / )1  2 R R 200    63º . Problem 2. In an a.c. circuit, V & I are given by V = 100 sin (100 t) volt. I = 100 sin (100 t +  /3) mA. The power dissipated in the circuit is : (a) 104 watt (b) 10 watt (c) 2.5 watt (d) 5 watt. Ans. (c) Solution: P  Vrms Irms cos  100 . 100 103 cos 60º    10  1  2.5 watt . 2 2 Problem 3. Resonance frequency of a circuit is f. If the capacitance is made 4 times the initial value, then the resonance frequency will become : (a) f/2 (b) 2f (c) f (d) f/4. Ans. (a) Solution : f  i.e. f    1 2 time Problem 4. The current flowing in a coil is 3 A and the power consumed is 108 W. If the a.c. source is of 120 V, 50 Hz, the resistance of the circuit is : (a) 24  (b) 10  (c) 12  (d) 6  . Ans. (c) Solution: From P  I2R, R  P  108  12 I2 32 Problem 5. In an L-R circuit, the value of L is (0.4 / ) henry and the value of R is 30 ohm. If in the circuit, an alternating emf of 200 volt at 50 cycles per second is connected, the impedance of the circuit and current will be : (a) 11.4 ohm, 17.5 ampere (b) 30.7 ohm, 6.5 ampere (c) 40.4 ohm, 5 ampere (d) 50 ohm, 4 ampere. Ans. (d) Solution: Here XL  L  2fL  2 50  0.4  40  R = 30   Z    50 Irms  Vrms  200  4A . Z 50 Problem 6. In the circuit shown in figure, what will be the read- ings of voltmeter and ammeter ? (a) 800 V, 2 A (b) 220 V, 2.2 A (c) 300 V, 2 A (d) 100 V, 2 A. Ans. (b) L C 100 Solution: As and  VL  VC  300 V, V  VR  V  220 V Also I  V  220  2.2 A. R 100 Problem 7. In an A.C. circuit, maximum value of voltage is 423 volt. Its effective voltage is : (a) 323 V (b) 340 V (c) 400 V (d) 300 V. Ans. (d) Solution: Erms  E0  423  300 V. 1.41 Problem 8. In a transformer, nP  500, nS  5000 . Input voltage is 20 V and frequency is 50 Hz. Then in the output, we have (a) 200 V, 500 Hz (b) 200 V, 50 Hz (c) 20 V, 50 Hz (d) 2 C, 5 Hz. Ans. (b) Solution: E  nS  E nP  5000  20  200 500 Frequency is not affected by transformer. Problem 9. The p.d. across an instrument in an a.c. circuit of frequency f is V and the current flowing through it is I such that V = 5 cos (2 ft ) volt and I = 2 sin (2 ft ) amp. The power dissipate in the instrument is : (a) zero (b) 10 watt (c) 5 watt (d) 2.5 watt. Ans. (a) Solution: As V = 5 cos (2  ft ) =5 sin (2 And I = 2 sin (2 ft )  ft +  / 2 )  phase difference between V and I is    2 Average power P = Vrms Irms  cos  0 Problem 10. The power factor of the circuit shown in figure is : (a) 0.2 (b) 0.4 XC=40  (c) 0.8 (d) 0.6. Ans. (c) Solution: R = 40 + 40 = 80  , XL  XC  100  40  60  Z  220V 50Hz X=L 100  r=40   100 Power factor, cos   R  80  0.8 . Z 100

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