An inductor 20 mH, a capacitor 50 and a resistor 40 are connected in series across a source of emf V = 10 340t. The power loss in A.C. circuit is
- 0.76 W
- 0.89 W
- 0.46 W
- 0.67 W
Answer
(3) 0.46 W
Solution
= 340 rad s-1, V0 = 10 V
XL = L = 340 20 10-3 = 6.8
XC = 1 C = 1340 50 10-6 58.8
Z = R2 + (XC - XL)2
= 402 + (58.8 - 6.8)2
= 1600 + 2704 65.6
Power is lost only in the resistor:
P = Irms2 R = Vrms2 RZ2
= (10/ 2)2 40(65.6)2
= 50 404304 0.46 W
Answer: (3)