An inductor 20 mH, a capacitor 100 and a resistor 50 are connected in series across a source of emf, V = 10 314 t. The power loss in the circuit is
- 0.79 W
- 0.43 W
- 2.74 W
- 1.13 W
Answer
(1) 0.79 W
Solution
XL = = 314 20 10-3 = 6.28
XC = 1 = 1314 100 10-6 = 31.85
Z2 = R2 + (XC - XL)2 = 2500 + (25.57)2 = 3153.9 2
Power loss (only in R):
P = Irms2 R = V022Z2R
= 100 502 3153.9
= 0.79 W
Answer: (1)