99 Percentile Qs Bank for JEE MainPhysicsAlternating Current
A coil has inductance of 0.4 H and resistance of 8 . It is connected to an AC source with peak emf 4 ~V and frequency 30 Hz . The average power dissipated in the circuit is
Options
- A1 ~W
- B0.5 ~W
- C0.3 ~W
- D0.1 ~W
Correct answer
D. 0.1 ~W
Step-by-step solution
Average power dissipated is aligned & P_ avg =V_ rms I_ rms & = V_ 2 ( V_ 2 ) 1 Z R Z & = V_ ^2 2 R Z^2 = V_ ^2 2 R ( X_L^2+R^2 ) & = (4)^2 2 8 . (0.4 60)^2+8^2 )^2 & = 16 8 2 (24^2+8^2 ) = 64 640 = 1 10 =0.1 ~W aligned