An inductor of inductance L and resistor R are joined together in series and connected by a source of frequency `omega`. The power dissipated in the circuit is

Options

A. `(V)/(R^(2) + omega^(2)L^(2))`
B. `(R^(2) + omega^(2)L^(2))/(V)`
C. `(V^(2)R)/(sqrt(R^(2)+ omega^(2)L^(2)))`
D. `(V^(2)R)/(R^(2) + omega^(2)L^(2))`

Answer

D

Explanation

(d): Power = VI `cos phi`
`:. P = V. (V)/(sqrt(omega^(2)L^(2)+R^(2)))xx(R)/(sqrt(omega^(2)L^(2)+R^(2)))`
= `(V^(2)R)/(omega^(2)L^(2)+R^(2))`