An inductor `20 mH`, a capacitor `100 muF` and a resistor `50 Omega` are connected in series across a source of emf, `V=10 sin 314 t`. The power loss in the circuit is

Options

A. 0.79 W
B. 0.43 W
C. 2.74 W
D. 1.13 W

Answer

A

Explanation

Impedance Z in an ac circuit is
` Z = sqrt(R^2 + (X_C - X_L)^2) `, where `X_C` = capacitive reactance and `X_L` = inductive reactance
Also , `X_C = (1)/(omega C) `and `X_L = omega L`
` therefore Z = sqrt((50)^2 ( (1)/(314 xx 100 xx 10^(-6)) - 314 xx 20 xx 10^(-3))^2)`
`Z = 56 Omega`
The power loss in the circuit is `P_(av) = ((V_(rms))/(Z))^2 R`
` therefore P_(av) = ((10)/((sqrt2)56) )^2 xx 50 = 0.76 W`