An L-C-R series circuit with a resistance of 100 Ω is connected to 200 V (AC source) and angular frequency 300 rad/s. When only the capacitor is removed, then the current lags behind the voltage by 60°. When only the inductor is removed the current leads the voltage by 60°. The average power dissipated in original L-C-R circuit is

50 W

100 W

200 W

400 W

Solution

400 W

Given:-

R = 100 Ω,

V = 200 V,

θ = 60°,

ω = 300 rad/s

∴ Phase angle

tanθ = $\frac{X_{L}}{R}$ = $\frac{X_{C}}{R}$

⇒ tan 60^o = $\frac{X_{L}}{X_{R}}$ = $\frac{X_{C}}{R}$

∴ X_L = X_C = $\sqrt{3} R$

Z = $\sqrt{R^{2} + {(\sqrt{3} R - \sqrt{3} R)}^{2}}$

Z = R

So average power

P = $\frac{V^{2}}{R}$

= $\frac{200 \times 200}{100}$

P = 400 W

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