An L C R series circuit with a resistance of 100 ohm is connected to an AC source of 200 V (r.m.s.) and angular frequency 300 rad / s. When only the capacitor is removed, 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 is

Question

An L C R series circuit with a resistance of 100 ohm is connected to an AC source of 200 V (r.m.s.) and angular frequency 300 rad / s. When only the capacitor is removed, 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 is (A) 50 W (B) 100 W (C) 200 W (D) 400 W

Tardigrade Admin · Accepted Answer

tan φ=(XL/R)=(XC/R) ⇒ tan 60°=(XL/R)=(XC/R) ⇒ XL=XC=√3 R i.e., Z=√R2+(XL-XC)2=R So average power, P=(V2/R) =(200 × 200/100)=400 W

50 W

100 W

200 W

400 W

tanϕ=RXL​​=RXC​​ ⇒tan60∘=RXL​​=RXC​​ ⇒XL​=XC​=3​R i.e., Z=R2+(XL​−XC​)2​=R So average power, P=RV2​ =100200×200​=400W

$tan ϕ = \frac{X _{L}}{R} = \frac{X _{C}}{R}$
$\Rightarrow tan 6 0^{\circ} = \frac{X _{L}}{R} = \frac{X _{C}}{R}$
$\Rightarrow X_{L} = X_{C} = 3 R$
i.e., $Z = R^{2} + (X_{L} - X_{C})^{2} = R$
So average power, $P = \frac{V ^{2}}{R}$
$= \frac{200 \times 200}{100} = 400 W$