In a series LR circuit, a power of $400$ W is dissipated from a source of $250$ V, $50$ Hz. The power factor of the circuit is $0.8$. In order to bring the power factor to unity, a capacitor of value $C$ is added in series to the $L$ and $R$. Taking the value of $C$ as $\left(\dfrac{n}{3\pi}\right)\ \mu\text{F}$, the value of $n$ is ______.
Numerical value type. Enter your answer.
Answer: 400
$P = I^2R$ with $I = \dfrac{V\cos\phi}{R}$ (since $Z = R/\cos\phi$), so
$$R = \frac{V^2\cos^2\phi}{P} = \frac{250^2\times0.64}{400} = 100\ \Omega$$
$\tan\phi = \dfrac{0.6}{0.8} = \dfrac34 \Rightarrow X_L = 75\ \Omega$.
Unity power factor needs $X_C = X_L = 75\ \Omega$:
$$C = \frac{1}{2\pi\times50\times75} = \frac{1}{7500\pi}\ \text{F} = \frac{400}{3\pi}\ \mu\text{F}$$
So $n = 400$.
Solution by Sreeraj P, M.Sc Physics