Question

In a series $LR$ circuit, power of $400\text{W}$ is dissipated from a source of $250\text{V},50\text{Hz}.$ 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 $C$ as $\left(\frac{n}{3\pi }\right)\mu \text{F},$ then value of $n$ is ________.

Answer 1

Given : Power $P=400\text{W},$ Voltage $V=250\text{V}$

$P=V_{rms}{I}_{rms}.\cos \varphi$

$\Rightarrow 400=250\times I_{rms}\times 0.8\Rightarrow I_{rms}=2\text{A}$

Using $P=I_{rms}^{2}R$

$(I_{rms}{)}^{2}R=P\Rightarrow 4\times R=400$

$\Rightarrow R=100\Omega$

Power factor is, $\cos \varphi =\frac{R}{\sqrt{R^2+X_L^2}}$

$\Rightarrow 0.8=\frac{100}{\sqrt{{100}^2+X_L^2}}\Rightarrow {100}^2+X_L^2={\left(\frac{100}{0.8}\right)}^2$

$\Rightarrow X_L=\sqrt{-{100}^2+{\left(\frac{100}{0.8}\right)}^2}\Rightarrow X_L=75\Omega$

When power factor is unity,
$X_C=X_L=75\Rightarrow \frac{1}{\omega C}=75$

$\Rightarrow C=\frac{1}{75\times 2\pi \times 50}=\frac{1}{7500\pi }\text{F}$

$=(\frac{{10}^6}{2500}\times \frac{1}{3\pi })\mu F=\frac{400}{3\pi }\mu F$

$\therefore n=400$