Question

In a series LR circuit, power of $400W$ is dissipated from a source of $250V,50Hz.$ The power factor of the circuit is $\mathrm{0.8.}$ 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 $\left(\frac{n}{3\pi }\right)\mu F,$ then value of $n$ is ________.

Answer 1

Answer: 400

Given in $1^{st}$ case power factor of LR circuit
$cos\varphi =0.8=\frac{R}{\sqrt{R^2+X_L^2}}$
where $\sqrt{R^2+X_L^2}=Z\dots \left(1\right)$
$\therefore P=VIcos\varphi$
$\Rightarrow 400={\left(250\right)}^2\times \frac{R}{Z^2}$
$400={\left(250\right)}^2\times \frac{0.8}{Z}$
$Z=\mathrm{125...}\left(2\right)$
$R=\frac{{\left(250\right)}^2\times 0.8\times 0.8}{400}\Rightarrow 100\Omega ...\left(3\right)$
from $\left(1\right),\left(2\right)$ and $\left(3\right)$
$(100+X_L^2=(125$
$X_L^2=15625-10000$
$X_L^2=5625$
$X_L=\mathrm{75...}\left(4\right)$
In $2^{nd}$ case given,
Power factor $=1$ that means $X_L=X_c$ (Resonance condition)
$X_L=\frac{1}{{\left(\omega \right)}_C}\Rightarrow 75=\frac{1}{\left(2\pi F\right)C}$
$C=\frac{1}{2\pi \times F\times 75}$
$C=\frac{1}{2\pi \times 50\times 75}F\dots \left(5\right)$
$C=\frac{n}{3\pi }\mu F$ given
From $\left(5\right)\&\left(6\right)$
$\frac{1}{2\pi \times 50\times 75}=\frac{n\times 10^{-6}}{3\pi }$
$n=\frac{10^6}{7500}\Rightarrow \frac{3\times 10^4}{75}\Rightarrow \frac{30000}{75}$
$n=400$