Question

The power factor of the circuit shown in the figure is $\frac{1}{\sqrt{2}}$. The capacitance of the circuit is equal to

• A. $400\mu \text{F}$
• B. $300\mu \text{F}$
• C. $500\mu \text{F}$
• D. $200\mu \text{F}$

Answer 1

The correct option is C $500\mu \text{F}$

Given:
Power factor, $\cos \varphi =\frac{1}{\sqrt{2}}$

$V=2\sin \left(100t\right)$

$\Rightarrow \omega =100$

$\Rightarrow X_L=\omega L=100\times 0.1=10\Omega$

$\Rightarrow X_C=\frac{1}{\omega C}=\frac{1}{100C}$

As, $\cos \varphi =\frac{R}{Z}=\frac{R}{\sqrt{R^2+(X_L-X_C{)}^2}}$

$\Rightarrow \frac{1}{\sqrt{2}}=\frac{R}{\sqrt{R^2+(X_L-X_C{)}^2}}$

$\Rightarrow \frac{R^2}{R^2+(X_L-X_C{)}^2}=\frac{1}{2}$

$\Rightarrow R^2=(X_L-X_C{)}^2$

$\Rightarrow \pm (X_L-X_C)=R$

$\Rightarrow \pm (10-\frac{1}{100C})=10$

$\Rightarrow 10-\frac{1}{100C}=-10$

$\Rightarrow 20=\frac{1}{100C}$

$\Rightarrow C=\frac{1}{2000}=500\times {10}^{-6}\text{F}=500\mu \text{F}$

Hence, option $\left(C\right)$ is correct.