Question

An alternating emf of $E=2\sin \left(100t\right)$ is applied across an $LCR$ series circuit, as shown in the diagram. The power factor of the circuit is $\frac{1}{\sqrt{2}}$. The capacitance of the circuit is equal to,

• A. $100\mu F$
• B. $300\mu F$
• C. $500\mu F$
• D. $200\mu F$

Answer 1

The correct option is C $500\mu F$

The power factor is given by,
$\cos \varphi =\frac{R}{Z}$

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

$10\sqrt{2}={\sqrt{(10{)}^2+(X_C-X_L)}}^2$

$200=100+(X_C-X_L{)}^2$

$\therefore X_C-X_L=10\Omega$

Here, $X_L=\omega L=100\times 0.1=10\Omega$

$\therefore X_C=20\Omega$

Now, $X_C=\frac{1}{\omega C}$
$\Rightarrow C=\frac{1}{X_C\times \omega }=\frac{1}{20\times 100}$

$=500\times {10}^{-6}\text{F}=500\mu F$

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