Question

The power factor of the given circuit is $1/\sqrt{2}$. The capacitance of the circuit (in $\mu \text{F}$) is equal to

Answer 1

We know that,
Power Factor $=\cos \theta =\frac{R}{Z}=\frac{1}{\sqrt{2}}$
where $Z$ is the impedance of the circuit.

In the given circuit,
$R=10\Omega ,L=0.1\text{H},\omega =100\text{rad/s}$

We know,
$Z=\sqrt{R^2+(X_L-X_C{)}^2}=\sqrt{(10{)}^2+(X_L-X_C{)}^2}$
where
$X_L=\omega L=100\times 0.1=10\Omega$
$X_C=\frac{1}{\omega C}=\frac{1}{100C}$

Therefore,
$\cos \theta =\frac{R}{\sqrt{(R{)}^2+{(\omega L-\frac{1}{\omega C})}^2}}=\frac{1}{\sqrt{2}}$
$\Rightarrow \frac{10}{\sqrt{(10{)}^2+{(10-\frac{1}{100C})}^2}}=\frac{1}{\sqrt{2}}$

Cross-multiplying and squaring both sides,
$200=100+{(10-\frac{1}{100C})}^2$
$\Rightarrow 100={(10-\frac{1}{100C})}^2$
$\Rightarrow \pm 10=10-\frac{1}{100C}$
$\Rightarrow -10=10-\frac{1}{100C}$
[for $+10,C\to \infty$ which is not possible]

$\Rightarrow 20=\frac{1}{100C}$
$\Rightarrow C=\frac{1}{2000}=500\mu F$