Question

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

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

Answer 1

The correct option is C $500\mu F$

Given that
$V=2\sin 100t$
Comparing this with general form
$V=V_0\sin \omega t$
we get
$V_0=2\omega =100\text{rad/s}$
$R=10\Omega L=0.1\text{H}$
Power factor $\cos \varphi =\frac{1}{\sqrt{2}}$
We know that
$\cos \varphi =\frac{R}{Z}=\frac{R}{\sqrt{R^2+(X_L-X_C{)}^2}}$ ......(i)
So
$X_L=\omega L=100\times 0.1=10\Omega X_C=\frac{1}{\omega C}=\frac{1}{100C}$
Putting all value in eq(i) and squaring both sides we get
$\frac{1}{2}=\frac{{10}^2}{{10}^2+(10-\frac{1}{100C}{)}^2}$
${10}^2+(10-\frac{1}{100C}{)}^2=200$
$(10-\frac{1}{100C}{)}^2={10}^2$
$10-\frac{1}{100C}=\pm 10$ (cannot be +10 because that would give indefinite C)
$10-\frac{1}{100C}=-10$
$C=500\mu F$