Question

The value of inductance which should be connected in series with capacitance of 0.5F, resistance of 10 and an A.C. source of 50cps so that the power factor of the circuit is unity, will be

• A. 10.13 H
• B. 32.15 H
• C. 15.26 H
• D. 20.28 H

Answer 1

The correct option is D 20.28 H

Solution:-

Given:-

Z = $\sqrt{R^2+{(\omega L-\frac{1}{\omega C})}^2}$

$\omega =2\pi \left(50\right)$

$C=0.5\mu F$

for ac circuit

Power factor = $\cos \varphi =\frac{R}{Z}$

For power factor = unity

R=Z i.e.

purely resistive i.e.

$\omega L=\frac{1}{\omega C}$

L = $\frac{1}{{\omega }^{2}C}$

L = $\frac{{10}^6}{[2\pi \left(50\right){]}^2\times 0.5}$

L = 20.28H