Question

A circuit draws 330W from a 110V, 60Hz AC line. The power factor is 0.6 and the current lags the voltage. The capacitance of a series capacitor that will result in a power factor of unity is equal to:

• A. $31\mu F$
• B. $54\mu F$
• C. $151\mu F$
• D. $201\mu F$

Answer 1

The correct option is B $54\mu F$

Resistance of circuit
$R=\frac{V^2}{P}=\frac{110\times 110}{330}=\frac{110}{3}\Omega$
Current lags the voltage, \therefore it is a R - L circuit
$cos\varphi =0.6=\frac{R}{\sqrt{R^2+X_L^2}}$
Squaring both sides
$0.36=\frac{R^2}{R^2+X_L^2}$
$0.36R^2+0.36X_L^2=R^2$
$0.36X_L^2=R^2-0.36R^2$
$0.36X_L^2=0.64R^2$
$X_L=\frac{4R}{3}$ ...(1)
Now addition of capacitor will result in power factor to unity
$cos\varphi =\frac{R}{Z}cos\varphi =1$ (given)
R = Z
It is condition of resonance
$X_L=X_C\frac{4R}{3}=\frac{1}{2\pi fc}\because \frac{X}{L}=\frac{4R}{3}\frac{4}{3}\times \frac{110}{3}=\frac{1}{2\times 3.14\times 60C}C=54\mu F$