Question

A circuit draws $330W$ from a $110V$, $60HzAC$ 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. $\text{54 }\mu F$
• C. $\text{151 }\mu F$
• D. $\text{201 }\mu F$

Answer 1

Answer: (B)

Resistance of circuit,
$R=\frac{V^2}{P}=\frac{110\times 110}{330}=\frac{110}{3}\Omega$
1st case: Power factor $\cos \varphi =0.6$
Since, current lags the voltage thus, the circuit contains resistance and inductance.
$\therefore$ $\cos \varphi =\frac{R}{\sqrt{R^2+X_L^2}}=0.6$
Or $R^2+X_L^2={\left(\frac{R}{0.6}\right)}^2$
Or $X_L^2=\frac{R^2}{{(0.6)}^2}-R^2$
Or $X_L^2=\frac{R^2\times 0.64}{0.36}$
$\therefore$ $X_L=\frac{0.8R}{0.6}=\frac{4R}{3}...(i)$
II nd case : Now $\cos \varphi =1$ (given)
Therefore, circuit is purely resistive, i.e., it contains only resistance. This is the condition of resonance in which
$X_L=X_c$
$\therefore$ $X_C=\frac{4R}{3}=\frac{4}{3}\times \frac{110}{3}=\frac{440}{9}\Omega$ [from Eq. (i)]
Or $\frac{1}{2\pi fc}=\frac{440}{9}\Omega$
$\therefore$ $C=\frac{9}{2\times 3.14\times 60\times 440}$
$=0.000054F$
$=54\mu F$