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

Answer: (B)

Ist Case : From formula
$R=\frac{V^2}{P}$
$=\frac{110\times 110}{330}$
$=\frac{110}{3}\Omega$
Since, current lags the voltage thus, the circuit contains resistance and inductance.
Power factor $\cos \varphi =0.6$
$\frac{R}{\sqrt{R^2+X_L^2}}=0.6$
$\Rightarrow R^2+X_L^2={\left(\frac{R}{0.6}\right)}^2$
$\Rightarrow X_L^2=\frac{R^2}{(0.6{)}^2}-R^2$
$\Rightarrow X_L^2=\frac{R^2\times 0.64}{0.36}$
$\therefore X_L=\frac{0.8R}{0.6}=\frac{4R}{3}$
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 $\frac{1}{2\pi fC}=\frac{440}{9}\Omega$
$\therefore C=\frac{9}{2\times 3.14\times 60\times 440}$
$=0.000054F=54\mu F$