Question

A circuit draws $330\, W$ from a $110\, V$, $60 \,Hz\, AC$ 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\text{ }\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 \phi =0.6 $
Since, current lags the voltage thus, the circuit contains resistance and inductance.
$ \therefore $ $ \cos \phi =\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 \phi =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}$
$=0.000054\,F $
$=54\mu F $