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 \, \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 $\cos\, \phi=0.6$
$ \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} $
$\Rightarrow X_{L}^{2} =\frac{R^{2} \times 0.64}{0.36}$
$\therefore X_{L}=\frac{0.8 R}{0.6}=\frac{4 R}{3}$
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} $
$\therefore X_{C}=\frac{4 R}{3}=\frac{4}{3} \times \frac{110}{3}$
$=\frac{440}{9}\, \Omega$
[from Eq. (i)]
or $\frac{1}{2 \pi f C}=\frac{440}{9} \Omega$
$\therefore C =\frac{9}{2 \times 3.14 \times 60 \times 440} $
$=0.000054\, F =54 \,\mu F$