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 series capacitor that will result in a power factor of unity equal to:
(A) $31\,\mu F$
(B) $54\,\mu F$
(C) $151\,\mu F$
(D) $201\,\mu F$

Answer 1

Hint
First by using the power formula the resistance can be determined, and the by using the power factor and the resistance, then the inductive reactance can be determined but the relation of the power factor, resistance and inductive reactance. By using the second condition of the power face factor given in the question, the capacitive reactance is determined. And by using the relation of capacitive reactance, frequency and capacitance, then the final capacitance is determined.
The power is given by,
$\Rightarrow P = \dfrac{{{V^2}}}{R}$
Where, $P$ is the power, $V$ is the voltage and $R$ is the resistance.
The relation of the power factor, resistance and inductive reactance is given by,
$\Rightarrow \cos \phi = \dfrac{R}{{\sqrt {{R^2} + X_L^2} }}$
Where, $\cos \phi$ is the power factor, $R$ is the resistance, ${X_L}$ is the inductive reactance.
The relation of capacitive reactance, frequency and capacitance is given by,
$\Rightarrow {X_C} = \dfrac{1}{{2\pi fC}}$
Where, ${X_C}$ is the capacitive reactance, $f$ is the frequency and $C$ is the capacitance.

Complete step by step answer
Given that, The power of the circuit is, $P = 330\,W$
The voltage of the circuit is, $V = 110\,V$
The frequency of the circuit is, $f = 60Hz$
The first power factor of the circuit is, $\cos \phi = 0.6$
The second power factor of the circuit is, $\phi = 1$
Now, The power is given by,
$\Rightarrow P = \dfrac{{{V^2}}}{R}\,................\left( 1 \right)$
By substituting the voltage and power in the above equation, then
$\Rightarrow 330 = \dfrac{{{{110}^2}}}{R}$
By rearranging the terms, then
$\Rightarrow R = \dfrac{{110 \times 110}}{{330}}$
By cancelling the terms, then
$\Rightarrow R = \dfrac{{110}}{3}\,\Omega$.
Now, The relation of the power factor, resistance and inductive reactance is given by,
$\Rightarrow \cos \phi = \dfrac{R}{{\sqrt {{R^2} + X_L^2} }}\,.....................\left( 2 \right)$
By substituting the power factor in the above equation (2), then
$\Rightarrow 0.6 = \dfrac{R}{{\sqrt {{R^2} + X_L^2} }}$
By rearranging the terms, then
$\Rightarrow \sqrt {{R^2} + X_L^2} = \dfrac{R}{{0.6}}$
Squaring on both sides, then
$\Rightarrow {R^2} + X_L^2 = {\left( {\dfrac{R}{{0.6}}} \right)^2}$
By rearranging the terms, then
$\Rightarrow X_L^2 = \dfrac{{{R^2}}}{{{{0.6}^2}}} - {R^2}$
By taking the common term, then
$\Rightarrow X_L^2 = {R^2}\left( {\dfrac{1}{{{{0.6}^2}}} - 1} \right)$
By squaring the terms, then
$\Rightarrow X_L^2 = {R^2}\left( {\dfrac{1}{{0.36}} - 1} \right)$
By cross multiplying, then
$\Rightarrow X_L^2 = {R^2}\left( {\dfrac{{1 - 0.36}}{{0.36}}} \right)$
On further simplification, then
$\Rightarrow X_L^2 = \dfrac{{0.64{R^2}}}{{0.36}}$
By taking square root on both sides, then
$\Rightarrow {X_L} = \dfrac{{0.8R}}{{0.6}}$
The above equation is also written as,
$\Rightarrow {X_L} = \dfrac{{8R}}{6} = \dfrac{{4R}}{3}$
The second power factor value is equal to unity, so that the inductive reactance is equal to the capacitive reactance. Then
$\Rightarrow {X_L} = {X_C}$
The capacitive reactance is equal to,
$\Rightarrow {X_C} = \dfrac{{4R}}{3}$
Now, The relation of capacitive reactance, frequency and capacitance is given by,
$\Rightarrow {X_C} = \dfrac{1}{{2\pi fC}}\,................\left( 3 \right)$
Now by equating the ${X_C}$, then
$\Rightarrow \dfrac{{4R}}{3} = \dfrac{1}{{2\pi fC}}$
By keeping the capacitance in one side, then
$\Rightarrow C = \dfrac{3}{{2\pi f \times 4R}}$
By substituting the values, then
$\Rightarrow C = \dfrac{3}{{2 \times \pi \times 60 \times 4 \times \dfrac{{110}}{3}}}$
By rearranging the terms, then
$\Rightarrow C = \dfrac{9}{{2 \times \pi \times 60 \times 4 \times 110}}$
By multiplying the terms, then
$\Rightarrow C = \dfrac{9}{{165876}}$
On dividing the terms, then
$\Rightarrow C = 5.4 \times {10^{ - 5}}\,F = 54 \mu F$.
Option (B) is correct.

Note
When the power factor is equal to the unity, the circuit is purely resistive. That means the circuit contains only resistance. This condition is known as the resonance of the circuit, where the inductive reactance and the capacitive reactance are equal.