Question

A circuit draws a power of 550W from a source of 220V, 50Hz. The power factor of the circuit is $0.8$ and the current lags in phase behind the potential difference. To make the power factor of the circuit as $1.0$, what capacitance will have to be connected to it?
A) $\dfrac{1}{{42\pi }} \times {10^{ - 2}}F$
B) $\dfrac{1}{{41\pi }} \times {10^{ - 2}}F$
C) $\dfrac{1}{{5\pi }} \times {10^{ - 2}}F$
D) $\dfrac{1}{{84\pi }} \times {10^{ - 2}}F$

Answer 1

Hint: Resistance is the measure of the opposition offered to the flow of current in a circuit. In an A.C. circuit the amplitudes constantly change over time. Also the power factor in an a.c. circuit determines how effectively the power is utilised by the load.

Complete step by step answer:
Given that power is $P = 550Watt$
${V_{rms}} = 220V$
$f = 50Hz$
$\cos {\Phi _1} = 0.8$
It is said that the power factor of the circuit should be 1. That is $\cos {\Phi _2} = 1$
The real power in an a.c. circuit defines the power consumed by the resistive part of the circuit. This power is actually transferred to the load. It is also known as true or active power. The formula for the real power for an a.c. circuit is written as
$P = {V_{rms}}.{I_{rms}}.\cos \Phi $
$\Rightarrow P = \dfrac{{V_{rms}^2}}{Z}\cos \Phi $
$\Rightarrow Z = \dfrac{{V_{rms}^2}}{P}\cos {\Phi _1}$
Substituting all the given values,
$\Rightarrow Z = \dfrac{{{{(220)}^2}}}{{550}} \times 0.8$
$\Rightarrow Z = 70.4\Omega $
The average power in an a.c. circuit is determined in terms of rms voltage and current. Given that the current lags in phase difference behind the potential. So the power factor is given by
$\cos {\Phi _1} = \dfrac{R}{Z}$
$R = Z\cos {\Phi _1}$
$\Rightarrow R = (70.4)(0.8)$
$\Rightarrow R = 56.32\Omega $

Inductors provide inductance to a circuit and store energy temporarily in the form of magnetic field. The total impedance of the circuit is given by
$Z = \sqrt {{L^2}{\omega ^2} + {R^2}} $
Where Z is the impedance
R is the reactance
L is the inductance
$\Rightarrow 70.4 = \sqrt {{{(\omega L)}^2} + {{(56.32)}^2}} $
$\Rightarrow \omega L = 42.2\Omega $

But according to the given condition in the question, the power factor of the circuit should be 1. It is possible when
${X_L} = {X_C}$
Where ${X_L}$ is the inductive reactance $ = \omega L$
And ${X_C}$ is the capacitive reactance$ = \dfrac{1}{{\omega C}}$
Therefore,
$\omega L = \dfrac{1}{{\omega C}}$
$\Rightarrow C = \dfrac{1}{{\omega (\omega L)}}$

When the capacitor is connected in the circuit, then
$\Rightarrow C = \dfrac{1}{{2\pi f(42.2)}}$
$\Rightarrow C = \dfrac{1}{{2\pi \times 50(42.2)}}$
$\Rightarrow C = \dfrac{1}{{42\pi }} \times {10^{ - 2}}F$

Therefore Option A is the correct answer.

Note:
It is to be noted that the terms impedance, reactance and resistance are different. Resistance is the opposition to flow of current, but reactance is the opposition to the change in flow of current due to inductance and capacitance only. Impedance is the sum of both resistance and reactance.