Question

The instantaneous values of current (in ampere) and potential (in volt) in an AC circuit are $i=4\sin \omega t$and $V=100\cos \left( \omega t-\dfrac{\pi }{3} \right)$ respectively. The power factor of the circuit is
a) $\dfrac{1}{2}$
b) 1
c) $\dfrac{1}{\sqrt{2}}$
d) $\dfrac{\sqrt{3}}{2}$

Answer 1

Hint: In the above question we are given the Current and potential difference in an Ac circuit and we have been asked to find out the power factor. The cosine of phase difference between the source voltage and current is the electrical power factor. The phase difference can simply be evaluated by looking at the sine or cosine terms of the current and voltage.

Complete step-by-step Solution:
The electrical power factor represents the useful work obtained from the total power applied. When the electrical circuit is made up of only resistors, the electrical energy will always be dissipated. But when a circuit contains inductors, capacitors, and resistors, some energy will also be stored. The capacitor will store electrical energy in the form of charge while the inductor in the form of magnetic energy. As the circuit consists of a resistor, capacitor, and inductor, some phase difference will be produced between the source voltage and current.
From the given problem,
$I=4\sin \omega t$
And $V=100\cos \left( \omega t-\dfrac{\pi }{3} \right)$

To evaluate the phase difference, α, I and V should have phase in the same trigonometric form i.e. sine or cosine. Hence, current, I in cosine form will be:
$V=4\cos \left( \dfrac{\pi }{2}-\omega t \right)$

Finally, we can evaluate the phase difference, α:
$\dfrac{\pi }{2}-\omega t-\omega t+\dfrac{\pi }{3}=\dfrac{5\pi }{6}$

The power factor$=\cos \alpha =\cos \dfrac{5\pi }{6}=\dfrac{1}{2}$
P.F. = $\dfrac{1}{2}$

Hence, we can conclude that option (a) is the correct answer.

Note:
One should always remember to convert the phase value of both current and voltage in the same trigonometric form that is sine or cosine. Avoiding this transition is the most common mistake while attempting questions like these. Also this power factor will always be between -1 to 1 as $\left( -1<\cos \alpha <1 \right)$.