Question

A pure capacitor is connected to an AC circuit. The power factor of the circuit will be
(A) $ 1 $
(B) Infinity
(C) Zero
(D) $ 0.5 $

Answer 1

Hint : To solve this question, we have to use the formula for the power factor of an AC circuit. For that we need to find out the phase, which has to be substituted in the formula to get the final answer.

Formula used: The formula which is used to solve this question is given by
 $ \varphi = {\tan ^{ - 1}}\left( {\dfrac{{{X_L} - {X_C}}}{R}} \right) $
 $ PF = \cos \varphi $
Here $ \varphi $ is the phase of the current with respect to voltage, $ {X_L} $ is the inductive reactance, $ {X_C} $ is the capacitive reactance, $ R $ is the resistance, and $ PF $ is the power factor.

Complete step by step answer
An AC circuit is made up of three elements. These three elements are; the resistor, the inductor, and the capacitor. When any of these is present in an AC circuit individually, then the prefix pure is used behind that element. Therefore a pure capacitor means that we have a capacitance only in our AC circuit. Now, we know that the phase of an AC circuit is given by
 $ \varphi = {\tan ^{ - 1}}\left( {\dfrac{{{X_L} - {X_C}}}{R}} \right) $
We know that $ {X_L} = \omega L $ , $ {X_C} = \dfrac{1}{{\omega C}} $ . Substituting these above, we get
 $ \varphi = {\tan ^{ - 1}}\left( {\dfrac{{\omega L - \dfrac{1}{{\omega C}}}}{R}} \right) $
 $ \varphi = {\tan ^{ - 1}}\left( {\dfrac{{{\omega ^2}LC - 1}}{{\omega CR}}} \right) $
As the circuit is pure capacitive, so we have $ L = R = 0 $ . Substituting these above, we get
 $ \varphi = {\tan ^{ - 1}}\left( {\dfrac{{{\omega ^2}\left( 0 \right)C - 1}}{{\omega \left( 0 \right)\left( 0 \right)}}} \right) $
 $ \Rightarrow \varphi = {\tan ^{ - 1}}\left( { - \infty } \right) $
We know that $ {\tan ^{ - 1}}\left( { - \infty } \right) = - \dfrac{\pi }{2} $ . Therefore the phase is
 $ \varphi = - \dfrac{\pi }{2} $ (1)
Now, the power factor of the AC circuit is equal to the cosine of the phase, that is,
 $ PF = \cos \varphi $
From (1)
 $ PF = \cos \left( { - \dfrac{\pi }{2}} \right) $
 $ \Rightarrow PF = 0 $
Thus, the power factor of the given AC circuit is equal to zero.
Hence, the correct answer is option C.

Note
We could also attempt this question by drawing the phase diagram of this circuit. From that we could directly get the phase of the circuit from which the power factor of the circuit would be evaluated.