Question

The average power dissipation in pure inductance in AC circuit is
(A) $ 1/2L{i^2} $
(B) $ 2L{i^2} $
(C) $ \dfrac{{L{i^2}}}{4} $
(D) Zero

Answer 1

Hint: In a purely inductive circuit, the current lags behind the voltage by 90 degrees. The average power can be given as the product of the root mean square voltage and the root mean square current and the power factor.

Formula used: In this solution we will be using the following formula;
 $\Rightarrow {P_{avg}} = {v_{rms}}{i_{rms}}\cos \theta $ where $ {P_{avg}} $ is the average power dissipated, $ {v_{rms}} $ is the root means square voltage and $ {i_{rms}} $ is the root means square current.
 $ \cos \theta $ is known as the power factor and $ \theta $ is the phase shift between the instantaneous current and the instantaneous voltage

Complete step by step answer
In general, when only an inductor is connected to a circuit, the current passing through it lags behind the voltage by exactly 90 degrees.
Also, generally, in calculating the power dissipated in an ac circuit, we use the instantaneous current and the instantaneous voltage, and that is called the instantaneous power. But the average power can be given as
 $\Rightarrow {P_{avg}} = {v_{rms}}{i_{rms}}\cos \theta $ where $ {v_{rms}} $ is the root means square voltage and $ {i_{rms}} $ is the root means square current. $ \cos \theta $ is known as the power factor and $ \theta $ is the phase shift between the instantaneous current and the instantaneous voltage.
Now, since the phase shift between the current and voltage is 90 degrees, we have that
 $\Rightarrow {P_{avg}} = {v_{rms}}{i_{rms}}\cos 90^\circ $
 $ \Rightarrow {P_{avg}} = 0 $

Hence the correct answer is option D.

Note
Alternatively, without the knowledge of the 90 degree phase shift between instantaneous current and voltage, the power factor can be given as
 $ \cos \theta = \dfrac{R}{Z} $ where $ R $ is the resistance of the circuit and $ Z $ is the impedance.
Now, since it is purely inductive $ R = 0 $ and hence
 $ \cos \theta = 0 $
Also, note that although it is assumed in the question, a purely inductive circuit is nearly impossible to create (unless we’re dealing with superconductors) since the inductor coil itself will have a resistance.