Question

The average power dissipated in a pure inductor is

A. $\dfrac{1}{2}VI$

B. $V{I^2}$

C. $\dfrac{{V{I^2}}}{4}$

D. Zero

Answer 1

Hint: In an A.C. circuit of an inductor the current lags behind the applied voltage by a phase of $\dfrac{\pi }{2}$ radian. So first establish the equation of Work done in terms of instantaneous voltage and current. Then later integrate and take its ratio with respect to time T. This will give the equation for average power dissipated.

Complete step by step answer:

An ideal inductor or a pure inductor is defined as an inductor which has zero or no internal resistance within its coil. But this is possible only theoretically, because in real life a lot of energy is lost due to heating of the coil from its own internal resistance.Average power dissipated is the amount of energy lost due to unwanted means in an electronic device or circuit. This waste occurs because energy gets converted to another form which is an unnecessary by-product, for example heat, sound etc.

When Alternating current is applied to an ideal inductor, current lags behind the voltage in phase by $\dfrac{\pi }{2}$ radian. So we can write the instantaneous values of voltage and current as follows:

$V = {V_ \circ }\sin \omega t$

and

$I = {I_ \circ }\sin \left( {\omega t - \dfrac{\pi }{2}} \right) $

$\Rightarrow I = - {I_ \circ }\sin \left( {\dfrac{\pi }{2} - \omega t} \right) $

$\Rightarrow I = - {I_ \circ }\cos \omega t $

Work done in small time $dt$ is

$dW = Pdt = - {V_ \circ }{I_ \circ }\sin \omega t\cos \omega tdt $

$\Rightarrow dW = - \dfrac{{{V_ \circ }{I_ \circ }}}{2}\sin 2\omega tdt $

The average power dissipated per cycle in the inductor is:

${P_{av}} = \dfrac{W}{T} = \dfrac{1}{T}\int\limits_0^T {dW} $

$\Rightarrow {P_{av}} = - \dfrac{{{V_ \circ }{I_ \circ }}}{{2T}}\int\limits_0^T {\sin 2\omega tdt} $

$\Rightarrow {P_{av}}= + \dfrac{{{V_ \circ }{I_ \circ }}}{{2T}}\left[ {\dfrac{{\cos 2\omega t}}{{2\omega }}} \right]_0^T $

$\Rightarrow {P_{av}}= \dfrac{{{V_ \circ }{I_ \circ }}}{{4T\omega }}\left[ {\cos \dfrac{{4\pi }}{T}t} \right]_0^T $

$\Rightarrow {P_{av}}= \dfrac{{{V_ \circ }{I_ \circ }}}{{4T\omega }}\left[ {\cos 4\pi - \cos 0} \right] $

$\Rightarrow {P_{av}}= \dfrac{{{V_ \circ }{I_ \circ }}}{{4T\omega }}\left[ {1 - 1} \right] $

$\therefore {P_{av}} = 0 $

Thus, the average energy dissipated per cycle in an ideal inductor is zero. Therefore, the correct option is (D).

Note: Another way to solve this problem is that: we know in an A.C. circuit the current lags behind the applied voltage by $\dfrac{\pi }{2}$ radian i.e. by ${90^ \circ }$. In other words the phase difference between the voltage and current is ${90^ \circ }$

The average power dissipated by an A.C. circuit is given by:

${P_{av}} = \dfrac{{{V_ \circ }{I_ \circ }}}{2}\cos \phi $

Therefore for a pure inductive circuit it will be:

${P_{av}} = \dfrac{{{V_ \circ }{I_ \circ }}}{2}\cos {90^ \circ }$

But cos${90^ \circ }$= 0

$ \Rightarrow {P_{av}} = 0$