Question: Q. 2. An $a c$ voltage $V=V_{0} \sin \omega t$ is applied to a pure inductor $L$. Obtain an expression for the current in the circuit. Prove that the average power supplied to an inductor over one complete cycle is zero.
A [SQP I 2017]
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Solution:
Ans. Expression for current 3
Proof 2
(i) If an alternating voltage $V=V_{0} \sin \omega t$ is applied across pure inductor of inductance $L$, then the magnitude of induced emf will be equal to applied voltage,
$$ E=L \frac{d I}{d t} $$
For the circuit, magnitude of induced emf
$=$ applied voltage
$$ \begin{aligned} L \frac{d I}{d t} & =V_{0} \sin \omega t \ d \mathrm{I} & =\frac{V_{0}}{L} \sin \omega t d t \end{aligned} $$
Integrating both the sides, we get
$$ \begin{aligned} & I=\frac{V_{0}}{L} \int \sin \omega t d t=\frac{V_{0}}{L}\left(\frac{-\cos \omega t}{\omega}\right) \ & I=-\frac{V_{0}}{\omega L} \cos \omega t=-\frac{V_{0}}{\omega L} \sin \left(\frac{\pi}{2}-\omega t\right) \ & I=\frac{V_{0}}{X_{L}} \sin \left(\omega t-\frac{\pi}{2}\right) \quad \text { if } X_{L}=\omega L \ & I=I_{0} \sin \left(\omega t-\frac{\pi}{2}\right) \end{aligned} $$
(ii) The average power supplied by the source over a complete cycle is
$$ P_{a v}=E_{r m s} I_{r m s} \cos \phi $$
When the circuit carries an ideal inductor, then the phase difference between the current and voltage is $\frac{\pi}{2}$
In case of pure inductive circuit,
$$ \begin{aligned} \phi & =\frac{\pi}{2} \ \text { But } \quad \cos \frac{\pi}{2} & =0 \end{aligned} $$
So power dissipated $=0$
Hence, when an ac source is connected to an ideal inductor, the average power supplied by the source over a complete cycle is zero.
[CBSE Marking Scheme 2017]
