Question

The figure represents the voltage applied across a pure inductor. The diagram which correctly represents the variation of current $i$ with time $t$ is given by

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

The given voltage function is a cosine function
$\therefore V=V_0\cos \omega t$

For a pure inductor

$i_{peak}=\frac{V_{peak}}{\omega L}$

But, current lags behind the voltage by $\frac{\pi }{2}$

$\therefore i=\frac{V_0}{\omega L}\cos (\omega t-\frac{\pi }{2})$

$\Rightarrow i=\frac{V_0}{\omega L}\sin \omega t$

Considering the $T$ as time period, we have

At, $t=\frac{T}{4}$, $\omega =\frac{2\pi }{T}$

$\Rightarrow i=\frac{V_0}{\omega L}\sin \frac{2\pi }{T}\times \frac{T}{4}=\frac{V_0}{\omega L}$

And at $t=\frac{T}{2}$, $\omega =\frac{2\pi }{T}$

$\Rightarrow i=\frac{V_0}{\omega L}\sin \frac{2\pi }{T}\times \frac{T}{2}=0$

Using the above points, we can conclude that, the graph $\left(C\right)$ is the correct option.