Question

A sinusoidal voltage $V(t) = 100\,\sin (500 t)$ is applied across a pure inductance of $L = 0.02 H$. The current through the coil is :

• A. 10 cos (500t)
• B. —10 cos (500t)
• C. 10 sin (500t)
• D. -10 sin (500t)

Answer 1

Answer: (B)

In a pure inductive circuit current always lags behind the emf by $\frac{\pi}{2}.$
If $v\left(t\right)=v_{0}\,\sin\,\omega t$
then $I=I_{0}\,\sin\left(\omega t-\frac{\pi}{2}\right)$
Now, given $v\left(t\right) = 100 \,sin \left(500\,t\right)$
and $I_{0}=\frac{E_{0}}{\omega L}=\frac{100}{500\times0.02}\left[\because L=0.02H\right]$
$I_{0}=10\,\sin\left(500t-\frac{\pi}{2}\right)$
$I_{0}=-10\,\cos\left(500t\right)$