Question

If an alternative voltage $V=V_0\sin \omega t$ is applied across an inductance $L$, the current through the inductance will be :

• A. $I=I_0\sin \omega t$
• B. $I=I_0\sin (\omega t-\pi /2)$
• C. $I=I_0\sin (\omega t+\pi /2)$
• D. $I=I_0\sin (\omega t+\pi )$

Answer 1

Answer: (B)

$I=I_0\sin (\omega t-\pi /2)$

The voltage-current relation for a inductor with inductance $L$ is given by,

$V=L\frac{dI}{dt}$

or, $I=\frac{1}{L}\int Vdt$

So, here, $I=\frac{V_0}{\omega L}(-cos\omega t)=\frac{V_0}{\omega L}sin(\omega t-\frac{\pi }{2})$

So current sinusoid lags the voltage sinusoid by $\pi /2$ in phase.

Option B is currect.