Question

Which sine-form expression gives the current through the pure inductor?

पर्याय

• \[i=i_m\sin\omega t\]

• \[i=i_m\cos\left(\omega t-\frac{\pi}{2}\right)\]

• \[i=i_m\sin\left(\omega t-\frac{\pi}{2}\right)\]

• \[i=i_m\sin\left(\omega t+\frac{\pi}{2}\right)\]

Answer 1

The current is \[i=i_m\sin\left(\omega t-\frac{\pi}{2}\right)\]. Its phase term differs from the applied voltage by \[-\frac{\pi}{2}\].