An inductor of inductance \( 100 \) mH is connected to an ac source of peak emf \( 211 \) V and frequency \( 50 \) Hz. When the current is at its peak value, the instantaneous voltage of the source will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → zero
In a pure inductor the current lags the voltage by \( \pi/2 \).
\( v = v_{m}\sin\omega t \)
\( i = i_{m}\sin(\omega t - \pi/2) = -i_{m}\cos\omega t \)
The current has its peak value when \( \cos\omega t = \pm1 \), that is when \( \sin\omega t = 0 \). At that instant
\( v = v_{m}\sin\omega t = 0 \)
The inductance, the peak emf and the frequency are not needed; the result follows from the phase difference alone.