The instantaneous current from an a.c. source is \( I=10\sin314t \). The frequency of the source would be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 50 Hz
An alternating current is written \( I = I_0\sin\omega t \), where \( \omega \) is the angular frequency, related to the frequency by \( \omega = 2\pi f \). Comparing with \( I = 10\sin314t \), \( \omega = 314\,\mathrm{rad\,s^{-1}} \).
\( \displaystyle f = \frac{\omega}{2\pi} = \frac{314}{2\times3.14}\,\mathrm{Hz} = 50\,\mathrm{Hz} \)
Shortcut: \( 314\approx100\pi \), so \( f = \frac{100\pi}{2\pi} = 50\,\mathrm{Hz} \) at once. This is the mains frequency used in India.