Q 12-07-015NEETJEE MainMedium
The average value of an alternating current $I = I_0\sin\omega t$ over half a cycle (from $0$ to $\pi/\omega$) is
Answer: (B) $\dfrac{2I_0}{\pi}$
Concept. Average over half a cycle $= \dfrac{1}{T/2}\displaystyle\int_0^{T/2} I\,dt$.
Step 1. Integrate.
$$\bar I = \frac{\omega}{\pi}\int_0^{\pi/\omega}I_0\sin\omega t\,dt = \frac{\omega}{\pi} \cdot \frac{2I_0}{\omega} = \frac{2I_0}{\pi}$$
(Over a full cycle the average is zero.)
Answer. $\dfrac{2I_0}{\pi}$ (Option B).
Solution by Sreeraj P, M.Sc Physics