Concept:The mean square value of a sinusoidal alternating current is the average of the squared instantaneous current over a complete cycle.
Formula / Reaction:$$\langle i^2 \rangle = \frac{1}{T} \int_{0}^{T} I_0^2 \sin^2(\omega t)\, dt = \frac{I_0^2}{2}$$
Solution:- The average value of \(\sin^2(\omega t)\) over one full cycle is \(\frac{1}{2}\).
- Therefore, the mean square current is \(i_{\text{ms}} = \langle i^2 \rangle = \frac{i_0^2}{2}\).
- Taking the square root gives the root-mean-square value: \(I_{\text{rms}} = \sqrt{\frac{i_0^2}{2}} = \frac{i_0}{\sqrt{2}}\).
Why other options are incorrect:- Opt_B: Divides by \(\sqrt{2}\), which confuses the mean square with the root mean square.
- Opt_C: Multiplies by 2 instead of dividing by 2.
- Opt_D: Multiplies by \(\sqrt{2}\).
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