Derive an expression for root-mean square value of an alternating current.

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Accepted Answer

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> [!TLDR]
> The rms (root-mean-square) value is found by squaring the instantaneous current, averaging over one cycle, then taking the square root — it gives $i_{rms}=i_0/\sqrt2$ for a sinusoidal current.
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> Let the alternating current be $i = i_0\sin\omega t$, where $i_0$ is the peak (maximum) value and $T=2\pi/\omega$ is the time period.
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> **Step 1 — Mean of $i^2$ over one cycle:**
> $$\langle i^2\rangle = \frac{1}{T}\int_0^T i_0^2\sin^2\omega t\,dt$$
> Using $\sin^2\omega t = \dfrac{1-\cos 2\omega t}{2}$,
> $$\langle i^2\rangle = \frac{i_0^2}{T}\int_0^T \frac{1-\cos 2\omega t}{2}\,dt = \frac{i_0^2}{2T}\left[T - 0\right] = \frac{i_0^2}{2}$$
> (the integral of $\cos 2\omega t$ over a full number of its own cycles is zero).
>
> **Step 2 — Root of the mean square:**
> $$i_{rms} = \sqrt{\langle i^2\rangle} = \sqrt{\frac{i_0^2}{2}} = \frac{i_0}{\sqrt2} \approx 0.707\,i_0$$
> …

Q.Derive an expression for root-mean square value of an alternating current.

Concept understanding — RMS and Peak Value

Why We Need a New Measure

The Intuition: Squaring Fixes the Sign Problem

The Physical Meaning

Peak Value

Q.Derive an expression for root-mean square value of an alternating current.

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