Derive an expression for the average value of alternating current.

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Accepted Answer

>

> [!TLDR]
> Over a full cycle the mean AC is zero; over a half-cycle it is $I_{avg} = 2I_0/\pi \approx 0.637\,I_0$.
>
> Let the instantaneous alternating current be
> $$I = I_0\sin\omega t,$$
> where $I_0$ is the peak value and $\omega = 2\pi/T$ the angular frequency.
>
> **Average over a full cycle.** The average value is
> $$I_{avg} = \frac{\int_0^{T} I\,dt}{\int_0^{T} dt} = \frac{1}{T}\int_0^{T} I_0\sin\omega t\,dt.$$
> Since $\sin\omega t$ is positive for the first half-cycle and equally negative for the second, the integral over a full cycle is zero:
> $$I_{avg}\,(	ext{full cycle}) = 0.$$
> Hence the average value is defined over the positive half-cycle.
>
> **Average over the positive half-cycle** (from $t = 0$ to $t = T/2$):
> $$I_{avg} = \frac{\int_0^{T/2} I_0\sin\omega t\,dt}{\int_0^{T/2} dt} = \frac{I_0}{T/2}\int_0^{T/2}\sin\omega t\,dt.$$
>
> Evaluate the integral: …

Q.Derive an expression for the average value of alternating current.

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Q.Derive an expression for the average value of alternating current.

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