Question
State the rms value of an alternating current? Write the relation between the rms value and peak value of an alternating current that varies with time.
✓
Answer
i. The constant current which produces the same amount of heat in a given resistance in a given time as is produced by an alternating current, when flowing through the same resistance for the same time is called the root mean square ( $r m s$ ) value of current.
It is given by, $i _{ rms }=\frac{ I _0}{\sqrt{2}}=0.707 i _0$.
ii. The relation between the rms value and peak value of alternating current is given by,
$ i _{\text {rms }}^2=\frac{\int_0^{2 \pi} i ^2 d \theta}{2 \pi}=\frac{1}{2 \pi} \int_0^{2 \pi} i _0^2 \sin ^2 \theta d \theta$
$=\frac{ i _0^2}{2 \pi} \int_0^{2 \pi} \frac{(1-\cos 2 \theta)}{2} d \theta \ldots \ldots \ldots \ldots . .(\because \cos 2 \theta=1-$
$\left.\sin ^2 \theta\right)$
$=\frac{ i _0^2}{2 \times 2 \pi}\left[\left(\theta-\frac{\sin 2 \theta}{2}\right)\right]_0^{2 \pi}$
$=\frac{ i _0^2}{2}$
$\therefore i _{ rms }=\frac{ i _0}{\sqrt{2}}=0.707 i _0 $
It is given by, $i _{ rms }=\frac{ I _0}{\sqrt{2}}=0.707 i _0$.
ii. The relation between the rms value and peak value of alternating current is given by,
$ i _{\text {rms }}^2=\frac{\int_0^{2 \pi} i ^2 d \theta}{2 \pi}=\frac{1}{2 \pi} \int_0^{2 \pi} i _0^2 \sin ^2 \theta d \theta$
$=\frac{ i _0^2}{2 \pi} \int_0^{2 \pi} \frac{(1-\cos 2 \theta)}{2} d \theta \ldots \ldots \ldots \ldots . .(\because \cos 2 \theta=1-$
$\left.\sin ^2 \theta\right)$
$=\frac{ i _0^2}{2 \times 2 \pi}\left[\left(\theta-\frac{\sin 2 \theta}{2}\right)\right]_0^{2 \pi}$
$=\frac{ i _0^2}{2}$
$\therefore i _{ rms }=\frac{ i _0}{\sqrt{2}}=0.707 i _0 $
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