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Q.
In an AC circuit, the rms value of the current, $I_{rms}$, is related to the peak current $I_0$ as
Alternating Current
Solution:
The alternating current is represented as
$I=I_{0} \sin (\omega t)$ ................(1)
where, $I_{0}$ is the peak value of current
Now, formula for a continuous function (or waveform) $f ( t )$ defined over the interval $t _{1} \leq t \leq t _{2}$ is
$f _{ rms }=\sqrt{\frac{1}{ t _{2}- t _{1}} \int\limits_{ t _{1}}^{ t _{2}} f ( t )^{2} d t }$
Therefore,
$I _{ rms }=\sqrt{\frac{1}{ t _{2}- t _{1}} \int\limits_{ t _{1}}^{ t _{2}}\left( I _{0} \sin (\omega t )\right)^{2} d t }$
since, $I_{0}$ is positive, we can write
$I _{ rms }= I _{0} \sqrt{\frac{1}{ t _{2}- t _{1}} \int\limits_{ t _{1}}^{ t _{2}}\left(\sin ^{2}(\omega t )\right) dt }$
Using a trigonometric identity to eliminate squaring of trigonometric function
$I _{ rms }= I _{0} \sqrt{\frac{1}{ t _{2}- t _{1}} \int\limits_{ t _{1}}^{ t _{2}} \frac{1}{2}(1-\cos (2 \omega t )) dt }$
$I _{ rms }= I _{0} \sqrt{\frac{1}{ t _{2}- t _{1}}\left[\frac{ t }{2}-\frac{\sin (2 \omega t )}{4 \omega}\right]_{ t _{1}}^{ t _{2}}}$
but since the interval is a whole number of complete cycles (as per definition of RMS), the $\$ \$ \backslash \sin \$ \$$ terms will cancel out, leaving
$I _{ rms }= I _{0} \sqrt{\frac{1}{ t _{2}- t _{1}}\left[\frac{ t }{2}\right]_{ t _{1}}^{ t _{2}}}$
$I _{ rms }= I _{0} \sqrt{\frac{1}{ t _{2}- t _{1}}\left[\frac{ t }{2}\right]_{ t _{1}}^{ t _{2}}}$
$I _{ rms }= I _{0} \sqrt{\frac{1}{ t _{2}- t _{1}} \frac{ t _{2}- t _{1}}{2}}$
$I _{ rms }=\frac{ I _{0}}{\sqrt{2}}$