The r.m.s. value of current for a variable cu | vedclass

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(C) The given current is $i = i_1 \cos \omega t + i_2 \sin \omega t$.The $r.m.s.$ value of current is defined as $i_{rms} = \sqrt{\frac{1}{T} \int_0^T

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}} (i_1^2 + i_2^2)^{1/2}$

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(C) The given current is $i = i_1 \cos \omega t + i_2 \sin \omega t$.The $r.m.s.$ value of current is defined as $i_{rms} = \sqrt{\frac{1}{T} \int_0^T i^2 dt}$.First,calculate $i^2 = (i_1 \cos \omega t + i_2 \sin \omega t)^2 = i_1^2 \cos^2 \omega t + i_2^2 \sin^2 \omega t + 2 i_1 i_2 \sin \omega t \cos \omega t$.Integrating over a full time period $T = \frac{2\pi}{\omega}$:$\int_0^T \cos^2 \omega t dt = \frac{T}{2}$,$\int_0^T \sin^2 \omega t dt = \frac{T}{2}$,and $\int_0^T \sin \omega t \cos \omega t dt = 0$.Thus,the mean square value is $i_{rms}^2 = \frac{1}{T} [i_1^2 (\frac{T}{2}) + i_2^2 (\frac{T}{2}) + 0] = \frac{i_1^2 + i_2^2}{2}$.Therefore,$i_{rms} = \sqrt{\frac{i_1^2 + i_2^2}{2}} = \frac{1}{\sqrt{2}} (i_1^2 + i_2^2)^{1/2}$.

The $r.m.s.$ value of current for a variable current $i = i_1 \cos \omega t + i_2 \sin \omega t$ is:

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