An alternating current is given by i = 2 sin | vedclass

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Question

(A) The given equation is $i = 2 \sin \omega t + 6 \cos \omega t$.This is of the form $i = a \sin \omega t + b \cos \omega t$,where $a = 2$ and $b = 6

Vedclass · Accepted Answer

$2 \sqrt{5}$

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(A) The given equation is $i = 2 \sin \omega t + 6 \cos \omega t$.This is of the form $i = a \sin \omega t + b \cos \omega t$,where $a = 2$ and $b = 6$.The peak current $i_0$ is given by $i_0 = \sqrt{a^2 + b^2}$.$i_0 = \sqrt{2^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40} = 2 \sqrt{10} \ A$.The $R.M.S.$ current $i_{rms}$ is related to the peak current $i_0$ by the formula $i_{rms} = \frac{i_0}{\sqrt{2}}$.$i_{rms} = \frac{2 \sqrt{10}}{\sqrt{2}} = 2 \sqrt{5} \ A$.

An alternating current is given by $i = 2 \sin \omega t + 6 \cos \omega t$. The $R.M.S.$ current in amperes is,

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