In a circuit, current varies with time as&nbs | vedclass

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Question

$\mathrm{I}^{2}=4 \mathrm{t}$

$<\mathrm{I}^{2}>=\int_{2}^{4} \mathrm{I}^{2} \mathrm{dt}=\int_{2}^{4} 4 \mathrm{t} \mathrm{dt}=\frac{4\left(4^

Vedclass · Accepted Answer

$2\sqrt 3 A$

Vedclass · Answer

$\mathrm{I}^{2}=4 \mathrm{t}$ $<\mathrm{I}^{2}>=\int_{2}^{4} \mathrm{I}^{2} \mathrm{dt}=\int_{2}^{4} 4 \mathrm{t} \mathrm{dt}=\frac{4\left(4^{2}-2^{2} ight)}{2 imes 2}=12$ $\mathrm{I}_{\mathrm{rms}}=\sqrt{<\mathrm{I}^{2}>}=\sqrt{12}=2 \sqrt{3} \mathrm{\,A}$

In a circuit, current varies with time as $i = 2\sqrt t $ . Root mean square value of current for interval $t = 2\,s$ to $t = 4\,s$ is

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