JEE MAIN - Physics (2023 - 30th January Evening Shift - No. 8)
In the given circuit, rms value of current $\left(I_{\mathrm{rms}}\right)$ through the resistor $R$ is:
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$ 2 \sqrt{2} \mathrm{~A}$
$2 \mathrm{~A}$
$\frac{1}{2} \mathrm{~A}$
$20 \mathrm{~A}$
Explanation
$${I_{rms}} = {{{V_{rms}}} \over z} = {{200\sqrt 2 } \over {\sqrt {{{100}^2} + {{(200 - 100)}^2}} }}$$
$$ = {{200\sqrt 2 } \over {100\sqrt 2 }}$$
= 2 A
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