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(B) The total current $I$ flowing through the wire is given by the sum of the direct current and the alternating current: $I = 5 + 10 \sin \omega t$.T

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$5\sqrt{3}\,A$

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(B) The total current $I$ flowing through the wire is given by the sum of the direct current and the alternating current: $I = 5 + 10 \sin \omega t$.The effective value ($RMS$ value) of the current is defined as $I_{	ext{eff}} = \sqrt{\frac{1}{T} \int_0^T I^2 dt}$.Substituting the expression for $I$: $I_{	ext{eff}} = \sqrt{\frac{1}{T} \int_0^T (5 + 10 \sin \omega t)^2 dt}$.Expanding the square: $(5 + 10 \sin \omega t)^2 = 25 + 100 \sin^2 \omega t + 100 \sin \omega t$.Integrating over one complete cycle $(T)$: $1$. The integral of $25$ over $T$ is $25T$.$2$. The integral of $100 \sin^2 \omega t$ over $T$ is $100 	imes (T/2) = 50T$.$3$. The integral of $100 \sin \omega t$ over $T$ is $0$.Thus,$I_{	ext{eff}} = \sqrt{\frac{1}{T} (25T + 50T + 0)} = \sqrt{25 + 50} = \sqrt{75} = 5\sqrt{3}\,A$.

$(A) \ x_0 \sin \omega t$	$(i) \ x_0$
$(B) \ x_0 \sin \omega t \cos \omega t$	$(ii) \ \frac{x_0}{\sqrt{2}}$
$(C) \ x_0 \sin \omega t + x_0 \cos \omega t$	$(iii) \ \frac{x_0}{2 \sqrt{2}}$

$A$ direct current of $5\,A$ is superposed on an alternating current $I = 10 \sin \omega t$ flowing through a wire. The effective value of the resulting current will be

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