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(C) The given source voltage is $V = 100\sqrt{2} \sin(\omega t) \, V$.Comparing this with $V = V_0 \sin(\omega t)$,the peak voltage is $V_0 = 100\sqrt

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$80$

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(C) The given source voltage is $V = 100\sqrt{2} \sin(\omega t) \, V$.Comparing this with $V = V_0 \sin(\omega t)$,the peak voltage is $V_0 = 100\sqrt{2} \, V$.The $RMS$ voltage is $V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{100\sqrt{2}}{\sqrt{2}} = 100 \, V$.In an $RL$ series circuit,the total $RMS$ voltage is given by $V_{rms}^2 = V_R^2 + V_L^2$.Given $V_R = 60 \, V$ and $V_{rms} = 100 \, V$.Substituting the values: $(100)^2 = (60)^2 + V_L^2$.$V_L^2 = 10000 - 3600 = 6400$.$V_L = \sqrt{6400} = 80 \, V$.

In the given circuit,the voltage across the inductor will be ..... $V$.

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