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(D) Given the applied voltage equation: $\varepsilon = 500 \sqrt{2} \sin \omega t$.Comparing this with $\varepsilon = \varepsilon_0 \sin \omega t$,the

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$400 \sqrt{2}$

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(D) Given the applied voltage equation: $\varepsilon = 500 \sqrt{2} \sin \omega t$.Comparing this with $\varepsilon = \varepsilon_0 \sin \omega t$,the peak voltage $\varepsilon_0 = 500 \sqrt{2} \,V$.The $r.m.s.$ voltage is $\varepsilon_{rms} = \frac{\varepsilon_0}{\sqrt{2}} = \frac{500 \sqrt{2}}{\sqrt{2}} = 500 \,V$.In a series $RLC$ circuit,the relationship between $r.m.s.$ voltages is given by $\varepsilon_{rms} = \sqrt{V_R^2 + (V_L - V_C)^2}$.Given $V_R = 400 \,V$ and $V_L = 700 \,V$,we substitute these values:$(500)^2 = (400)^2 + (700 - V_C)^2$$250000 = 160000 + (700 - V_C)^2$$(700 - V_C)^2 = 90000$$700 - V_C = \pm 300$.This gives two possibilities: $V_C = 700 - 300 = 400 \,V$ or $V_C = 700 + 300 = 1000 \,V$.Assuming the standard case where $V_C = 400 \,V$ is the $r.m.s.$ voltage,the peak voltage across the capacitor is $V_{C,peak} = V_C \sqrt{2} = 400 \sqrt{2} \,V$.

In a series $RLC$ circuit,the $r.m.s.$ voltages across the resistor and the inductor are respectively $400 \,V$ and $700 \,V$. If the equation for the applied voltage is $\varepsilon = 500 \sqrt{2} \sin \omega t$,then the peak voltage across the capacitor is ........... $V$.

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