In a series RLC circuit, the r.m.s. voltage across the resistor and the inductor are respectively 400V and 700V. If the equation for the applied voltage is $ε = 500 \sqrt{2} \sin ωt$ , then the peak voltage across the capacitor if the circuit behave as an inductive circuit is

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$1200 \sqrt{2} V$

$400 \sqrt{2} V$

1200V

400V

answer is D.

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Detailed Solution

$\begin{aligned} E_{rms} = \frac{E_{0}}{\sqrt{2}} = \frac{500 \sqrt{2}}{\sqrt{2}} = 500 V \\ E_{rms}^{2} = E_{R}^{2} + {(E_{L} - E_{C})}^{2} \\ 250000 = 160000 + {(700 - E_{C})}^{2} \\ => {(700 - E_{C})}^{2} = 90000 \\ 700 - E_{C} = \sqrt{90000} \\ E_{c} = 700 - 300 \\ E_{C} = 400 \\ E_{rms} = \frac{E_{0}}{\sqrt{2}} \\ => E_{O} = \sqrt{2} E_{rms} \\ = \sqrt{2} \times 400 V \\ E_{o} = 400 \sqrt{2} V \end{aligned}$