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In LCR series circuit, source voltage is $120 \mathrm{~V}$ and voltage across inductor $50 \mathrm{~V}$. If voltage drop across resistance is $40 \mathrm{~V}$, then determine voltage in capacitor.
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The correct answer is:
$10(5+8 \sqrt{2}) \mathrm{V}$
$\begin{aligned} & 120^2=40^2+\left(V_C-50\right)^2 \\ & V_C=40 \sqrt{8}+50=10(5+8 \sqrt{2}) \mathrm{V}\end{aligned}$
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