Question

In the series $L-C-R$ circuit shown in the figure, the rms voltage across the resistor and inductor are $400 \, V$ and $700 \, V$ respectively. If the applied voltage is $E=500\sqrt{2 \, }sin\left(\omega t\right)$ , then the peak voltage across the capacitor is

.

• A. $1200 \, V$
• B. $1200\sqrt{2}V$
• C. $400 \, V$
• D. $400\sqrt{2}V$

Answer 1

Answer: (D)

$V_{R}^{2}+(V_{L}-V_{C}^{2}=E_{r m s}^{2}$
$\left(V_{L} \, - \, V_{C}\right)^{2}=\left(500\right)^{2} \, - \, \left(400\right)^{2}$
$V_{L} \, - \, V_{C}=\sqrt{( 500^{2} - ( 400 ^{2}}=300$
$V_{C}=V_{L} \, - \, 300=700 \, - \, 300=400$
$\therefore \left(\text{V}\right)_{\text{C}} \left(\text{peak}\right) = \sqrt{2} \left(\text{V}\right)_{\text{c}} = \sqrt{2} \times 4 0 0 \text{volts}$