Question

In series $R-L-C$ circuit, the root mean square voltage across the resistor and inductor are respectively $400 \, V$ and $700 \, V$ . If the equation for the applied voltage is $E=500\sqrt{2}sin \left(\omega t\right)V$ , 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)

$V_{R}^{2}+\left(V_{L} - V_{C}\right)^{2}=E_{r m s}^{2}$
$\left(V_{L} - V_{C}\right)^{2}=\left(500\right)^{2}-\left(400\right)^{2}$
$V_{L}-V_{C}=\sqrt{\left(\right. 500 \left(\left.\right)^{2} - \left(\right. 400 \left(\left.\right)^{2}}=300$
$V_{C}=V_{L}-300=700-300=400\text{ V}$
$\therefore \, \, \, V_{C \left(p e a k\right)}=V_{C}\sqrt{2}=400\sqrt{2} \, V$