Question

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 $\epsilon =500\sqrt{2}\sin \omega t$, then the peak voltage across the capacitor is

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

Answer 1

Answer: (D)

$\epsilon =500\sqrt{2}\sin \omega t$
$V_R=400V,V_L=700V$
${\epsilon }_{\text{rms }}=\frac{{\epsilon }_0}{\sqrt{2}}=\frac{500\sqrt{2}}{\sqrt{2}}=500V$
${\epsilon }_{\text{rms }}=\sqrt{V_R^2+{\left(V_L-V_c\right)}^2}$
$(500{)}^2=(400{)}^2+{\left(V_L-V_c\right)}^2$
$250000-160000={\left(V_L-V_c\right)}^2$
$90000={\left(V_L-V_c\right)}^2$
$V_L-V_c=300$
$V_c=700-300$
$V_c=400V$
$V_0=V_{\text{rms }}\sqrt{2}$
$=400\sqrt{2}V$