Question

In a series RLC circuit, the r.m.s. voltage across the resistor and the inductor are respectively $400\, V$ and $700\, V$. If the equation for the applied voltage is $\varepsilon=500 \sqrt{2} \sin \omega t$, 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)

$\varepsilon=500 \sqrt{2} \sin \omega t$
$V_{R}=400 \,V , V_{L}=700\, V$
$\varepsilon_{ \text{rms }}=\frac{\varepsilon_{0}}{\sqrt{2}}=\frac{500 \sqrt{2}}{\sqrt{2}}=500 \,V$
$\varepsilon_{ \text{rms }}=\sqrt{V_{R}^{2}+\left(V_{L}-V_{c}\right)^{2}}$
$(500)^{2}=(400)^{2}+\left(V_{L}-V_{c}\right)^{2}$
$250000-160000=\left(V_{L}-V_{c}\right)^{2}$
$90000=\left(V_{L}-V_{c}\right)^{2}$
$V_{L}-V_{c}=300$
$V_{c}=700-300$
$V_{c}=400 \,V$
$V_{0}=V_{ \text{rms }} \sqrt{2}$
$=400 \sqrt{2} \,V$