Question

In a series LCR circuit, the voltage across resistance, capacitance and inductance is the same, each equal to 80 V. If the capacitor is short circuited, the voltage across the inductor becomes

• A. zero
• B. 40 V
• C. 80 V
• D. $40\sqrt{2}V$

Answer 1

The correct option is D $40\sqrt{2}V$

Given $V_R=V_L=V_c$. Therefore, $R=X_L=X_C$. When the capacitance is short circuited, $X_C=0$ and the impedance is
$Z=\sqrt{R^2+X_L^2}=\sqrt{R^2+R^2}=\sqrt{2}R$
The voltage of the a.c. source is given by $(\because R+X_L)$
$V=\sqrt{V_R^2+(V_L-V_C{)}^2}=V_R=80V$
$\therefore$ Current in the circuit is
$I=\frac{V}{Z}=\frac{80}{\sqrt{2}R}$
Hence, $V_L=I{X}_L=\frac{80}{\sqrt{2}R}\times R$ $\because X_L=R)$
=$40\sqrt{2}$volt.