Question

In a series L-C-R circuit, voltage across resistor, capacitor and inductor is $20$V each. If capacitor is short circuited the voltage across inductor is?

• A. $10\sqrt{2}$V
• B. $5\sqrt{2}$V
• C. $20\sqrt{2}$V
• D. $\frac{5}{\sqrt{2}}$V

Answer 1

The correct option is A $10\sqrt{2}$V

The voltage across Inductor lags behind the current through the resistor. The voltage across the capacitor leads the current through R.

When$V_R=V_L=V_C$ = potentials across resistor, inductor and capacitors are all equal, then$I\times X_L=I\times X_C$, and they are in opposite phase. So they cancel. The collective impedance is$X_L-X_C=0$ The net voltage across the $3$ elements is only $10V$. So the voltage applied across these elements, is $10V$.

Now if the capacitance is shorted, only inductor and resistor are there in the circuit. So$Z=\sqrt{[R^2+X_l^2]}=\sqrt{2}R$ There is an external voltage source of $10V.$

So current $I=\frac{10}{\sqrt{2}R}$ amp.

P.d. across L :$I{X}_L=IR=\frac{10}{\sqrt{2}}V$

or,

The voltages across Resistance and voltage across inductor have a phase difference of 90 deg. Hence, the $netvoltage$ =${\sqrt{V}}_L^2+{\sqrt{V}}_R^2=10V$

=>$\sqrt{2}{V}_R=10V$

=>$V_R=V_L=\frac{10}{\sqrt{2}}volts$