Question

In a series $LCR$ circuit, the voltage across the resistance, capacitance and inductance is $10V$ each. If the capacitance is short circuited, the voltage across the inductance will be :

• A. $10V$
• B. $\frac{10}{\sqrt{2}}V$
• C. $\frac{10}{3}V$
• D. $20V$

Answer 1

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

given ,
$V_R=\frac{R}{\left|Z\right|}{V}_s=10\Rightarrow R=10\left|Z\right|/V_s$

$V_C=\frac{X_C}{\left|Z\right|}{V}_s=10\Rightarrow X_C=10\left|Z\right|/V_s=R$

$V_L=\frac{X_L}{\left|Z\right|}{V}_s=10\Rightarrow X_L=10\left|Z\right|/V_s=R$

Therefore,
$\left|Z\right|=|R+j(X_L-X_C\left)\right|=R$

$\Rightarrow V_R=\frac{R}{\left|Z\right|}{V}_s=\frac{R}{R}{V}_s=V_s=10$

Now, voltage across inductor when capacitor is shorted,
$V_L^{\prime \prime }=\frac{X_L}{\sqrt{R^2+X_L^2}}{V}_s=\frac{R}{\sqrt{R^2+R^2}}10=10/\sqrt{2}V$