Q: In a series LCR circuits, the voltage across the resistance, capacitance and inductance is 10 V each. If the capacitance is short circuited then the voltage across the inductance ω i l l be

Sol: As V_{R} = V_{L} = V_{C} ; R = X_{L} = X_{C}

Z = R ; V = IR = 10 volts

When capacitor is short circuited,

$\large Z = \sqrt{R^2 + X_L^2}$

$\large Z = \sqrt{R^2 + R^2} = \sqrt{2} R$

New current, $\large I’ = \frac{V}{\sqrt{2} R} = \frac{10}{\sqrt{2} R}$

Potential drop across inductance

$\large = I’ X_L = I’ R =\frac{10}{\sqrt{2} R} \times R $

= 10/√2 volt