Question

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

• A. 10 V
• B. $10\sqrt{2}V$
• C. $\left( 10\sqrt{2} \right)V$
• D. 20 V

Answer 1

Answer: (C)

$\left( 10\sqrt{2} \right)V$

Answer 2

: Here, $R = X_{L} = X_{C}$

($\because$voltage across them is same

Total voltage in the circuit,

$V = I\lbrack R^{2} + (X_{L} - X_{C})^{2}\rbrack^{1/2} = IR = 10V$

When capacitor is short circuited,

$I' = \frac{10}{(R^{2} + X_{L}^{2})^{1/2}} = \frac{10}{\sqrt{2}R}$

$\therefore$ Potential drop across inductance

$= I'X_{L} = I'R = \frac{10}{\sqrt{2}}V$