Question

In a series $LCR$ circuit, the voltage across the resistor, capacitor and inductor is $10 \, V$ each. If the capacitor is short-circuited, the voltage across the inductor will be

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

Answer 1

Answer: (B)

Here, $R=X_{L}=X_{C}$ , because voltage across them is the same.
When the capacitor is short circuited,
$I=\frac{10}{\left(R^{2} + X_{L}^{2}\right)^{\frac{1}{2}}}=\frac{10}{\sqrt{2} R}$
Hence, potential drop across inductor $=IX_{L}=IR=\frac{10}{\sqrt{2}}V$