Question

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

​​​​​​​$[$1 Mark$]$

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

Answer 1

The correct option is C $\left(\frac{10}{\sqrt{2}}\right)\text{V}$

As per the given condition,

$R=X_L=X_C$

$\because$ voltage across them is same.

Total voltage in the circuit,

$V=I[R^2+(X_L-X_C{)}^2{]}^{\frac{1}{2}}=IR=10\text{V}$

When the capacitor is short-circuited,

$I^{\prime }=\frac{10}{(R^2+X_L^2{)}^{\frac{1}{2}}}=\frac{10}{\sqrt{2}R}$

$\therefore$ Potential drop across inductor = $I^{\prime }{X}_L=I^{\prime }R=\frac{10}{\sqrt{2}}\text{V}$