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

• A. \(10\ \text{V}\)
• B. \(5\sqrt{2}\ \text{V}\)
• C. \(\dfrac{5}{\sqrt2}\ \text{V}\)
• D. \(10\sqrt2\ \text{V}\)

Hint:

If \(V_L=V_C\), the circuit is in resonance and \[ R=X_L. \] After removing the capacitor, treat the circuit as an \(RL\) circuit and use \[ Z=\sqrt{R^2+X_L^2}. \]

Answer 1

The Correct Option is D

Concept: In a series LCR circuit, the applied voltage is the phasor sum of the voltages across the resistor, inductor and capacitor. The resultant voltage is \[ V=\sqrt{V_R^2+(V_L-V_C)^2}. \] When the inductive and capacitive voltages are equal, the circuit is in resonance and the impedance becomes purely resistive.

Step 1: Calculate the source voltage in the original circuit.
Given, \[ V_R=10\ \text{V}, \qquad V_L=10\ \text{V}, \qquad V_C=10\ \text{V}. \] Therefore, \[ V=\sqrt{10^2+(10-10)^2} \] \[ V=10\ \text{V}. \] Hence, the applied voltage is \(10\) V.

Step 2: Determine the circuit after the capacitor is short circuited.
When the capacitor is short circuited, \[ X_C=0. \] The circuit now becomes a series \(RL\) circuit. Since originally \[ V_R=V_L, \] we have \[ IR=IX_L \] which implies \[ R=X_L. \]

Step 3: Calculate the new voltage across the inductor.
The impedance of the \(RL\) circuit is \[ Z=\sqrt{R^2+X_L^2} \] Since \[ R=X_L, \] \[ Z=\sqrt{2R^2} =\sqrt2\,R. \] Hence, \[ I'=\frac{V}{Z} = \frac{10}{\sqrt2\,R}. \] The voltage across the inductor becomes \[ V_L'=I'X_L = \frac{10}{\sqrt2 R}\times R = \frac{10}{\sqrt2} = 5\sqrt2\ \text{V}. \] Therefore, \[ \boxed{V_L'=5\sqrt2\ \text{V}} \]