Question

An LCR series ac circuit is at resonance with \(10\,V\) each across \(L\), \(C\) and \(R\). If the resistance is halved, the respective voltage across \(R\), \(C\) and \(L\) are

• A. \(5\,V, 10\,V, 10\,V\)
• B. \(10\,V, 5\,V, 5\,V\)
• C. \(5\,V, 5\,V, 5\,V\)
• D. \(10\,V, 20\,V, 20\,V\)

Hint:

At resonance in series LCR circuit, current depends only on resistance. If resistance is halved, current doubles, so \(V_L\) and \(V_C\) double while \(V_R\) remains equal to applied voltage.

Answer 1

The Correct Option is D

Step 1: Understand resonance condition in LCR circuit.
At resonance in a series LCR circuit, inductive reactance and capacitive reactance are equal.
\[ X_L = X_C \]

Step 2: Voltage across resistance at resonance.
At resonance, impedance of the circuit is only resistance. Therefore, applied voltage appears across resistance.
Initially, voltage across \(R\) is:
\[ V_R = 10\,V \]

Step 3: Effect of halving resistance on current.
At resonance, current is given by:
\[ I = \frac{V}{R} \]
If resistance is halved, current becomes double.
\[ I' = 2I \]

Step 4: New voltage across resistance.
New resistance is \(R' = \frac{R}{2}\).
\[ V_R' = I'R' = (2I)\left(\frac{R}{2}\right) \]
\[ V_R' = IR = 10\,V \]

Step 5: New voltage across capacitor.
Voltage across capacitor is:
\[ V_C = IX_C \]
Since \(X_C\) remains unchanged and current doubles, capacitor voltage doubles.
\[ V_C' = 2V_C = 20\,V \]

Step 6: New voltage across inductor.
Similarly, voltage across inductor is:
\[ V_L = IX_L \]
Since \(X_L\) remains unchanged and current doubles, inductor voltage also doubles.
\[ V_L' = 2V_L = 20\,V \]

Step 7: Final conclusion.
\[ \boxed{10\,V, 20\,V, 20\,V} \] Hence, correct answer is option (D).