Question

A LCR circuit is at resonance for a capacitor C, inductance L and resistance R. Now the value of resistance is halved keeping all other parameters same. The current amplitude at resonance will be now:

• A. Zero
• B. double
• C. same
• D. halved

Answer 1

The Correct Option is B

To solve the problem, let's analyze the behavior of an LCR circuit at resonance.

Concept Explanation: At resonance, an LCR circuit has its inductive reactance equal to its capacitive reactance, and the circuit impedance is purely resistive. The impedance \(Z\) is given by the resistance \(R\) alone, thus:

\(Z = R\)

The current amplitude \(I_0\) at resonance is given by:

\(I_0 = \frac{V}{R}\)

where \(V\) is the peak voltage applied across the circuit.

Problem Statement Analysis:

Initially, for a resistance \(R\), the current at resonance is:

\(I_0 = \frac{V}{R}\)

Now, the resistance is halved to \(R/2\), while the inductance \(L\) and capacitance \(C\) remain the same. Hence, the new current amplitude \(I'_0\) is:

\(I'_0 = \frac{V}{R/2} = \frac{2V}{R} = 2I_0\)

Conclusion: The current amplitude at resonance will be double the initial current amplitude when the resistance is halved, assuming all other parameters remain constant.

Therefore, the correct answer is double.

Answer 2

At resonance, impedance is given by:

\[ Z = R. \]

The current in the circuit is:
\[ I = \frac{V}{R}. \]

When the resistance \(R\) is halved:
\[ R \to \frac{R}{2}, \quad I \to 2I. \]

Thus, the current amplitude becomes double.

Final Answer: Double.