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 address the problem, we will examine the behavior of an LCR circuit during resonance.

Concept Explanation: Resonance in an LCR circuit occurs when the inductive reactance equals the capacitive reactance. At this point, the circuit's impedance becomes purely resistive, defined solely by the resistance \(R\). The impedance \(Z\) is thus:

\(Z = R\)

The peak current amplitude \(I_0\) at resonance is calculated as:

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

where \(V\) represents the peak voltage applied to the circuit.

Problem Statement Analysis:

Initially, with a resistance \(R\), the resonant current is:

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

The resistance is subsequently reduced by half to \(R/2\), while the inductance \(L\) and capacitance \(C\) remain unchanged. The new peak current amplitude, denoted as \(I'_0\), is therefore:

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

Conclusion: Assuming all other parameters are constant, halving the resistance at resonance will result in a doubling of the initial current amplitude.

Consequently, the correct answer is double.