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:
- Zero
- double
- same
- halved
The Correct Option is B
Approach Solution - 1
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.
Approach Solution -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.
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