Question:

Given below are two statements:
Statement I: In an LCR series circuit, current is maximum at resonance. 
Statement II: Current in a purely resistive circuit can never be less than that in a series LCR circuit when connected to the same voltage source. 
In the light of the above statements, choose the correct from the options given below.

Updated On: Nov 1, 2025
  • Statement I is true but Statement II is false
  • Statement I is false but Statement II is true
  • Both Statement I and Statement II are true
  • Both Statement I and Statement II are false
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The Correct Option is C

Approach Solution - 1

Statement-I: \[ I_m = \frac{V_m}{\sqrt{R^2 + (X_L - X_C)^2}} \] At resonance, \(X_L = X_C\), so: \[ I_m = \frac{V_m}{R} \]

Thus, the impedance is minimum, and therefore, \(I\) is maximum at resonance.

Statement-II: \[ I = \frac{V}{R} \] In a purely resistive circuit.

Hence, in a purely resistive circuit, the current cannot be less than that in a series LCR circuit.

Thus, both Statement I and Statement II are true.

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Approach Solution -2

To determine the correctness of the given statements, let's analyze each statement based on the principles of electrical circuits.

  1. Statement I: In an LCR series circuit, current is maximum at resonance.
    • An LCR series circuit is composed of an inductor (L), a capacitor (C), and a resistor (R) in series with an AC source.
    • At resonance, the inductive reactance (\(X_L = \omega L\)) and capacitive reactance (\(X_C = \frac{1}{\omega C}\)) are equal, causing them to cancel each other out. Thus, the impedance of the circuit is purely resistive and given by the resistance \(R\) alone.
    • The formula for impedance at resonance: \(Z = R\)
    • Since impedance is minimized, the current reaches its maximum value based on Ohm's law, \(I = \frac{V}{Z}\), where \(I\) is the current, \(V\) is the applied voltage, and \(Z\) is the impedance.
    • Hence, Statement I is true.
  2. Statement II: Current in a purely resistive circuit can never be less than that in a series LCR circuit when connected to the same voltage source.
    • In a purely resistive circuit, the impedance is equal to the resistance (\(Z = R\)), and according to Ohm's law, the current is \(I = \frac{V}{R}\).
    • For an LCR series circuit at resonance, the impedance is also equal to the resistance (\(Z_{res} = R\)), hence the current is \(I_{res} = \frac{V}{R}\).
    • Therefore, under the same voltage, the current at resonance in an LCR circuit cannot exceed the current in a purely resistive circuit.
    • Thus, Statement II is true since the current in a purely resistive circuit is equal to or greater than the current in an LCR circuit (except at resonance where it would be equal).

In conclusion, both statements are accurate based on their descriptions of the principles of electrical circuits. Therefore, the correct answer is that Both Statement I and Statement II are true.

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