Question

A series LCR circuit is connected to an alternating source of emf \( E \). The current amplitude at resonance frequency is \( I_0 \). If the value of resistance \( R \) becomes twice of its initial value, then amplitude of current at resonance will be:

• A. \( \frac{I_0}{2} \)
• B. \( 2I_0 \)
• C. \( I_0 \)
• D. \( \frac{I_0}{\sqrt{2}} \)

Hint:

In a series LCR circuit, the current at resonance is inversely proportional to the resistance. Doubling the resistance reduces the current by half.

Answer 1

The Correct Option is A

In an LCR circuit, the current amplitude \( I \) at resonance is given by: \[ I = \frac{E}{R} \] where \( R \) is the resistance, \( E \) is the emf, and \( R \) is the total resistance in the circuit. If the resistance \( R \) is doubled, the current will decrease as the current is inversely proportional to the resistance. Therefore, if the resistance becomes twice the initial value, the current will be half of its initial value: \[ I_{\text{new}} = \frac{I_0}{2}. \] Thus, the correct answer is \( \boxed{\frac{I_0}{2}} \).

Answer 2

Step 1: Recall the formula for current amplitude in a series LCR circuit.
The current amplitude is given by: \[ I = \frac{E}{Z}, \] where \( Z \) is the impedance of the circuit: \[ Z = \sqrt{R^2 + (X_L - X_C)^2}. \] 

Step 2: At resonance condition.
At resonance, \( X_L = X_C \). Thus, impedance becomes minimum: \[ Z = R. \] Therefore, current amplitude at resonance is: \[ I_0 = \frac{E}{R}. \]

Step 3: When resistance is doubled.
If resistance becomes \( 2R \), then the new impedance at resonance is: \[ Z' = 2R. \] Hence, the new current amplitude: \[ I' = \frac{E}{Z'} = \frac{E}{2R}. \]

Step 4: Relate new current with the original.
\[ I_0 = \frac{E}{R} \quad \Rightarrow \quad I' = \frac{I_0}{2}. \]

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Final Answer:

\[ \boxed{I' = \dfrac{I_0}{2}} \]