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

At resonance in an LCR circuit, the current amplitude \( I \) is determined by \( I = \frac{E}{R} \), where \( E \) is the electromotive force and \( R \) is the total circuit resistance. An increase in resistance leads to a proportional decrease in current. Specifically, if the resistance \( R \) is doubled, the new current amplitude \( I_{\text{new}} \) will be half of the initial current \( I_0 \): \( I_{\text{new}} = \frac{I_0}{2} \). The correct answer is therefore \( \boxed{\frac{I_0}{2}} \).