A series LCR circuit is connected to an alternating source of emf E. The current amplitude at resonant frequency is $I_0$. If the value of resistance R becomes twice of its initial value then amplitude of current at resonance will be
Solution
<p>At resonance in a series LCR circuit, the reactive components (inductive and capacitive) cancel each other out, leaving only the resistance. Therefore, the amplitude of the current is given by:</p>
<p>$I_0 = \frac{E}{R}$</p>
<p>Now, if the resistance is doubled, the new resistance becomes $2R$. The current amplitude at resonance then becomes:</p>
<p>$I = \frac{E}{2R} = \frac{1}{2}\left(\frac{E}{R}\right) = \frac{I_0}{2}$</p>
<p>Thus, the amplitude of current at resonance when the resistance is doubled is $ \frac{I_0}{2} $.</p>
<p>The correct option is A.</p>
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
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