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:

<p>To solve this problem, we need to understand the relationship between the current amplitude in a series LCR circuit at resonance and the resistance $ R $. At resonance, the impedance $ Z $ of the series LCR circuit is equal to the resistance $ R $, and thus:</p> <p>$Z = R$</p> <p>The amplitude of the current $ I_0 $ at resonance is given by Ohm's law:</p> <p>$I_0 = \frac{V_0}{R}$</p> <p>where $ V_0 $ is the amplitude of the voltage supplied.</p> <p>Now, if the resistance $ R $ is halved while keeping the voltage amplitude $ V_0 $, capacitance $ C $, and inductance $ L $ the same, the new resistance becomes:</p> <p>$R_{new} = \frac{R}{2}$</p> <p>The new current amplitude $ I_0' $ at resonance is given by the modified Ohm's law:</p> <p>$$ I_0' = \frac{V_0}{R_{new}} = \frac{V_0}{\frac{R}{2}} = \frac{2V_0}{R} = 2 \times I_0 $$</p> <p>Therefore, the current amplitude at resonance will be doubled. Hence, the correct answer is:</p> <p><strong>Option D: double</strong></p>