Question 138: AC fundamentals

Question

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:

Accepted Answer

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:
$$ Z = R $$
The amplitude of the current $ I_0 $ at resonance is given by Ohm's law:
$$ I_0 = \frac{V_0}{R} $$
where $ V_0 $ is the amplitude of the voltage supplied.
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:
$$ R_{new} = \frac{R}{2} $$
The new current amplitude $ I_0' $ at resonance is given by the modified Ohm's law:
$$ I_0' = \frac{V_0}{R_{new}} = \frac{V_0}{\frac{R}{2}} = \frac{2V_0}{R} = 2 	imes I_0 $$
Therefore, the current amplitude at resonance will be doubled. Hence, the correct answer is:
Option D: double

Answer

halved

Answer

same

Answer

Zero

Answer

double

AC fundamentals

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:

Choose the correct answer: