Question 135: AC fundamentals

Question

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

Accepted Answer

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:
$$ I_0 = \frac{E}{R} $$
Now, if the resistance is doubled, the new resistance becomes $2R$. The current amplitude at resonance then becomes:
$$ I = \frac{E}{2R} = \frac{1}{2}\left(\frac{E}{R}\right) = \frac{I_0}{2} $$
Thus, the amplitude of current at resonance when the resistance is doubled is $ \frac{I_0}{2} $.
The correct option is A.

Answer

$\frac{\mathrm{I}_0}{2}$

Answer

$\frac{\mathrm{I}_0}{\sqrt{2}}$

Answer

$2 \mathrm{I}_0$

Answer

$\mathrm{I_0}$

AC fundamentals

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

Choose the correct answer: