A series LCR circuit with L = 5.0 H, C = 80 μF and R = 40 Ω is connected to a variable frequency 200 V source. The amplitude of current at the resonant frequency is

7.1 A

The correct answer is Option (3) → 7.1 A

At resonance in a series LCR circuit, the reactances cancel, so the impedance becomes purely resistive:

$Z = R$

The given 200 V is RMS voltage, but the question asks for amplitude (peak current).

So first convert RMS voltage to peak voltage:

$\text{Peak voltage} = \sqrt{2} \times \text{RMS voltage}$

$= \sqrt{2} \times 200$

$= 200\sqrt{2} \text{ V}$

$\text{Current amplitude} = \frac{\text{Peak voltage}}{\text{Resistance}}$

$= \frac{200\sqrt{2}}{40}$

$= 5\sqrt{2}$

$\approx 7.1 \text{ A}$