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
Answer & explanation
Correct answer: option 3
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:
$\text{Peak voltage} = \sqrt{2} \times \text{RMS voltage}$
$= \sqrt{2} \times 200$
$= 200\sqrt{2} \text{ V}$
$\text{Current amplitude} = \frac{\text{Peak voltage}}{\text{Resistance}}$
$= \frac{200\sqrt{2}}{40}$
$= 5\sqrt{2}$
$\approx 7.1 \text{ A}$