A series LCR circuit has a capacitor ‘C’, an inductance ‘L’ and a resistance ‘R’ connected across an ac source of voltage V. At resonance, the amplitude of current will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( V\sqrt{2}/R \)
Note: The given answer is as per the official NTA Answer Key. However, the question can have more than one correct answer, i.e. both Option (2) and Option (3), as explained below. An ac voltage given as a single value, without the word peak or amplitude, is normally the rms value; this is probably the reading behind the NTA answer, Option (3).
At resonance \( X_L = X_C \), so the reactances cancel and the impedance equals R, its minimum value.
\( \displaystyle Z = \sqrt{R^{2} + (X_L - X_C)^{2}} = R \)
The current amplitude is therefore the voltage amplitude divided by R. The question does not say whether the source voltage V is its rms value or its peak value, and the answer depends on this.
Option (1) Zero — Incorrect. At resonance a series LCR circuit has its minimum impedance, so the current is maximum, not zero.
Option (2) \( V/R \) — Correct. If V is the peak (amplitude) value of the source voltage, the amplitude of current is simply \( I_0 = V/R \).
Option (3) \( V\sqrt{2}/R \) — Correct. If V is the rms value, then \( I_{\mathrm{rms}} = V/R \) and the amplitude is \( I_0 = \sqrt{2}\, I_{\mathrm{rms}} = V\sqrt{2}/R \).
Option (4) \( V^{2}/2R \) — Incorrect. \( V^{2}/2R \) has the unit of power, not of current.
NCERT (2026-27):
Resonance – “… So if ω is varied, then at a particular frequency ω0, Xc = XL, and the impedance is minimum
\( \displaystyle Z = \sqrt{R^2 + 0^2} = R \)”
Resonance – “… At resonant frequency, the current amplitude is maximum; im = vm/R.”