In LCR series circuit with resistance R, inductive reactance \( X_L \) and capacitive reactance \( X_C \), the maximum current ‘\( I_0 \)’ at resonance is given as:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( I_0=\displaystyle\frac{V_0}{R} \)
The peak current in a series LCR circuit is \( I_0 = V_0/Z \), where the impedance is \( Z = \sqrt{R^{2}+(X_L-X_C)^{2}} \). At resonance \( X_L = X_C \), so the reactances cancel and the impedance is at its minimum value, \( Z = R \).
\( \displaystyle I_0 = \frac{V_0}{Z} = \frac{V_0}{R} \)
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Resonance – “… So if ω is varied, then at a particular frequency ω0, Xc = XL, and the impedance is minimum \( \left(Z = \sqrt{R^2 + 0^2} = R\right) \). … At resonant frequency, the current amplitude is maximum; im = vm/R.”