A series LCR circuit connected to an AC source with voltage of the source \( v = v_m\sin\omega t \).
If ‘q’ is the charge on the capacitor and ‘i’ is the current, from Kirchhoff’s loop rule :
\( L\displaystyle\frac{di}{dt} + iR + \frac{q}{c} = V \)
The current in the circuit is given by \( i = I_m\sin(\omega t + \phi) \) where \( \phi \) is the phase difference between the voltage across the source and current in the circuit.
We know \( V_{Rm} = I_m R \); \( V_{Lm} = I_m X_L \); \( V_{Cm} = I_m X_C \); and
\( X_L = \omega L \); \( X_C = \displaystyle\frac{1}{\omega C} \)
Total impedance in the circuit regulates current. At resonance frequency of the LCR circuit current in the circuit is maximum.
A pure inductor of 0.25 mH, a pure capacitor of \( 250\,\mu\mathrm{F} \) and a resistor of \( 350\,\Omega \) are connected in series. An ac source of amplitude 210 V is connected across this series combination of L, C and R. The impedance and current amplitude in the circuit at resonance will be :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( Z_{\min} = 350\,\Omega \) and \( I_{\max} = 0.60\,\mathrm{A} \)
At resonance the inductive and capacitive reactances are equal, \( X_L = X_C \), so they cancel and the impedance is minimum and equal to R.
\( \displaystyle Z = \sqrt{R^{2} + (X_L - X_C)^{2}} \;\Rightarrow\; Z_{\min} = R = 350\,\Omega \)
The current amplitude is then the largest possible:
\( \displaystyle I_{\max} = \frac{V_m}{Z_{\min}} = \frac{210\,\mathrm{V}}{350\,\Omega} = 0.60\,\mathrm{A} \)
The values of L and C fix the resonant frequency, but they are not needed for the impedance or the current at resonance.
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
