A series LCR circuit is connected to a variable frequency 230 V source. L = 5H, C = 80 µF & R = \( 40\,\Omega \). The potential drop across R at resonating frequency is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 230\,\mathrm{V} \)
At resonance of a series LCR circuit the inductive and capacitive reactances are equal, \( X_L = X_C \), so they cancel and the impedance is only the resistance, \( Z = R \).
\( \displaystyle \omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{5\times80\times10^{-6}}} = \frac{1}{\sqrt{4\times10^{-4}}} = 50\,\mathrm{rad\,s^{-1}} \)
\( X_L = \omega_0 L = 50\times5 = 250\,\Omega,\quad X_C = \frac{1}{\omega_0 C} = \frac{1}{50\times80\times10^{-6}} = 250\,\Omega \)
With \( Z = R = 40\,\Omega \), the current and the potential drop across R are
\( \displaystyle I = \frac{V}{Z} = \frac{230\,\mathrm{V}}{40\,\Omega} = 5.75\,\mathrm{A} \)
\( V_R = IR = 5.75\times40 = 230\,\mathrm{V} \)
Shortcut: At resonance the voltages across L and C are equal and opposite, so the whole source voltage appears across R. \( V_R = 230\,\mathrm{V} \) follows at once, without using the values of L, C or 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):
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