Read the following passage and answer the questions.
A student studies a series LCR circuit like the tuning circuit of a radio. She connects a 50 Ω resistor, a 20 mH inductor and a 50 μF capacitor in series with an ac source. The rms voltage of the source stays at 10 V. Its angular frequency \(\omega\) can be changed over a wide range. An ac ammeter in the circuit reads the rms current. An ac voltmeter can be joined across each element in turn. She starts with \(\omega\) at a low value and raises it slowly. At each step she notes the ammeter reading and the three voltmeter readings. She then repeats the whole run with a 100 Ω resistor in place of the 50 Ω one. The inductor, the capacitor and the source stay the same.
At the resonant frequency, the current in the circuit
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → is in phase with the source voltage
In a series LCR circuit, the phase difference \( \phi \) between the source voltage and the current is given by \( \tan\phi=(X_C-X_L)/R \). At resonance \( X_C=X_L \), so \( \tan\phi=0 \).
Option (1) is in phase with the source voltage — Correct. With \( \phi=0 \) the current and the source voltage reach their maximum values at the same instant. The voltages across the inductor and the capacitor cancel, and the circuit behaves like the resistor alone.
Option (2) leads the source voltage by \( \pi/2 \) — Incorrect. The current leads the voltage only when \( X_C \gt X_L \). This happens below the resonant frequency. Because of the resistor, the lead is then less than \( \pi/2 \).
Option (3) lags behind the source voltage by \( \pi/2 \) — Incorrect. A lag appears only when \( X_L \gt X_C \), above the resonant frequency. Because of the resistance, it also stays below \( \pi/2 \).
Option (4) leads the source voltage by \( \pi/4 \) — Incorrect. A lead of \( \pi/4 \) needs \( X_C-X_L=R \). This happens at a frequency below \( \omega_0 \). At the resonant frequency the phase difference is zero.
NCERT (2026-27):
Phasor-diagram solution – “… we have”
\( \displaystyle \tan\phi=\frac{X_C-X_L}{R} \qquad (7.27) \)
Power in AC Circuit: the Power Factor – “Case (iv) Power dissipated at resonance in LCR circuit: At resonance Xc – XL = 0, and φ = 0. …”