A series LCR circuit is connected to an ac source. At resonance, the phase difference between the applied voltage and the current in the circuit is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0
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 the two reactances are equal, \( X_L=X_C \).
\( \begin{aligned} \tan\phi &= \frac{X_C-X_L}{R} \\ &= \frac{0}{R} \\ &= 0 \end{aligned} \)
\( \Rightarrow\; \phi=0 \)
Hence, at resonance the applied voltage and the current are in phase, and the circuit behaves as a pure resistance.
NCERT (2026-27):
Phasor-Diagram Solution – “From the circuit … we see that the resistor, inductor and capacitor are in series. Therefore, the ac current in each element is the same at any time, having the same amplitude and phase. Let it be
\( i=i_m\sin(\omega t+\phi) \qquad (7.21) \)
where φ is the phase difference between the voltage across the source and the current in the circuit.”
Power in AC Circuit: The Power Factor – “Case (iv) Power dissipated at resonance in LCR circuit: At resonance Xc – XL = 0, and φ = 0.”