A series LCR circuit has \(R=40\ \Omega\), \(X_L=60\ \Omega\) and \(X_C=20\ \Omega\) at the frequency of the source. The current in the circuit
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → lags behind the source voltage by \(45^\circ\)
In a series LCR circuit, the phase angle \( \phi \) between the source voltage and the current is given by \( \tan\phi=(X_C-X_L)/R \). A negative \( \phi \) means that the current lags behind the voltage.
\( \begin{aligned} \tan\phi &= \frac{X_C-X_L}{R} \\ &= \frac{20-60}{40} \\ &= -1 \end{aligned} \)
\( \phi=-45^\circ \)
From the above, \( \phi \) is negative because \( X_L \gt X_C \). The circuit is mainly inductive. Hence, the current lags behind the source voltage by \( 45^\circ \).
NCERT (2026-27):
Phasor-diagram solution – “… we have”
\( \displaystyle \tan\phi=\frac{X_C-X_L}{R} \)\( \qquad (7.27) \)
Phasor-diagram solution – “… If XC < XL, φ is negative and the circuit is predominantly inductive. Consequently, the current in the circuit lags the source voltage.”