Which of the following statements is/are correct?
I. In an LCR series ac circuit, as the frequency of the source increases, the impedance of the circuit first decreases and then increases.
II. If the inductive reactance of an LCR series ac circuit exceeds its capacitive reactance by an amount equal to its resistance, then the current lags behind the voltage by \( 45^\circ \).
III. Below resonance, the voltage leads the current, while above it, the current leads the voltage.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → I and II
The impedance of a series LCR circuit is \( Z=\sqrt{R^2+(X_C-X_L)^2} \) and the phase angle is given by \( \tan\phi=(X_C-X_L)/R \), with \( X_L=\omega L \) and \( X_C=1/\omega C \).
I. In an LCR series ac circuit, as the frequency of the source increases, the impedance of the circuit first decreases and then increases. — True. At low frequency \( X_C \) is large and Z is large. As \( \omega \) rises, \( X_C \) falls and \( X_L \) grows. At the resonant frequency \( X_C=X_L \), so Z has its minimum value R. Beyond it \( X_L \) keeps growing and Z increases again.
II. If the inductive reactance of an LCR series ac circuit exceeds its capacitive reactance by an amount equal to its resistance, then the current lags behind the voltage by \( 45^\circ \). — True. Here \( X_L-X_C=R \), so
\( \begin{aligned} \tan\phi &= \frac{X_C-X_L}{R} \\ &= \frac{-R}{R} \\ &= -1 \end{aligned} \)
\( \Rightarrow\; \phi=-45^\circ \)
A negative \( \phi \) means the circuit is predominantly inductive, so the current lags the voltage by \( 45^\circ \).
III. Below resonance, the voltage leads the current, while above it, the current leads the voltage. — False. Below resonance \( X_C\gt X_L \), the circuit is predominantly capacitive and the current leads the voltage. Above resonance \( X_L\gt X_C \), the circuit is inductive and the voltage leads the current. The statement has the two cases the wrong way round.
NCERT (2026-27):
Resonance – “… So if ω is varied, then at a particular frequency ω0, Xc = XL, and the impedance is minimum
\( \left(Z=\sqrt{R^2+0^2}=R\right) \)”
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 positive and the circuit is predominantly capacitive. Consequently, the current in the circuit leads the source voltage. If XC < XL, φ is negative and the circuit is predominantly inductive. Consequently, the current in the circuit lags the source voltage.”