In a series LCR circuit, the resistance is 40 Ω. At a source frequency of 50 Hz, the inductive reactance is 30 Ω and the capacitive reactance is 120 Ω. When the frequency is raised to 100 Hz, the impedance of the circuit is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(40\ \Omega\)
The inductive reactance \( X_L=\omega L \) is proportional to the frequency. The capacitive reactance \( X_C=1/\omega C \) is inversely proportional to the frequency. So when the frequency is doubled, \( X_L \) becomes twice and \( X_C \) becomes half.
At 100 Hz:
\( \begin{aligned} X_L^{\prime} &= 2\times X_L \\ &= 2\times30 \\ &= 60\ \Omega \end{aligned} \)
\( \begin{aligned} X_C^{\prime} &= \frac{X_C}{2} \\ &= \frac{120}{2} \\ &= 60\ \Omega \end{aligned} \)
\( \begin{aligned} Z &= \sqrt{R^{2}+(X_C^{\prime}-X_L^{\prime})^{2}} \\ &= \sqrt{40^{2}+(60-60)^{2}} \\ &= 40\ \Omega \end{aligned} \)
The two reactances are now equal, so the circuit is at resonance. Hence, its impedance is equal to the resistance, 40 Ω.
NCERT (2026-27):
Phasor-diagram solution – “By analogy to the resistance in a circuit, we introduce the impedance Z in an ac circuit:”
\( \displaystyle i_m=\frac{v_m}{Z} \qquad [7.25(b)] \)
“where”
\( \displaystyle Z=\sqrt{R^{2}+(X_C-X_L)^{2}} \)\( \qquad (7.26) \)
AC Voltage Applied to a Series LCR Circuit – “… So if ω is varied, then at a particular frequency ω0, Xc = XL, and the impedance is minimum (Z = \( \sqrt{R^{2}+0^{2}} \) = R). This frequency is called the resonant frequency…”