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 reactance of the inductor is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 20 Ω
At resonance, the reactance of the inductor equals the reactance of the capacitor. This happens at the angular frequency \( \omega_0=1/\sqrt{LC} \). The reactance of the inductor at this frequency is \( X_L=\omega_0L \).
Given:
\( L=20\ \mathrm{mH}=20\times10^{-3}\ \mathrm{H} \)
\( C=50\ \mu\mathrm{F}=50\times10^{-6}\ \mathrm{F} \)
\( \begin{aligned} \omega_0 &= \frac{1}{\sqrt{LC}} \\ &= \frac{1}{\sqrt{20\times10^{-3}\times50\times10^{-6}}} \\ &= \frac{1}{\sqrt{10^{-6}}}=1000\ \mathrm{rad\,s^{-1}} \end{aligned} \)
\( \begin{aligned} X_L &= \omega_0L \\ &= 1000\times20\times10^{-3} \\ &= 20\ \Omega \end{aligned} \)
Therefore, the reactance of the inductor at resonance is 20 Ω. The capacitor has the same reactance at this frequency.
NCERT (2026-27):
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 … This frequency is called the resonant frequency:”
\( \displaystyle X_c=X_L \quad \text{or} \quad \frac{1}{\omega_0C}=\omega_0L \quad \text{or} \quad \)\( \displaystyle \omega_0=\frac{1}{\sqrt{LC}} \)\( \qquad (7.28) \)
Summary – “3. … XL = ωL is called inductive reactance.”