L = 2 H, C = 5 mF and R = 12 Ω are connected in series to an a.c. generator of frequency 50 Hz. Then (a) at resonance, impedance of the circuit is zero. (b) at resonance, impedance of the circuit is 12 Ω. (c) the resonant frequency of the circuit is 1/2π. (d) the inductive reactance is less than the capacitive reactance.

Answer: (b). At resonance XL = XC, so Z = R = 12 Ω (not zero). So (a) is wrong and (b) is right. ω₀ = (1)/(√(LC)) = 1√2 × 5×10⁻³ = 10 rad/s, so f₀ = (10)/(2π) = 5/π Hz, not 1/2π: (c) is wrong. At 50 Hz: XL = 2π(50)(2) ≈ 628 Ω, XC = 12π(50)(5×10⁻³) ≈ 0.64 Ω, so XL > XC: (d) is wrong. So the answer is (b) 12 Ω.

L = 2 H, C = 5 mF and R = 12 Ω are connected in series to an a.c… | MHT CET 2024 (2 May, Shift 2)

Q53MHT CET 2024 · 2 MAY · SHIFT 2Medium

Resonance, Bandwidth & Q Factor

$L = 2\text{ H}, C = 5\text{ mF}$ and $R = 12\ \Omega$ are connected in series to an a.c. generator of frequency 50 Hz. Then

Show answer and solution

Answer: (b)

At resonance $X_L = X_C$ , so $Z = R = 12\ \Omega$ (not zero). So (a) is wrong and (b) is right.
$\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{2 \times 5\times10^{-3}}} = 10\text{ rad/s}$ , so $f_0 = \frac{10}{2\pi} = \frac{5}{\pi}\text{ Hz}$ , not $\frac{1}{2\pi}$ : (c) is wrong.
At 50 Hz: $X_L = 2\pi(50)(2) \approx 628\ \Omega$ , $X_C = \frac{1}{2\pi(50)(5\times10^{-3})} \approx 0.64\ \Omega$ , so $X_L > X_C$ : (d) is wrong.
So the answer is (b) $12\ \Omega$ .

Report a mistake

Resonance, Bandwidth & Q Factor question from MHT CET 2024 (2 May, Shift 2)

More questions on Resonance, Bandwidth & Q Factor