Target Exam

CUET

Subject

Physics

Chapter

Alternating Current

Question:

A sinusoidal voltage with peak value 220 V and variable frequency is applied to a series LCR circuit in which R = 10Ω, L = 4 mH, and C = 100 mF. Consider the following:

(A) Resonant angular frequency $(ω_0) = 50\, rad/s$
(B) Resonant frequency $(v_0) = 25/π\, Hz$
(C) Resonant frequency $(v_0) = 50/π\, Hz$
(D) The rms current at resonance is 15.5 A

Choose the correct answer from the options given below:

Options:

(A) and (C) only

(A), (B) and (D) only

(A), (C) and (D) only

(B) and (D) only

Correct Answer:

(A), (B) and (D) only

Explanation:

The correct answer is Option (2) → (A), (B) and (D) only

Given:

$R = 10\ \Omega,\ L = 4\ \text{mH} = 4\times 10^{-3}\ H,\ C = 100\ \text{mF} = 0.1\ F,\ V_0 = 220\ V$ (peak)

Resonant angular frequency:

$\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{(4\times10^{-3})(0.1)}} = \frac{1}{\sqrt{4\times10^{-4}}} = \frac{1}{0.02} = 50\ \text{rad/s}$

Resonant frequency:

$v_0 = \frac{\omega_0}{2\pi} = \frac{50}{2\pi} = \frac{25}{\pi}\ \text{Hz}$

At resonance, impedance $Z = R$, so rms current:

$I_{\text{rms}} = \frac{V_{\text{rms}}}{R} = \frac{V_0/\sqrt{2}}{R} = \frac{220/\sqrt{2}}{10} \approx 15.56\ \text{A}$