Question

A series LCR circuit with inductance 10 H, capacitance 10 μF, resistance 50 Ω is connected to an ac source of voltage, V =200 sin(100 t) volt. If the resonant frequency of the LCR circuit is v_o and the frequency of the ac source is v, then:

• A. v₀ = v = 50 Hz
• B. v₀ = v = \(\frac{50}{π}\) Hz
• C. v₀ = \(\frac{50}{π}\) Hz, v = 50 Hz
• D. v₀ = 100 Hz, v₀ = \(\frac{100}{π}\) Hz

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In a series LCR circuit, resonance occurs when the inductive reactance equals the capacitive reactance.
The angular frequency of the source is obtained from the standard equation \(V = V_0 \sin(\omega t)\).
Key Formula or Approach:
1. Source angular frequency: \(\omega = 2\pi\nu\)
2. Resonant angular frequency: \(\omega_0 = \frac{1}{\sqrt{LC}}\)
3. Resonant frequency: \(\nu_0 = \frac{1}{2\pi\sqrt{LC}}\)
Step 2: Detailed Explanation:
1. Find source frequency (\(\nu\)):
From \(V = 200 \sin(100t)\), we have \(\omega = 100 \text{ rad/s}\).
\[ 2\pi\nu = 100 \implies \nu = \frac{100}{2\pi} = \frac{50}{\pi} \text{ Hz} \]
2. Find resonant frequency (\(\nu_0\)):
Given \(L = 10 \text{ H}\) and \(C = 10 \text{ \(\mu\)F} = 10 \times 10^{-6} \text{ F} = 10^{-5} \text{ F}\).
\[ \omega_0 = \frac{1}{\sqrt{10 \times 10^{-5}}} = \frac{1}{\sqrt{10^{-4}}} = \frac{1}{10^{-2}} = 100 \text{ rad/s} \]
Since \(\omega_0 = 100 \text{ rad/s}\), the resonant frequency is:
\[ \nu_0 = \frac{\omega_0}{2\pi} = \frac{100}{2\pi} = \frac{50}{\pi} \text{ Hz} \]
3. Comparison:
Both \(\nu\) and \(\nu_0\) are equal to \(\frac{50}{\pi}\) Hz.
Step 3: Final Answer:
Therefore, \(\nu_0 = \nu = \frac{50}{\pi}\) Hz.