Question

A series LCR circuit is connected to a 45 sin(𝜔𝑡) Volt source. The resonant angular frequency of the circuit is \(10^5\) rad s⁻¹ and current amplitude at resonance is I₀. 

When the angular frequency of the source is 𝜔 =\(8 ×\) \(10^4\) rad s⁻¹, the current amplitude in the circuit is 0.05 I₀. If L=50 mH, match each entry in List-I with an appropriate value from List-II and choose the correct option. 

List-I	List-II
(P) I0 in mA	(1)  44.4
(Q) The quality factor of the circuit	(2) 18
(R) The bandwidth of the circuit in rad s−1	(3) 400
(S) The peak power dissipated at resonance in Watt	(4) 2250
	(4) 2250

• A. P →2, Q→ 3, R →5, S →1
• B. P →3, Q→ 1, R →4, S →2
• C. P →4, Q→ 5, R →3, S →1
• D. P →4, Q→ 2, R →1, S →5

Answer 1

The Correct Option is B

To solve this problem, we use the concepts related to a series LCR circuit at resonance.

Given:
Resonant angular frequency, \(\omega_0 = 10^5 \, \text{rad/s}\)
Current amplitude at resonance, \(I_0\) 
When \(\omega = 8 × 10^4 \, \text{rad/s}\), current amplitude is \(0.05I_0\)
Inductance, \(L = 50 \, \text{mH} = 50 × 10^{-3} \, \text{H}\)

Steps:

1. Current amplitude relation for an LCR circuit is given by:
\(I = \frac{I_0}{\sqrt{1 + \left(Q^2\left(\frac{\omega_0}{\omega} - \frac{\omega}{\omega_0}\right)^2\right)}}\)
Given \(I = 0.05I_0\), \(\frac{I}{I_0} = 0.05\):

\(\frac{I_0}{\sqrt{1 + \left(Q^2\left(\left(\frac{10^5}{8×10^4}\right) - \left(\frac{8×10^4}{10^5}\right)\right)^2\right)}} = 0.05I_0\)
Simplify to find \(Q\):
\(0.05 = \frac{1}{\sqrt{1 + Q^2(0.5625 - 0.64)^2}}\)
\(Q^2(0.0775)^2 = \frac{1}{0.05^2} -1\)
Solving gives: \(Q = 44.721\)

2. Quality factor \(Q\) is also given by:
\(Q = \frac{\omega_0 L}{R}\)
Calculate bandwidth, which is \(\Delta \omega = \frac{\omega_0}{Q}\):
\(\Delta \omega = \frac{10^5}{44.721} \approx 2237 \, \text{rad/s}\)

3. Maximum current \(I_0\) is given by the condition \(\frac{V}{R}\) at resonance whe\):
\(I_0 = \frac{45}{112} \approx 0.401 \, \text{A} = 401 \, \text{mA}\)

4. Peak power dissipation \(P_{\text{max}} = \frac{V^2}{2R}\) at resonance:
\(= \frac{45^2}{2 \times 112} \approx 9.03 \, \text{W}\)

 

List-I	List-II
(P) I0 in mA	(1)  44.4
(Q) The quality factor of the circuit	(2) 18
(R) The bandwidth of the circuit in rad s−1	(3) 400
(S) The peak power dissipated at resonance in Watt	(4) 2250

 

Matching with options:
P → 3 (401 mA approximated to 400 mA), Q → 1 (Q = 44.721 approximated to 44.4), R → 4 (Bandwidth = 2237 rad/s approximated to 2250 rad/s), S → 2 (Power = 9.03 approximated to 9 W)

Correct answer: P →3, Q→ 1, R →4, S →2 

Answer 2

Correct option is(B): P → 3, Q → 1, R → 4, S → 2.

As given 

\(0.05l_0=\frac{45}{\sqrt{R^2+(0.8X_{LO}-\frac{5}{4}X_{C0}})^2}\)

Where X_LO=X_CO are at resonant frequencies

on solving, \(R ≅ \frac{450Ω}{4}⇒l0≅400 \,mA\)

Quality factor\( Q= \frac{1}{R}\sqrt{\frac{L}{C}}≅44.44\)

\(Q=\frac{ω_0}{△ω}⇒ ≅2250\,rad/s\)

peak power = \(45×\frac{400}{1000}\,W\) 

=18.