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

Answer: (B)

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 XLO=XCO 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.