Question

A series LCR circuit is connected to a 45 sin (ωt) volt source. The resonant angular frequency of the circuit is 10⁵ rad/sec and the current amplitude at resonance is I₀. When the angular frequency of the source is ω = 8 x 10⁴ rad/sec, the current amplitude in the circuit is 0.05 I₀. If m = 50 mH, match each entry in the list - I with an approximate value from list - II and choose the 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/sec	(3)	400
(S)	The peak power dissipated at resonance in watt	(4)	2250
		(5)	500

• A.

  P	Q	R	S
  4	3	2	1

• B.

  P	Q	R	S
  2	1	3	4

• C.

  P	Q	R	S
  3	1	2	5

• D.

  P	Q	R	S
  3	1	4	2

Answer 1

The Correct Option is D

The correct option is: (D)
\(E=45\,sin\omega t\)
\(\omega_rL=\frac{1}{\omega_rC}\)
\(\omega_r^2=\frac{1}{LC}\)
\((10^5)^2=\frac{1}{50\times10^{-3}\times C}\)
\(10^{10}=\frac{1}{5\times10^{-2}C}\Rightarrow c=2\times10^{-9}F\)
at \(\omega=8\times10^4\,rad/sec\)
\(X_L=\omega_L=8\times10^4\times50\times10^{-3}=4000\Omega\)
\(X_C=\frac{1}{\omega_c}=\frac{1}{8\times10^4\times2\times10^{-9}}=6250\Omega\)
\(X=X_C-X_L=2250\Omega\)

Also,

\(0.05I_0=\frac{45}{Z}\)

\(0.5\times \frac{45}{R}=\frac{45}{Z}\)

\(Z=\frac{R}{0.05}\)
\(\sqrt{R^2+x^2}=20R\)

\(R^2+x^2=400R^2\)

\(\Rightarrow R=112.6\Omega\,\,\,\,(as\, x=2250\Omega)\)

\((P)I_0=\frac{45}{112.6}A=\frac{45\times1000}{112.6}mA\approx400mA\)

\((Q)Q_{factor}=\frac{\Omega_r\times L}{R}=\frac{10^5\times50\times10^{-3}}{112.6}=44.4\)

\((R)\) B and width = \(\frac{R}{L}=2250\,rad/sec\)

\((S)\) Peek power at resonance = \(\frac{(45)^2}{R}=\frac{45^2}{112.6}\approx 18\omega\)