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

Answer: (D)

P	Q	R	S
3	1	4	2

Answer 2

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 = $\frac{(45)^2}{R}=\frac{45^2}{112.6}\approx 18\omega$