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QJEEAdv23Ph5e9a: JEE Advanced 2023 Physics - Class 12 JEEAdv23Ph5e9a Solution: A series LCR circuit is connected to a Volt source. The…

A series LCR circuit is connected to a 45sin⁡(ωt)45 \sin (\omega t) Volt source. The resonant angular frequency of the circuit is 105 rad s−110^5 ~\mathrm{rad}~ \mathrm{s}^{-1} and current amplitude at resonance is I0I_0 . When the angular frequency of the source is ω=8×104 rad s−1\omega=8 \times 10^4 ~\mathrm{rad} ~\mathrm{s}^{-1} , the current amplitude in the circuit is 0.05I00.05 I_0 . If L=50 mHL=50 ~\mathrm{mH} , match each entry in List-I with an appropriate value from List-II and choose the correct option.

List - I List - II
(P) I0I_0 in mA\mathrm{mA} (1) 44.4
(Q) The quality factor of the circuit (2) 18
(R) The bandwidth of the circuit in rad s−1\mathrm{rad}~ \mathrm{s}^{-1} (3) 400
(S) The peak power dissipated at resonance in Watt (4) 2250
(5) 500

Multiple Choice Options

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

Step-by-Step Verified Solution

amp;ω0=105rads−1,L=50×10−3Hamp;⇒XL=5000Ω=XC (at resonance) amp;1LC=ω0amp;⇒LC=1ω02amp;⇒C=2×10−9 Famp;I0=ε0R2+(XL−XC)2amp;I0=ε0R..........(i)\begin{aligned} & \omega_0=10^5 \mathrm{rad} \mathrm{s}^{-1}, L=50 \times 10^{-3} \mathrm{H} \\\\ & \Rightarrow X_L=5000 \Omega=X_C \text { (at resonance) } \\\\ & \frac{1}{\sqrt{L C}}=\omega_0 \\\\ & \Rightarrow L_C=\frac{1}{\omega_0^2} \\\\ & \Rightarrow C=2 \times 10^{-9} \mathrm{~F} \\\\ & I_0=\frac{\varepsilon_0}{\sqrt{R^2+\left(X_L-X_C\right)^2}} \\\\ & I_0=\frac{\varepsilon_0}{R} ..........(i) \end{aligned}

0.05I0amp;=ε0R2+(XL−XC)2..........(ii) At ωamp;=8×104rads−1XLamp;=8×104×50×10−3=4000ΩXCamp;=1ωC=6250Ω\begin{aligned}0.05 I_0 & =\frac{\varepsilon_0}{\sqrt{R^2+\left(X_L-X_C\right)^2}} ..........(ii) \\\\\text { At } \omega & =8 \times 10^4 \mathrm{rad} \mathrm{s}^{-1} \\\\X_L & =8 \times 10^4 \times 50 \times 10^{-3}=4000 \Omega \\\\X_C & =\frac{1}{\omega_C}=6250 \Omega\end{aligned}

From (i) and (ii), we get

amp;R2+(XL−XC)2=R0.05amp;R2+(6250−4000)2=400R2amp;⇒R2≃(2250)2400amp;R=225020=112.5Ωamp;I0=ε0R=45112.5amp;=400 m A\begin{aligned}& \sqrt{R^2+\left(X_L-X_C\right)^2} =\frac{R}{0.05} \\\\& R^2+(6250-4000)^2 =400 R^2 \\\\& \Rightarrow R^2 \simeq \frac{(2250)^2}{400} \\\\& R =\frac{2250}{20}=112.5 \Omega \\\\&I_0 =\frac{\varepsilon_0}{R}=\frac{45}{112.5} \\\\& =400 \mathrm{~m} \mathrm{~A}\end{aligned}

(P→3)(\mathrm{P} \rightarrow 3)

→Qamp;=1RLC⇒Qamp;=22250×50×10−32×10−7=104225amp;=44.4(Q→1)Qamp;=ω0BW\begin{aligned} \rightarrow Q & =\frac{1}{R} \sqrt{\frac{L}{C}} \\\\ \Rightarrow Q & =\frac{2}{2250} \times \sqrt{\frac{50 \times 10^{-3}}{2 \times 10^{-7}}}=\frac{10^4}{225} \\\\ & =44.4\\\\ (\mathrm{Q} \rightarrow 1) \\\\ Q & =\frac{\omega_0}{B W}\end{aligned}

⇒BW=ω0Q=105104225=2250\Rightarrow B W=\frac{\omega_0}{Q}=\frac{10^5}{\frac{10^4}{225}}=2250

(R→4)(\mathrm{R} \rightarrow 4)

P=ε02R (peak power) =452112.5=18 W\begin{array}{rrr}P =\frac{\varepsilon_0^2}{R} \text { (peak power) } \\\\=\frac{45^2}{112.5}=18 \mathrm{~W} \end{array}

(S→2)(S \rightarrow 2)

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