JEE Advance - Physics (2023 - Paper 1 Online - No. 16)
A series LCR circuit is connected to a $45 \sin (\omega t)$ Volt source. The resonant angular frequency of the circuit is $10^5 ~\mathrm{rad}~ \mathrm{s}^{-1}$ and current amplitude at resonance is $I_0$. When the angular frequency of the source is $\omega=8 \times 10^4 ~\mathrm{rad} ~\mathrm{s}^{-1}$, the current amplitude in the circuit is $0.05 I_0$. If $L=50 ~\mathrm{mH}$, match each entry in List-I with an appropriate value from List-II and choose the correct option.
| List - I | List - II |
|---|---|
| (P) $I_0$ in $\mathrm{mA}$ | (1) 44.4 |
| (Q) The quality factor of the circuit | (2) 18 |
| (R) The bandwidth of the circuit in $\mathrm{rad}~ \mathrm{s}^{-1}$ | (3) 400 |
| (S) The peak power dissipated at resonance in Watt | (4) 2250 |
| (5) 500 |
$P \rightarrow 2, Q \rightarrow 3, R \rightarrow 5, S \rightarrow 1$
$P \rightarrow 3, Q \rightarrow 1, R \rightarrow 4, S \rightarrow 2$
$P \rightarrow 4, Q \rightarrow 5, R \rightarrow 3, S \rightarrow 1$
$P \rightarrow 4, Q \rightarrow 2, R \rightarrow 1, S \rightarrow 5$
Explanation
$\begin{aligned} & \omega_0=10^5 \mathrm{rad} \mathrm{s}^{-1}, L=50 \times 10^{-3} \mathrm{H} \\\\ & \Rightarrow X_L=5000 \Omega=X_C \text { (at resonance) } \\\\ & \frac{1}{\sqrt{L C}}=\omega_0 \\\\ & \Rightarrow L_C=\frac{1}{\omega_0^2} \\\\ & \Rightarrow C=2 \times 10^{-9} \mathrm{~F} \\\\ & I_0=\frac{\varepsilon_0}{\sqrt{R^2+\left(X_L-X_C\right)^2}} \\\\ & I_0=\frac{\varepsilon_0}{R} ..........(i) \end{aligned}$
$$ \begin{aligned} 0.05 I_0 & =\frac{\varepsilon_0}{\sqrt{R^2+\left(X_L-X_C\right)^2}} ..........(ii) \\\\ \text { At } \omega & =8 \times 10^4 \mathrm{rad} \mathrm{s}^{-1} \\\\ X_L & =8 \times 10^4 \times 50 \times 10^{-3}=4000 \Omega \\\\ X_C & =\frac{1}{\omega_C}=6250 \Omega \end{aligned} $$
From (i) and (ii), we get
$$ \begin{aligned} & \sqrt{R^2+\left(X_L-X_C\right)^2} =\frac{R}{0.05} \\\\ & R^2+(6250-4000)^2 =400 R^2 \\\\ & \Rightarrow R^2 \simeq \frac{(2250)^2}{400} \\\\ & R =\frac{2250}{20}=112.5 \Omega \\\\ &I_0 =\frac{\varepsilon_0}{R}=\frac{45}{112.5} \\\\ & =400 \mathrm{~m} \mathrm{~A} \end{aligned} $$
$(\mathrm{P} \rightarrow 3)$
$\begin{aligned} \rightarrow Q & =\frac{1}{R} \sqrt{\frac{L}{C}} \\\\ \Rightarrow Q & =\frac{2}{2250} \times \sqrt{\frac{50 \times 10^{-3}}{2 \times 10^{-7}}}=\frac{10^4}{225} \\\\ & =44.4\\\\ (\mathrm{Q} \rightarrow 1) \\\\ Q & =\frac{\omega_0}{B W}\end{aligned}$
$\Rightarrow B W=\frac{\omega_0}{Q}=\frac{10^5}{\frac{10^4}{225}}=2250$
$(\mathrm{R} \rightarrow 4)$
$\begin{array}{rrr}P =\frac{\varepsilon_0^2}{R} \text { (peak power) } \\\\ =\frac{45^2}{112.5}=18 \mathrm{~W} \end{array}$
$(S \rightarrow 2)$
$$ \begin{aligned} 0.05 I_0 & =\frac{\varepsilon_0}{\sqrt{R^2+\left(X_L-X_C\right)^2}} ..........(ii) \\\\ \text { At } \omega & =8 \times 10^4 \mathrm{rad} \mathrm{s}^{-1} \\\\ X_L & =8 \times 10^4 \times 50 \times 10^{-3}=4000 \Omega \\\\ X_C & =\frac{1}{\omega_C}=6250 \Omega \end{aligned} $$
From (i) and (ii), we get
$$ \begin{aligned} & \sqrt{R^2+\left(X_L-X_C\right)^2} =\frac{R}{0.05} \\\\ & R^2+(6250-4000)^2 =400 R^2 \\\\ & \Rightarrow R^2 \simeq \frac{(2250)^2}{400} \\\\ & R =\frac{2250}{20}=112.5 \Omega \\\\ &I_0 =\frac{\varepsilon_0}{R}=\frac{45}{112.5} \\\\ & =400 \mathrm{~m} \mathrm{~A} \end{aligned} $$
$(\mathrm{P} \rightarrow 3)$
$\begin{aligned} \rightarrow Q & =\frac{1}{R} \sqrt{\frac{L}{C}} \\\\ \Rightarrow Q & =\frac{2}{2250} \times \sqrt{\frac{50 \times 10^{-3}}{2 \times 10^{-7}}}=\frac{10^4}{225} \\\\ & =44.4\\\\ (\mathrm{Q} \rightarrow 1) \\\\ Q & =\frac{\omega_0}{B W}\end{aligned}$
$\Rightarrow B W=\frac{\omega_0}{Q}=\frac{10^5}{\frac{10^4}{225}}=2250$
$(\mathrm{R} \rightarrow 4)$
$\begin{array}{rrr}P =\frac{\varepsilon_0^2}{R} \text { (peak power) } \\\\ =\frac{45^2}{112.5}=18 \mathrm{~W} \end{array}$
$(S \rightarrow 2)$
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