JEE Advanced · Physics · 22. AC Circuits

The circuit shown in the figure contains an inductor $L$, a capacitor $C_0$, a resistor $R_0$ and an ideal battery. The circuit also contains two keys $\mathrm{K}_1$ and $\mathrm{K}_2$. Initially, both the keys are open and there is no charge on the capacitor. At an instant, key $K_1$ is closed and immediately after this the current in $R_0$ is found to be $I_1$. After a long time, the current attains a steady state value $I_2$. Thereafter, $\mathrm{K}_2$ is closed and simultaneously $\mathrm{K}_1$ is opened and the voltage across $C_0$ oscillates with amplitude $V_0$ and angular frequency $\omega_0$.

Match the quantities mentioned in List-I with their values in List-II and choose the correct option.

A $\mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 5$
B $\mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 2 ; \mathrm{R} \rightarrow 3 ; \mathrm{S} \rightarrow 5$
C $\mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 4$
D $\mathrm{P} \rightarrow 2 ; \mathrm{Q} \rightarrow 5 ; \mathrm{R} \rightarrow 3 ; \mathrm{S} \rightarrow 4$

Verified Solution

Answer & Solution

Correct Answer

(A) $\mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 5$

Step-by-step Solution

Detailed explanation

(P) When $K_1$ is closed current in $R_0$ is $I_1$ At $\mathrm{t}=0$; circuit will be

$\begin{aligned} & \mathrm{I}_1=0 \\ & \mathrm{P} \rightarrow(1)\end{aligned}$
(Q) After long time inductor behave as a wire so $\mathrm{I}_2$

$\begin{aligned} & \mathrm{I}_2=\frac{20}{5}=4 \mathrm{~A} \\ & \mathrm{Q} \rightarrow(3)\end{aligned}$
(R) When $\mathrm{K}_2$ is closed and $\mathrm{K}_1$ open

$\begin{aligned} & \omega_0=\frac{1}{\sqrt{\mathrm{LC}}} \\ & \omega_0=\frac{1}{\sqrt{25 \times 10^{-3} \times 10 \times 10^{-6}}}=\frac{1}{5 \times 10^{-4}} \\ & \omega_0=2 \times 10^3 \mathrm{rad} / \mathrm{s} \\ & \omega_0=2 \text { kilo-radian } / \mathrm{s} \\ & \mathrm{R} \rightarrow(2)\end{aligned}$
(S) Now $\mathrm{K}_2$ is closed and $\mathrm{K}_1$ open

$\begin{aligned} & \frac{1}{2} \mathrm{LI}_2^2=\frac{1}{2} \mathrm{CV}_0^2 \\ & 25 \times 10^{-3} \times(4)^2=10 \times 10^{-6} \times \mathrm{V}_0^2 \\ & \mathrm{~V}_0^2=2500 \times 16 \\ & \mathrm{~V}_0=50 \times 4=200 \mathrm{~V} \\ & \mathrm{~S} \rightarrow(5)\end{aligned}$

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Match List I with List II and select the correct answer using the codes given below the lists:

	$\text{List I}$		$\text{List II}$
a.	Boltzmann constant	p.	$\left[M L^2 T^{-1}\right]$
b.	Coefficient of viscosity	q.	$\left[M L^{-1} T^{-1}\right]$
c.	Planck constant	r.	$\left[M L T^{-3} K^{-1}\right]$
d.	Thermal conductivity	s.	$\left[M L^2 T^{-2} K^{-1}\right]$

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One mole of a monatomic ideal gas is taken along two cyclic processes $E \to F \to G \to E and E \to F \to H \to E$ as shown in the PV diagram. The processes involved are purely isochoric, isobaric, isothermal or adiabatic.

Match the paths in List I with the magnitudes of the work done in List II and select the correct answer using the codes given below the lists.
$\begin{matrix} List I & List II \\ A. & G → E & P . & 160 P_{0} V_{0} ln 2 \\ B. & G → H & Q . & 36 P_{0} V_{0} \\ C. & F → H & R . & 24 P_{0} V_{0} \\ D. & F → G & S . & 31 P_{0} V_{0} \end{matrix}$

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x = 0, y = 0, z = 0

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\vec{E}

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1. Electron with $\vec{v} = 2 \frac{E_{0}}{B_{0}} \hat{x}$	(i) $\vec{E} = E_{0} \hat{z}$	(P) $\vec{B} = - B_{0} \hat{x}$
2. Electron with $\vec{v} = \frac{E_{0}}{B_{0}} \hat{y}$	(ii) $\vec{E} = - E_{0} \hat{y}$	(Q) $\vec{B} = B_{0} \hat{x}$
3.Proton with $\vec{v}$ =0	(iii) $\vec{E} = {- E}_{0} \hat{x}$	(R) $\vec{B} = - B_{0} \hat{y}$
4. Proton with $\vec{v} = 2 \frac{E_{0}}{B_{0}} \hat{x}$	(iv) $\vec{E} = E_{0} \hat{x}$	(S) $\vec{B} = - B_{0} \hat{z}$

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In Circuit- $1$ and Circuit- $2$ shown in the figures, $R_{1} = 1 Ω, R_{2} = 2 Ω$ and $R_{3} = 3 Ω$ .
$P_{1}$ and $P_{2}$ are the power dissipations in Circuit- $1$ and Circuit- $2$ when the switches $S_{1}$ and $S_{2}$ are in open conditions, respectively.
$Q_{1}$ and $Q_{2}$ are the power dissipations in Circuit- $1$ and Circuit- $2$ when the switches $S_{1}$ and $S_{2}$ are in closed conditions, respectively.

Which of the following statement(s) is(are) correct?

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	\(\text{List I}\)		\(\text{List II}\)
a.	Boltzmann constant	p.	\(\left[M L^2 T^{-1}\right]\)
b.	Coefficient of viscosity	q.	\(\left[M L^{-1} T^{-1}\right]\)
c.	Planck constant	r.	\(\left[M L T^{-3} K^{-1}\right]\)
d.	Thermal conductivity	s.	\(\left[M L^2 T^{-2} K^{-1}\right]\)