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The capacitor of capacitance C can be charged (with the help o

Question 1-3 are based on the following passage

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The capacitor of capacitance $C$ can be charged (with the help of a resistance $R$ ) by a voltage source $V$ , by closing switch $S_{I}$ while keeping switch $S_{2}$ open. The capacitor can be connected in series with an inductor $L$ by closing switch $S_{2}$ and opening $S_{1}$ .

Name: Video solution 1: The capacitor of capacitance C can be charged (with the help of a resistance R ) by a voltage source V, by closing switch SI while keeping switch S2 open. The capacitor can be connected in series with an inductor L by closing switch S2 and opening S1.
Uploaded: 2023-09-17T11:38:07.000Z
Duration: 4 min
Description: Video solution 1: The capacitor of capacitance C can be charged (with the help of a resistance R ) by a voltage source V, by closing switch SI while keeping switch S2 open. The capacitor can be connected in series with an inductor L by closing switch S2 and opening S1.

Updated on: Sep 17, 2023

Exam:

JEE Advanced 2018

Hard

Initially, the capacitor was uncharged. Nowwitch $S_{1}$ is closed and $S_{2}$ is kept open. If time constant of this circuit is $τ$ , then

after time interval

τ

, charge on the capacitor is

C V /2

after time interval

2 τ

, charge on the capacitor is

C V (1 - e^{- 2})

the work done by the voltage source will be half of the heat dissipated when the capacitor is fully charged

after time interval

2 τ

, charge on the capacitor is

C V (1 - e^{- 1})

Hard

After the capacitor gets fully charged, $S_{1}$ is opened and $S_{2}$ is closed so that the inductor is connected in series with the capacitor. Then,

t = 0

, energy stored in the circuit is purely in the form of magnetic energy.

at any time

t > 0

, current in the circuit is in the same direction.

t > 0

, there is no exchange of energy between the inductor and capacitor.

at any time

t > 0

, maximum instantaneous current in the circuit may be

V \frac{C}{L}

Hard

If the total charge stored in the $L C$ circuit is $Q_{0}$ , then for $t \geq 0$ ,

the charge on the capacitor is

Q = Q_{0} cos (\frac{π}{2} + \frac{t}{L C})

the charge on the capacitor is

Q = Q_{0} cos (\frac{π}{2} - \frac{t}{L C})

the charge on the capacitor is

Q = - L C \frac{d ^{2} Q}{d t ^{2}}

the charge on the capacitor is

Q = - \frac{1}{L C} \frac{d ^{2} Q}{d t ^{2}}

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Liam Discussed

The capacitor of capacitance

C

can be charged (with the help of a resistance

R

) by a voltage source

V

, by closing switch

S_{I}

while keeping switch

S_{2}

open. The capacitor can be connected in series with an inductor

L

by closing switch

S_{2}

and opening

S_{1}

9 mins ago

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Text solutionVerified

1. Charge on capacitor at time

t

q = q_{0} (1 - e^{- t / τ}) Here, q_{0} = C V and t = 2 τ ∴ q = C V (1 - e^{- 2 τ / τ}) = C V (1 - e^{- 2})

2. From conservation of energy,

\frac{1}{2} L I_{m a x}^{2} ∴ I_{m a x} = \frac{1}{2} C V^{2} = V \frac{C}{L}

3. Comparing the

L - C

oscillations with normal SHM, we get

\frac{d ^{2} Q}{d t ^{2}} Here, ω^{2} ∴ Q = - ω^{2} Q = \frac{1}{L C} = - L C \frac{d ^{2} Q}{d t ^{2}}

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Question Text	The capacitor of capacitance $C$ can be charged (with the help of a resistance $R$ ) by a voltage source $V$ , by closing switch $S_{I}$ while keeping switch $S_{2}$ open. The capacitor can be connected in series with an inductor $L$ by closing switch $S_{2}$ and opening $S_{1}$ .
Updated On	Sep 17, 2023
Topic	Alternating Current
Subject	Physics
Class	Class 12
Answer Type	Video solution: 1
Upvotes	75
Avg. Video Duration	4 min

The capacitor of capacitance $C$ can be charged (with the help of a resistance $R$ ) by a voltage source $V$ , by closing switch $S_{I}$ while keeping switch $S_{2}$ open. The capacitor can be connected in series with an inductor $L$ by closing switch $S_{2}$ and opening $S_{1}$ .

Initially, the capacitor was uncharged. Nowwitch $S_{1}$ is closed and $S_{2}$ is kept open. If time constant of this circuit is $τ$ , then

After the capacitor gets fully charged, $S_{1}$ is opened and $S_{2}$ is closed so that the inductor is connected in series with the capacitor. Then,

If the total charge stored in the $L C$ circuit is $Q_{0}$ , then for $t \geq 0$ ,

Filo tutor solutions (1)

Text solutionVerified

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The capacitor of capacitance C can be charged (with the help of a resistance R ) by a voltage source V, by closing switch SI ​ while keeping switch S2​ open. The capacitor can be connected in series with an inductor L by closing switch S2​ and opening S1​.

Initially, the capacitor was uncharged. Nowwitch S1​ is closed and S2​ is kept open. If time constant of this circuit is τ, then

After the capacitor gets fully charged, S1​ is opened and S2​ is closed so that the inductor is connected in series with the capacitor. Then,

If the total charge stored in the LC circuit is Q0​, then for t≥0,

Filo tutor solutions (1)

Text solutionVerified

Questions from JEE Advanced 2018 - PYQs

Practice questions from Alternating Current in the same exam

Practice more questions from Alternating Current

Practice questions on similar concepts asked by Filo students

Related books for questions on Alternating Current

The capacitor of capacitance $C$ can be charged (with the help of a resistance $R$ ) by a voltage source $V$ , by closing switch $S_{I}$ while keeping switch $S_{2}$ open. The capacitor can be connected in series with an inductor $L$ by closing switch $S_{2}$ and opening $S_{1}$ .

Initially, the capacitor was uncharged. Nowwitch $S_{1}$ is closed and $S_{2}$ is kept open. If time constant of this circuit is $τ$ , then

After the capacitor gets fully charged, $S_{1}$ is opened and $S_{2}$ is closed so that the inductor is connected in series with the capacitor. Then,

If the total charge stored in the $L C$ circuit is $Q_{0}$ , then for $t \geq 0$ ,