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An L-C-R circuit is equivalent to a damped pendulum. In an L-C

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An $L - C - R$ circuit is equivalent to a damped pendulum. In an $L - C - R$ circuit, the capacitor is charged to $Q_{0}$ and then connected to the $L$ and $R$ as shown.

If a student plots graphs of the square of maximum charge $(Q_{m a x}^{2})$ on the capacitor with time $(t)$ for two different values $L_{1}$ and $L_{2} (L_{1} > L_{2})$ of $L$ , then which of the following represents this graph correctly? (plots are schematic and not drawn to scale)
(2015 Main)

Name: Video solution 1: An L−C−R circuit is equivalent to a damped pendulum. In an L−C−R circuit, the capacitor is charged to Q0 and then connected to the L and R as shown.If a student plots graphs of the square of maximum charge (Qmax2) on the capacitor with time (t) for two different values L1 and L2(L1>L2) of L, then which of the following represents this graph correctly? (plots are schematic and not drawn to scale)(2015 Main)
Uploaded: 2022-12-14T16:58:39.000Z
Duration: 3 min

Views: 5,972 students

Updated on: Dec 14, 2022

Exam:

JEE Advanced 2015

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Charlotte Discussed

L - C - R

circuit is equivalent to a damped pendulum. In an

L - C - R

circuit, the capacitor is charged to

Q_{0}

and then connected to the

L

and

R

as shown.

If a student plots graphs of the square of maximum charge

(Q_{m a x}^{2})

on the capacitor with time

(t)

for two different values

L_{1}

and

L_{2} (L_{1} > L_{2})

L

, then which of the following represents this graph correctly? (plots are schematic and not drawn to scale)
(2015 Main)

13 mins ago

Discuss this question LIVE

13 mins ago

Text solutionVerified

At a general time

t

, suppose charge on capacitor is

q

and current in the circuit is

i

, then applying Kirchhoff's loop law, we have

\frac{q}{C} - i R - L \frac{d i}{d t} = 0

Putting

i = - \frac{d q}{d t} and \frac{d i}{d t} = - \frac{d ^{2} q}{d t ^{2}}

In the above equation, we have

\frac{d ^{2} q}{d t ^{2}} + \frac{R}{L} (\frac{d q}{d t}) + \frac{q}{L C} = 0

Comparing this equation with the standard differential equation of damped oscillation,

\frac{d ^{2} X}{d t ^{2}} + \frac{b}{m} \frac{d X}{d t} + \frac{K}{m} X = 0

Which has a general solution of amplitude,

A = A_{0} e^{\frac{- b t}{2 m}}

The general solution of Eq. (i) will be

Q_{m a x} = Q_{0} e^{\frac{- Rt}{2 L}} or Q_{m a x}^{2} = Q_{0}^{2} e^{\frac{- Rt}{L}}

Hence,

Q_{m a x}^{2}

versus time graph is exponentially decreasing graph. Lesser the value of self inductance, faster will be the damping.

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Question Text	An $L - C - R$ circuit is equivalent to a damped pendulum. In an $L - C - R$ circuit, the capacitor is charged to $Q_{0}$ and then connected to the $L$ and $R$ as shown. If a student plots graphs of the square of maximum charge $(Q_{m a x}^{2})$ on the capacitor with time $(t)$ for two different values $L_{1}$ and $L_{2} (L_{1} > L_{2})$ of $L$ , then which of the following represents this graph correctly? (plots are schematic and not drawn to scale) (2015 Main)
Updated On	Dec 14, 2022
Topic	Alternating Current
Subject	Physics
Class	Class 12
Answer Type	Text solution:1 Video solution: 2
Upvotes	354
Avg. Video Duration	8 min

Text solutionVerified

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