Home
Class 12
PHYSICS
In the circuit shown in fig, the time co...

In the circuit shown in fig, the time constant of the two braches are equal(=T). Then if the key S is closed at the instant t=0, the time in which the current in the circuit through the battery will rise to its final value of `(E//R)` will be, (assume that the internal resistance of the battery and of the connecting wire are negligible)

A

Instantly

B

`t=T/2`

C

`t=2T`

D

`t=oo`

Text Solution

Verified by Experts

The correct Answer is:
A
The (L-R) and (C-R) sections having the same time constant are in parallel.
The current through the R-L branch will be unaffacted by the parallel R-C branch
`i_(L)(t)=E/R [1-r^((-Rt)/(L))]` ...(i)
Similarly the current in the RC branch is given by
`i_(C)(t)=E/R e^((-t)/(RC))`...(ii)
Hence the current through the battery is
`i(t)=i_(L)(t)+i_(C)(t)=E/R[1-e^(-(R)/(C)t)+e^(-(t)/(RC))]`....(iii)
Time cosntant of RL branch is `tau_(L)=L/R` and the time constant of RC branch is `tau(C)=RC`.
If these are equal , `tau_(L)=tau_(C)=T`, equation (iii) becomes
`i(t)=E/R[1-e^(-(R)/(C)t)+e^(-(t)/(RC))]=E/R` for all value of tgto=0. Hence the result.
Promotional Banner

Topper's Solved these Questions

  • MISCELLANEOUS VOLUME 5

    CENGAGE PHYSICS|Exercise Multiple Correct|34 Videos

  • MISCELLANEOUS VOLUME 5

    CENGAGE PHYSICS|Exercise Linked Comprehension|84 Videos

  • MISCELLANEOUS VOLUME 3

    CENGAGE PHYSICS|Exercise True and False|3 Videos

  • Moving charges and magnetism

    CENGAGE PHYSICS|Exercise Question Bank|20 Videos

Similar Questions

Explore conceptually related problems

In adjacent circuit, switch S is closed at t = 0. The time at which current in the circuit becomes half of the steady current is

Consider the circuit shown in figure. The current through the battery a long time after the switch S is closed is:

In the circuit shown below, the key K is closed at t = 0. The current through the battery is

In the circuit shown in figure switch S is closed at time t=0 . The charge which passes through the battery in one time constant is

In the circuit shown below, the key K is closed at t = 0 . The current through the battery is

In th circuit the key (K) is closed at t=0 . Find the current in ampere through the key (K) at the instant t=10^(2)ln2 sec.

In the circuit shown in figure switch S is closed at time t=0 Current I from the battery at time t is given by

In the circuit shown, the switch 'S' is closed at t = 0 . Then the current through the battery steady state reached is

In the circuit shown in figure switch S is closed at time t=0, which statement is true after one time constant of L-R circuit?