Home
Class 12
PHYSICS
The time constant of an LR circuit is 4...

The time constant of an LR circuit is 40 ms. The circuit is connected at t= 0 and the steady state current is found ot be 2.0 A. Find the current at (a) t= 10 ms (b) t =20ms, (c ) t =100ms and (d) t = 1 s.

Text Solution

Verified by Experts

Here `tau= 40ms , i_0 = 2A `
` (a) t = 10ms,`
` Since I = i_0 (1-e^(-t/tau)) `
` = 2 (1- e(-10/40 ) ) `
`=2((1-e^(-1//4)`
`=2(1-0.7788)=2(0.2211)`
`(b)`When t=20 ms`
`t=i_0(1-e^(-t/tau)) `
` = 2 (1- e(-100/40 ) ) `
` = 2 (1- e(-10/4 ) ) `
` = 2 (1-0.082)`
`=1.835~= 1.8A`
(c ) When t=1s
i=i_0(1-e^(-t//tau)`
= 2 (1- e(-10/40 xx10^(-3)) ) `
` = 2 (1- e(-10/400) ) `
` = 2 (1- e(-25 ) )=2xx1=2A. `
`.
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • ELECTROMAGNETIC INDUCTION

    HC VERMA|Exercise EXERCISE|9 Videos

  • ELECTROMAGNETIC INDUCTION

    HC VERMA|Exercise Objective 2|10 Videos

  • ELECTRIC FIELD AND POTENTIAL

    HC VERMA|Exercise Exercises|75 Videos

  • ELECTROMAGNETIC WAVES

    HC VERMA|Exercise Exercises|9 Videos

Similar Questions

Explore conceptually related problems

Switch is closed at t = 0 then the current in the circuit at t = L/2R is

An LR circuit has L = 1.0 H R =20 Omega . It is connected across an emf of 2.0 V at t=0 . Find di/dt at (a) t =100ms, (b) t =200 ms and (c )t = 1.0 s .

Knowledge Check

  • 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

    A
    `tau ln 2`
    B
    `(ln2)/(tau)`
    C
    `2 tau ln 2`
    D
    `(tau)/(2)ln2`

  • In given LR circuit the switch S is closed at time t=0 then

    A
    The ratio of induced emfs in the inductors of inductance L and 2L will be constant
    B
    The ratio of induced emfs in the inductor of inductance L and 2L will be time varying
    C
    the potential difference `V_(A)-V_(B)` increases with time
    D
    The potential difference `V_(A)-V_(B)` will be constant

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

    A
    `2 A`
    B
    `4A`
    C
    `6A`
    D
    `8A`

    • Similar Questions

    Explore conceptually related problems

    Two coil A and B have inductances 1.0 H and 2.0 H respectively. The resistance of each coil is 10 Omega . Each coil is connected to an ideal battery of emf 2.0 V at t = 0 Let i_A and i_B be the current in the two circuit at time t. Find the raito i_A/ I_B at (a) t=100ms, (b) t= 200 ms and (c ) t = 1 s.

    Switch S is cloesd in the circuit at time t = 0 . Find the current through the capacitor and the inductor at any time t

    The electric current in a circuit is given by i = 3t. Find the rms current for the period t = 0 to t = 2 sec

    In the circuit shown, switch S is closed at time t = 0 . Find the current through the inductor as a function of time t .

    An inductor-coil of inductance 20 mH having resistance 10 Omega is joined to an ideal battery of emf 5.0 V. Find the rate of change of the induced emf at t=0, (b) t = 10 ms and (c ) t = 1.0s.

    Home
    Profile