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In the LR circuit shown, what is the var...

In the `LR` circuit shown, what is the variation of the current `I` as a function of time? The switch is closed at time `t=0` sec.

Text Solution

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The correct Answer is:
`-V/Re^(-Rt)/L`
Applying loop law for `CE_(2)DC`
`-2V+V+I_(2)R=0implies I_(2)=V/R …(1)`
Applying loop law for `ABCDE`
`-L(dI_(1))/(dt)-2V+3V-I_(1)R=0`

`L(dI_(1))/(dt)+I_(1)R=V`
`I_(1)=V/R(1-e^(-Rt//L))…(2)`
Applying junction law at `C`
`I_(1)=I+I_(2) implies I=I_(1)-I_(2)=- V/R e^(-(R//L)t)`
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Knowledge Check

  • 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

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

    A
    `(E)/(R_(1))`
    B
    `(E)/(R_(2))`
    C
    `(E)/(R_(1)+R_(2))`
    D
    `(E(R_(1)+R_(2)))/((R_(1)R_(2))`

  • In the circuit as shown in figure the switch is closed at t = 0 . At the instant of closing the switch

    A
    the battery delivers maximum current
    B
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    C
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    D
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