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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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