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I have worked through a question and I am unsure if my answer is correct. It is a series RL circuit with a switch that closes at t=0, supplied by a dc source, which I am trying to derive an expression for the current i, when the switch is closed.

Something like the circuit below:

enter image description here

My answer is: i(t) = V/R e^(t/T)

But does this not represent the discharging current as an exponentional decay rather than a growth? I can show all my workings out if it is not correct.

I thought the current when the switch is pressed should be represented by:

i(t) = V/R ( 1 - e^(t/T) )

EDIT

The image below shows the only way I know how to derive an equation for the instantaneous current in this circuit:

enter image description here

Thanks

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    \$\begingroup\$ Neither of your proposed answers is correct. One predicts the current going to infinity as t increases, the other predicts current going to negative infinity. You're looking for a solution where the current goes to V/R as t goes to infinity. \$\endgroup\$ – The Photon Feb 4 at 19:25
  • \$\begingroup\$ This is the simplest RL circuit possible with the solution readily available at request. \$\endgroup\$ – Eugene Sh. Feb 4 at 19:35
  • \$\begingroup\$ @The Photon But surely you can derive an equation to give a current value at any time after t = 0+ ? \$\endgroup\$ – David777 Feb 4 at 20:13
  • \$\begingroup\$ The sign of the signal it's a sin! \$\endgroup\$ – Dirceu Rodrigues Jr Feb 4 at 20:24
  • \$\begingroup\$ @David, yes you can. I'm trying to give you a hint how to work out that equation without just giving it to you. The point of my comment is if you think about the limit as t goes to infinity, you can see that neither of the two equations you showed can be the correct equation. \$\endgroup\$ – The Photon Feb 4 at 20:52

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