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If in the given circuit (L = 100 mH, R = 100 Ω, and E = 100 V), after being in steady state for a long time, A and B are connected, find the current flowing in the circuit after a millisecond.

To do this I used the current decay equation:

$$I=\frac{E}{R} (1-e^{\frac{-Rt}{L}})\rightarrow 2$$

Substituting into equation 2, I got \$\ I=1-\frac{1}{e}\$.

However, my book says that this is wrong (the answer is \$\frac{1}{e}\$).

I have looked online, and am unable to understand why. I'd really appreciate some help.

Edit:- As I have been asked to better describe the circuit at t<0 and t>0, I will. a t<0;A and B are connected to the voltage source, E. At t>0, A and B are directly connected to one another via a conducting wire.

math and physics forever
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  • What do you think the initial condition is just prior to connecting A to B? Does this connection bypass/short-out the voltage source that is showing? Or does it engage the voltage source where it wasn't, before? And what's the value of **E**? – jonk Oct 09 '22 at 06:32
  • Steady state. Short circuit. E= 100V. Sorry for leavinf that out didn't notice – math and physics forever Oct 09 '22 at 06:39
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    You are confusing the charge and decay equations. – Mattman944 Oct 09 '22 at 06:52
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    @mathandphysicsforever Just to be clear: So, at \$t=0^{-}\$ the current should be \$1\:\text{A}\$? – jonk Oct 09 '22 at 06:56
  • yes, if $\$0^{-}\$ means just before we short circuit@jonk – math and physics forever Oct 09 '22 at 06:58
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    @mathandphysicsforever How is it that you think that your current decay equation provides the value \$1\:\text{A}\$ at \$t=0\:\text{s}\$? It yields \$0\:\text{A}\$, doesn't it? Doesn't this violate the initial condition you know to be true? – jonk Oct 09 '22 at 07:25
  • @jonk, you're right, I forgot the one – math and physics forever Oct 09 '22 at 10:45
  • Sorry for my previous comments; I was plain wrong. @jonk is right, your “decay equation” yields \$I=0\$ at \$t=0\$, which is not consistent with your \$I=1A\$ at \$t=0^-\$. – user2233709 Oct 09 '22 at 12:12
  • Because equation 2 isn't the "current decay" equation. You need to think about why. Hint: As t increases, what happens to I? And is that "decay"? – Andrew Lentvorski Oct 10 '22 at 06:08

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