# Why did the current go follow this branch instead the path of least resistance?

1. Mar 26, 2006

### mr_coffee

Its me again! I listend to what you said about right when the switch is closed there is a huge change in current so the inductor is going to have massive resistance, but in this case, the current still seemed to go directly into the inductor and not branch off into the 5 OHM resistor toward the powersource of 10v. I also don't understand how the 10v isn't affect anything. ANy ideas?

Thanks! this is the correct answer in the back of the book

2. Mar 26, 2006

### nrqed

Something does not make sense in the things you wrote.

2-2 =0 so your first equation for i_l(t) gives 2 at all times!

Also, notice that using the final result, you should get i(0) = 2 A but i(infinity) = 4A.

Also, where is the switch? What happens at t=0??

Patrick

3. Mar 26, 2006

### mr_coffee

Sorry about that, the book only says:
Find iL in the circuit at t equals to (a)-.5s (b) .5s, (c) 1.5s
I dind't draw the switch because the professor wants me (i'm the only kid in the class lol) to be able to analyze the circuit without haveing to redraw it. But i'll stick to redrawing it. Its a unit step function, so i guess you assume there is a switch there.

The solution manual shows this:

Here is the redraw:

yeah what i did was definatley wrong, that isn't a legal operation when ifactored out that 2...
Whoops!

Okay i think i got it now!
So iL(0) = 0;
iL(infinity) = 2A
iL = 2A -2A*e^-t/T

Wait a tic,
if i plug IL(0) and IL(infinity) into the equation above, i get
iL(t) = 2-2e^-t/T
factoring out a 2 gives:
iL(t) = 2(1-e^-t/T)

but the book has an extra 2+ out there, hm...

Last edited: Mar 26, 2006
4. Mar 26, 2006

### nrqed

Ah ok.. sorry, I missed the u(t) factor as a step function. So it is as if there was a switch in the middle branch.

Ok. Let`s say the middle branch is not there and the system has been like this for a long time. Then the inductor acts as a wire and the current is just 10V/5 ohms = 2A.

Now, the current source in the middle is turned on at t=0.
When I said that then the inductor acts as an infinite resistance, I means that it acts like this to any *change* (sorry, maybe I shoudl have been more clear). If there was no inductor at all in the left branch, the current there would jump right away to 4 amps (2 from the current source, 2 from the battery. None of the current from the current source goes in the branch of the battery since it has the choice to go in a branch with no resistance at all). But the inductor resists the change and keeps the current in the left branch to 2 A (for an infinitesimal time). Now, after a very long time, the current in the inductor will be 4 A because it will act as an ideal wire.

So maybe it woukd be better to say the following: if there is an inductor and there is already a current in the inductor at t=0 (when a change to the circuit is made) the inductor will react in such a way that the current in it will no change in a discontinuous way (which would be the case if instead of an inductor there would be an ideal wire in the left branch). So i(t=0 minus) = i(t=+) in the inductor.

SO now you know that I(t=0 minus) = t(0) = t(0+) = 2 A and I(t=infinity)= 4 A (where I is the current in the inductor).

You know that $I_l(t) = C_1 + C_2 e^{-t/\tau}$. Imposing the two conditions above fixes C_1 = 4 A and C_2 = -2 A.

5. Mar 26, 2006

### mr_coffee

Ooo i c what your saying now!
to make sure i'm getting this right when u said:
But the inductor resists the change and keeps the current in the left branch to 2 A (for an infinitesimal time).

So when the switch is thrown the 2 Amps in the middle doesn't affect the 2 amps already going into the inductor at i(0-), because it doesn't let it change to 4 Amps, but once its at time infinity, it will act as an ideal wire, thus letting the 2 Amps from the middle join the already 2 Amps flowing from the right branch? so at i(infinity) = 4A

When i plug that in i get:
iL(t) = 4 + [2 - 4]e^(-t/T);
iL(t) = 4 -2e^(-t/T);
iL(t) = 2[2 -e^(-t/T)]

But the book has a slightly different answer did i msss somthing up algebrically?

Last edited: Mar 26, 2006
6. Mar 26, 2006

### nrqed

Yes, more or less.
The only thing that I want to correct is that we know that there is a current of 2 A flowing through the inductor a t=0 minus, and at t = 0+. The inductor does not let the current through it jump in a discontinuous way, as you correctly said.
However, one shoudl not try to guess where this current is coming from without a more detailed study (for example, here one should not conclude necessarily that there is no current from the current source at t=0+ and all the current from the battery. Actually, the current source is producing a 2 A flowing up at t=0+ and we know that there is a current of 2 A floring through the inductor. By the junction rule, we are forced to conclude that at t=0+, there is no current being drawn from the battery!. So what happens is that the battery can't push any current in the source (the way it is connected) and the battery ''sees'' an inductor acting like an infinite resistance, so the battery can't push any current out. Then, as t goes to infinity, the current drawn from the battery goes back to 2 A and these 2 A get combined with the 2 A from the current source to give 4A in the inductor.)

My point is that the inductor will ''insure'' that the current will not have any discontinuous jump but what happens in the rest of the circuit is not always obvious to see at first glance.

Patrick

7. Mar 26, 2006

### mr_coffee

awsome explanation! you explain it better then my professor! hah thanks again!

8. Mar 26, 2006

### nrqed

Thanks!!!

But notice that you gained alot from the explanatiosn because you already had done a lot of work and had thought about all this. And you understand quickly...So kuddos to you!

Patrick

9. Mar 26, 2006

### nrqed

Just write 2[2 - e^(-t/T)] = 4 - 2 e^(-t/T) = 2 + 2[1- e^(-t/T)]

That is the same as the book