# Finding the current Io for t>0 in a RL circuit

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1. Mar 22, 2015

### sugz

1. The problem statement, all variables and given/known data

For the circuit in the attached file, find io (current) for t>0

2. Relevant equations

3. The attempt at a solution

I have included an attachment which shows the solution. However, I will copy the solution here as well for reference:

When t<0, the switch is closed, the inductor acts as a short circuit to dc. The question I have in this step is what happens to the resistor that is in series with the inductor when the inductor becomes a short circuit?
The resulting circuit is one with the 24V source, 4 ohm resistor and the 4 ohm and 8 ohm resistors in parallel. The 4//8=8/3 ohms.

i1=24/(8/3+4)=3.6 A

We obtain i(t) from t1, using current division, by writing,

i(t)= 3.6*4/(4+8) = 1.2A, t<0

Since the current through the inductor cannot change instantaneously, i(0)=i(0-)=1.2A

When t=0, the switch is open and the voltage source is disconnected. Now we have a circuit with 4 ohm in series with the 4H inductor and the 8 ohm resistor.

Req=4+8=12 ohms
T(tow)=L/R=4/12=1/2 seconds
io(t)=io(0)e^(-t/tow)
io(t)=1.2e^-3t

The solution manual shows that io(0)=1.4118 A. I am not sure how they obtained it. I will post the solution for reference as well. Can someone tell me where I went wrong?

Last edited: Mar 22, 2015
2. Mar 22, 2015

### Staff: Mentor

Your first image shows a 4 Ω resistor in series with the 24 V supply, but the solution shows a 3 Ω resistor. This would make a difference...

3. Mar 22, 2015

### sugz

Wow, I am so sorry I missed that. Thanks a lot for pointing that out! Would all my steps be correct otherwise?

In addition, I have a question unrelated to this if I am allowed to ask this here. When a question states "initial charging current ic(0+)", what does this mean for an RL circuit? Is this when t=0 since this is the initial charging current? This is for a circuit where a switch closes, which provides a voltage source to a capacitor. Thanks again!

4. Mar 22, 2015

### SammyS

Staff Emeritus
Yes, if you re-do the problem using 3Ω rather than 4Ω, with the same steps you used, you get 1.41176... A for i0 .

5. Mar 23, 2015

### Staff: Mentor

If there's a current on the diagram labelled ic then it's that current the instant after t=0 (presumably the instant after some switch change or other event that alters the circuit at time t=0). If the "c" in "ic" refers to a capacitor, well there's no capacitor in an RL circuit...
If you have an RC circuit then ic(0+) would refer to the initial current that the capacitor sees after the switching event.

6. Mar 23, 2015

### sugz

Hello,
Sorry for the confusion. It is in fact an RC circuit. Thanks again for clarifying that! You are awesome!