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

  • #1
sugz
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Homework Statement



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

Homework Equations



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?

problem.jpg
20150322_210402.jpg
Capture.PNG
 
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  • #2
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...
 
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  • #3
gneill said:
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...
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
sugz said:
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!
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 .
 
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  • #5
sugz said:
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?
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...
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!
If you have an RC circuit then ic(0+) would refer to the initial current that the capacitor sees after the switching event.
 
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  • #6
gneill said:
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.
Hello,
Sorry for the confusion. It is in fact an RC circuit. Thanks again for clarifying that! You are awesome!
 

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