First order transient circuit

  • #1

Homework Statement


Captura de pantalla 2015-11-15 a las 14.01.12.png


It asks me to find io(t=0-), io(t=0+), and Vc(t=0-). C=100μF R= 2kΩ

Homework Equations


V=I*R, i(t)= i(∞)+[i(0+)-i(∞)]*e-t/τ, Vc(0-)=Vc(0+)

The Attempt at a Solution



[/B]I first tried to calculate Vc(0-) as it will be the same as Vc(0+), stating that at t=0- the capacitor acts as an open circuit because it's been fully charged (the switch has been off since -∞), and I think it's 24V but I'm not sure about that. I also calculated io(0-), which is 2A, but I couldn't go any further.

Thank you guys.
 

Answers and Replies

  • #2
gneill
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Can you show some details of your calculations for initial values of Vc and io?

To me, 24 V looks too low and 2 A looks to large...
 
  • #3
You're right about the io, I believe it should be 36-12=io(2000+3*2000+2*2000) io=2x10-3 A. For the Vc I just did 36-12, but now that I think of it maybe it's 32, as the voltage drops 4V in the first resistance. I struggle a lot to calculate voltages when the capacitor is in the middle of the circuit.
 
  • #4
gneill
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You're right about the io, I believe it should be 36-12=io(2000+3*2000+2*2000) io=2x10-3 A. For the Vc I just did 36-12, but now that I think of it maybe it's 32, as the voltage drops 4V in the first resistance.
Those are better values :smile: You should always lay out the equations, particularly to confirm any "intuitive" results.
 
  • #5
Those are better values :smile: You should always lay out the equations, particularly to confirm any "intuitive" results.
Yeah you're right, thanks. I'm now trying to calculate io(0+) and io(∞). For io(0+) I replaced the capacitor for a voltage source of 32V (assuming the other calculations were right) and applying KVL I figured that the intensity should be 32/6000= 5,3x10-3. But i'm not sure how the capacitor acts when t=∞ /:
 
  • #6
gneill
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At steady state, capacitor currents go to zero. Effectively they "look like" an open circuit to the rest of the circuit. So for t=∞ remove the capacitor and find the potential across the points where it was connected.

Fig1.png
 
  • #7
At steady state, capacitor currents go to zero. Effectively they "look like" an open circuit to the rest of the circuit. So for t=∞ remove the capacitor and find the potential across the points where it was connected.

View attachment 91864
Thank you very much @gneill, my teacher posted the solution and it turns out the second calculations I assumed (with your help) were right! :D Do you know any good books with this type of exercices so I can practice for my exam? Thank you again for your time.
 
  • #8
gneill
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Thank you very much @gneill, my teacher posted the solution and it turns out the second calculations I assumed (with your help) were right! :D Do you know any good books with this type of exercices so I can practice for my exam? Thank you again for your time.
Well done.

I don't know of any particular books, but a Google search on "first order circuits examples" will turn up lots of hits.
 

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