Thevenin equivalent circuit with dependent source

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  • Thread starter bnosam
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  • #1
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Homework Statement


http://oi57.tinypic.com/263tqvr.jpg

I decided to try using mesh currents for this


First mesh is IA, second IB, third, IC


So I have the equations

IA = 500 μA

1310 IB + 100(IA - IB) = - 4*10-5V2

IC = -80IB


Are these equations correct so far? I get the feeling they're not
 

Answers and Replies

  • #2
gneill
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Your second equation is not right given the current directions in the components in the loop. Consider the directions of the potential drops caused by each current in the components.

You should define your objective before starting to write equations. What is the goal of solving your loop equations? Are you heading for the Thevenin voltage or Thevenin resistance at this point?
 
  • #3
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I'm solving for voltage at this point so I can find the thevenin voltage.

For the second equation would it be:

1310 IB + 100(IB - IA) = -4*10-5 V2

Or did you mean I have the sign for V2 wrong
 
  • #4
gneill
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I'm solving for voltage at this point so I can find the thevenin voltage.

For the second equation would it be:

1310 IB + 100(IB - IA) = -4*10-5 V2

Or did you mean I have the sign for V2 wrong
No, your equation looks okay now. I was worried about the sign of the term for for the 100 Ω resistor. Keep going :)
 
  • #5
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1410 IB - 5000*10-6 = -4*10-5

So now I need an equation for V2:

(-80IB )*(50000)
= -4 000 000 IB


so i plug it into the equation:

1410 IB - 5000*10-6 = -4*10-5 *(4 000 000 IB)

IB = 4*10-6 I

Does my conclusion so far seem correct?

I appreciate your help :)
 
Last edited:
  • #6
gneill
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1410 IB - 5000*10-6 = -4*10-5
You seem to have lost an order of magnitude in the second term: 100 x 500 x 10-6 = 5 x 10-2 = 1/20 . You also dropped the v2 on the right hand side.
So now I need an equation for V2:

(-80IB )*(50000)
= -4 000 000 IB
Okay. Fix your second term above and re-solve.
 
  • #7
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1410 IB - 50000*10-6 = -4*10-5 V2

V2 = (-80IB)(50000 Ω)

so IB = 4 * 10-5
 
  • #8
gneill
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1410 IB - 50000*10-6 = -4*10-5 V2

V2 = (-80IB)(50000 Ω)

so IB = 4 * 10-5
Looks good. So What's v2?
 
  • #9
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V2 = - 160V
 
  • #10
gneill
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V2 = - 160V
Also looks good.

Now, how are you going to approach the Thevenin resistance?
 
  • #11
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Also looks good.

Now, how are you going to approach the Thevenin resistance?
Hook up a test charge on the outside of the circuit to find that out?
 
  • #12
gneill
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Hook up a test charge on the outside of the circuit to find that out?
You mean a voltage source? Yes, I suppose that would work.

I have another suggestion though. Do you know the way Thevenin and Norton models are related? If you could determine the short circuit current (at the output) fairly easily, would that help?
 
  • #13
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You mean a voltage source? Yes, I suppose that would work.

I have another suggestion though. Do you know the way Thevenin and Norton models are related? If you could determine the short circuit current (at the output) fairly easily, would that help?
Before dependent sources, the only way I've been shown how to do it was just by combining resistors. Then I was shown this whole Etest/Itest method and I have no idea how to grasp it really. I've had one example of it.
 
  • #14
gneill
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Okay, very briefly:
The Thevenin and Norton resistances for a given circuit have the same value. The Norton current is the current that flows when the output is shorted. The Thevenin voltage is the open circuit voltage at the output. You should be able to show, using simple Norton and Thevenin models, that the Thevenin (or Norton) resistance is given by Vth/IN. That is, the open circuit voltage divided by the short circuit current.

You have already determined the open circuit voltage. If you can find the short circuit current then you'll be in good shape. How does shorting the output simplify the circuit?
 

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