Engineering Thevenin equivalent circuit with dependent source

[bnosam]

Homework Statement

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

I decided to try using mesh currents for thisFirst mesh is I_A, second I_B, third, I_CSo I have the equations

I_A = 500 μA

1310 I_B + 100(I_A - I_B) = - 4*10^-5V₂

I_C = -80I_BAre these equations correct so far? I get the feeling they're not

 

[gneill]

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?

 

[bnosam]

I'm solving for voltage at this point so I can find the thevenin voltage.

For the second equation would it be:

1310 I_B + 100(I_B - I_A) = -4*10^-5 V₂

Or did you mean I have the sign for V₂ wrong

 

[gneill]

I'm solving for voltage at this point so I can find the thevenin voltage.

For the second equation would it be:

1310 I_B + 100(I_B - I_A) = -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 :)

 

[bnosam]

1410 I_B - 5000*10^-6 = -4*10^-5

So now I need an equation for V₂:

(-80I_B )*(50000)
= -4 000 000 I_Bso i plug it into the equation:

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

I_B = 4*10^-6 I

Does my conclusion so far seem correct?

I appreciate your help :)

 

[gneill]

1410 I_B - 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:

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

Okay. Fix your second term above and re-solve.

 

[bnosam]

1410 I_B - 50000*10-6 = -4*10-5 V₂

V₂ = (-80I_B)(50000 Ω)

so I_B = 4 * 10^-5

 

[gneill]

1410 I_B - 50000*10-6 = -4*10-5 V₂

V₂ = (-80I_B)(50000 Ω)

so I_B = 4 * 10^-5

Looks good. So What's v2?

 

[bnosam]

V₂ = - 160V

 

[gneill]

V₂ = - 160V

Also looks good.

Now, how are you going to approach the Thevenin resistance?

 

[bnosam]

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?

 

[gneill]

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?

 

[bnosam]

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.

 

[gneill]

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 V_th/I_N. 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?