Thévenin equivalent

1. Feb 20, 2014

Maylis

1. The problem statement, all variables and given/known data
Find the Thevenin equivalent circuit at terminals (a,b) of the circuit below:

2. Relevant equations

3. The attempt at a solution
my currents aren't adding up right. When I try to solve for Ix using my two equations that should find it, I get different answers.

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• 3.10 attempt 1.pdf
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2. Feb 20, 2014

The Electrician

When you add the 1 volt test voltage at terminals a-b, you no longer have 4 meshes; you now have 5.

Your top schematic shows 4 meshes, and this allows you to calculate Vth, because the 2Ω resistor has no effect when the output is open circuited. Vth is just the voltage across the 8Ω resistor.

But as soon as you add the 1 volt test voltage as in the bottom schematic, you have 5 meshes. You need the current in that 5th mesh to calculate Rth.

You could solve it with only 4 meshes by assuming a short across a-b; this then replaces the 8Ω resistor with the parallel combination of 2Ω and 8Ω. Solving your 4 equations would give you the current through that parallel combination. Then use the current divider rule to calculate the current through the 2Ω resistor; that would be the short circuit current through a-b.

If you solve this using the nodal method, you only have 3 equations to deal with. That's what I meant when I said "You usually can use any network analysis method. One method may be easier for a given circuit than another."

3. Feb 21, 2014

Maylis

I'm doing the proper amount of meshes, and now my voltage is high and the resistance seems way too high. I just put those equations into a 5x5 and 4x4 solver. Here is my 2nd attempt at it

Attached Files:

• 3.10 attempt 2.pdf
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210.2 KB
Views:
43
4. Feb 21, 2014

The Electrician

Your 4th equation is incorrect. Your units are wrong. I3 is a current and 8*I4 is a voltage.

What you need is to say I4-I3 = 2*Ix, then set Ix = I1-I2

Give that a try, and make the same correction in your 5 mesh equations.

Your 5th equation for the 5 mesh system should be -8*I4 + 10*I5 = -1

5. Feb 21, 2014

Maylis

With the dependent current source there I don't know how to express the voltage drop across it. I redid it and got 75.4 V. my first 3 equations are unchanged but the last two are now

2 i1 - 2 i2 + i3 - i4 = 0

then for the 5 mesh system similarily, I changed to the same and what you said about -8 i4 + 10i5 = -1

Last edited: Feb 21, 2014
6. Feb 21, 2014

The Electrician

The voltage across the dependent current source is the same as the voltage across the 8Ω resistor in the 4 mesh case; that would be 8*I4 in the 4 mesh case. Your last two equations look good now, but I don't get 75.4 volts for that voltage.

Show the 4 equations for the 4 mesh case and the values you get for the 4 currents, and I should be able to see what the difficulty is.

Also, what do you get for the currents in the 5 mesh case?