Find the Thevenin Equivalent Circuit

[zr95]

Homework Statement

Homework Equations

Node Voltage Method
V=IR

The Attempt at a Solution

KCL at Node 1:

KCL at Node 2:

i is the current moving towards "a"

I've simplified the two equations down but the problem I run into is that normally I'm able to write a third equation to define i₀. In this case there is no resistance. i₀=v1/R but since R=0 this would be undefined. Where do I move from here?
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[gneill]

Get creative

Insert a "temporary" resistor in the $i_0$ path, call it $R_x$. Solve the problem keeping $R_x$ as a variable. Then let $R_x → 0~Ω$.

By the way, your KCL for node 2 doesn't make sense with the v1 and v2 being summed as currents. I think you missed dividing by a resistance value...

 

[zr95]

I need to eliminate v1 when I'm solving the system. How can I do this with having v1/Rx now being involved? I can't eliminate this term or combine it with the other v1. If I find a common denominator to combine it I can't eliminate that whole thing.

 

[zr95]

If I were to solve this using KVL and I made 3 loops the first loop would give me a value i1. That value i1 would be the current across the 3kohm resistor. i2 would be the current across the 2kohm resistor. Can I consider that part of the wire with i₀ to be i1-i2? Or for the sake of my KCL equations (v1-15)/3k - (v2-v1)/2k = i₀?

 

[gneill]

I need to eliminate v1 when I'm solving the system. How can I do this with having v1/Rx now being involved? I can't eliminate this term or combine it with the other v1. If I find a common denominator to combine it I can't eliminate that whole thing.

Solve your v1 node equation for v1. Yes, it'll have both Rx and v2 variables in it. You'll substitute the whole lot into your v2 node equation to replace 'v1' there. Only after substitution will you let Rx go to zero.

 

[gneill]

If I were to solve this using KVL and I made 3 loops the first loop would give me a value i1. That value i1 would be the current across the 3kohm resistor. i2 would be the current across the 2kohm resistor. Can I consider that part of the wire with i₀ to be i1-i2? Or for the sake of my KCL equations (v1-15)/3k - (v2-v1)/2k = i₀?

Yes, if you're using mesh currents. Mesh currents sum through shared paths (i.e. on the "borders" of the loops.

 

[zr95]

I decided just to solve using mesh currents. My professor's supposed easy way to solve it without having to solve for Vth then Isc apparently wasn't all that much easier for this problem.

 

[gneill]

I decided just to solve using mesh currents. My professor's supposed easy way to solve it without having to solve for Vth then Isc apparently wasn't all that much easier for this problem.

I think that any way you approach this problem the algebra will not be pretty. The nodal analysis approach that I suggested works, but again the algebra will be a bit time consuming. If you show your complete solution I'll offer mine

 

[zr95]

My teacher mentioned in class the note I put in the top right corner which would make it easier to solve the whole thing in one go.

 

[The Electrician]

Your final result schematic showing a voltage Vth in series with Rth appears to have the sign of Vth reversed.

Using linear algebra techniques the problem can be solved in a compact manner using nodal analysis. Here's how it would be done if the wire carrying i0 is temporarily replaced with a resistor Rx. After the node voltages are found Rx is allowed to go to zero:

The problem can also be solved with mesh analysis. The wire carrying i0 can be treated like a current source, so the 3 meshes form a supermesh:

 

[gneill]

Okay, everyone's showing a solution and I did promise to show mine if the OP showed his. So here we go.

Insert $R_x$ in the $i_o$ path so that we can write node equation for node 1. Later we'll let this resistance go to zero. Also stick a load resistor $R_L$ on the output (across a-b). You'll see why in a bit.

Write the node equations:

Node 1:
$\frac{v1 - 15}{3000} + \frac{v1}{R_x} + \frac{v1 - v2}{2000} = 0$

and solving for v1: $~~~~v1 = \frac{3}{5} (10 + v2) \frac{R_x}{R_x + 1200}$

Node 2:
$\frac{v2 - v1}{2000} + 18\frac{v1}{R_x} + \frac{v2}{2000} + \frac{v2}{R_L} = 0$

Factor out v1, then substitute for v1 from the node 1 equation. Hit the whole thing with the algebra hammer until v2 is isolated. When the smoke clears:

$v2 = \frac{30 (R_x - 36000) R_L}{(7 R_x R_L + 120000 R_L + 10000 R_x + 12000000)}$

Now it's time to let $R_x$ go to zero:

$v2 = \frac{-1080000 R_L}{12000000 + 120000 R_L}$

Now make this look like a voltage divider equation. For a Thevenin model with a load it will resemble:

$V_{out} = V_{th} \frac{R_L}{R_L + R_{th}}$

Thus we have:

$v2 = \frac{-1080000}{120000} \frac{R_L}{R_L + \frac{12000000}{120000}}$

$v2 = -9 \frac{R_L}{R_L + 100}$

So the Thevenin voltage is -9 V and the Thevenin resistance is 100 Ω.