Why am I getting NaN and INF values for my Norton Equivalent Circuit

Question

I am given this question

I have 2 questions:

1) Is my work wrong? I am assuming so since I am getting some funky answers. If I am wrong, will you please point out what I should have done differently?

2) Since he didn't add a ground I figured I would have to consider the entire bottom wire as + 0.2I_0 because of the dependent source, but would I also have to consider my external voltage (V_x) in some way? If so, how would I add it into my work?

Problems I had with this question:

1) There is no ground, I am not familiar with problems like this.

2) I was not sure what equations I needed for all my unknowns. The equation that feels most like it doesn't belong is Equation 2, but I could not find any better approach.

Thanks!

Here is my attempt to solve for R_th:

Answer 1

First of all, you can have "ground" wherever you want. In this case, "ground" is just a theoretical reference point for all of your voltages, a point at which you say the voltage (or potential difference) is 0 and all other voltages are compared to. The best place would certainly be node b.

The fact that you have a voltage source positioned "upside down" has no effect on this, it just means that the voltage on the other side of the source will be negative.

Next, you should look into superposition. It makes things pretty easy to solve. I would start with the known voltage first:

• Treat the dependent voltage as a short
• Find out the voltages at all the different points

^{simulate this circuit – Schematic created using CircuitLab}

• Lump all the resistances together, giving a single 4/31 ohm resistor

• Get the current through this due to your source, which is 7.75A

  -Now find the voltages due to the dependent voltage

^{simulate this circuit}

• The second diagram is just a rearrangement of the first, making it obvious that R4 and R3 form a potential divider. This means v2 is -1/4*0.2Io, and v1 is -0.2Io.

• Now we can sum the superpositions, giving a V2 of ((-Io/20) + 0.25Vin)

• The current through R4 is equal to Io, so it would be useful to get this. We know the voltage on both sides, so we just do Io=(0-((-Io/20) + 0.25Vin))/0.1
• After a bit of maths, we get Io=-5Vin
• This means that V1=Vin