# Finding the Thevenin resistance

1. Oct 4, 2012

### uasaki

1. The problem statement, all variables and given/known data

The circuit diagram is attached below.
We're supposed to a) find the Thevenin equivalent circuit and b) find the value of Rl (the load resistor) so that the current flowing through it is 0.5A.

2. Relevant equations

3. The attempt at a solution

Since the circuit doesn't contain any dependent sources, we can zero out all the independent sources and calculate the Thevenin resistance directly. The 4A current source becomes an open circuit and the 48V voltage source becomes a short. I've attached a picture of what the resulting circuit should look like (it's the circuit2.png file). We can find the equivalent resistance by combining the resistances in series and parallel. Starting from the right, the 6-ohm and 10-ohm resistor are in series, so you can combine them to get a 16-ohm resistor. Then, that 16-ohm resistor is in parallel with the 8-ohm resistor, which is equivalent to a 5.3-ohm resistor. Finally, you can add that 5.3-ohm resistor with the two 4-ohm resistors in order to get an equivalent resistance of 13.3-ohms. Is this calculation correct?

Also, I tried to calculate the short-circuit current flowing from a to b, but I keep getting 0. Is it possible to get 0 for the Thevenin/Norton short-circuit current, or am I missing something?

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2. Oct 4, 2012

### Staff: Mentor

Oops! No, they are not in series so long as the output port 'a' is also connected at their junction. In fact the 10Ω resistor is in parallel with ab, and so is in parallel with the 'rest' of the network.

Often it's convenient to work from the 'back end' of a network towards the 'front' where the port that the resistance is being measured is located.

[snip]
It's possible to get any value, including zero, given the right circuit... in this case it's not.
You'll have to show your calculations if we're to see how to help.

3. Oct 4, 2012

### uasaki

Oh, ok. So, if we start at the end opposite the output ports, then the two 4-ohm resistors are in series, so we can combine them to get an 8-ohm resistor. Then, that 8-ohm resistor is in parallel with the 8-ohm resistor in the circuit, so we can combine them to get an 8-ohm resistor. That 8-ohm resistor is in series with the 6-ohm resistor, so we can combine them to get a 14-ohm resistor. That 14-ohm resistor is actually in parallel with the 10-ohm resistor to the right (because there's an output port connected at their junction), so we can combine them to get a
5.83-ohm resistor. Is this calculation correct?

As for the short-circuit current, I think I know what I did wrong. If we put a short between terminals a and b, then the 10-ohm resistor is bypassed, which means that we shouldn't include the 10-ohm resistor in our mesh calculations. (At least that's what I think I did incorrectly). I'm currently re-doing my calculations without the 10-ohm resistor.

4. Oct 4, 2012

### Staff: Mentor

Um, 8Ω || 8Ω ≠ 8Ω. Try once again; you're very close.

5. Oct 4, 2012

### uasaki

Oops, my bad. The two 8-ohm resistors in parallel should be produce a 4-ohm resistor. Was that the only mistake in my calculation?

6. Oct 4, 2012

### Staff: Mentor

Well, since it occurred early on it affected the following results. But otherwise your method is fine.

7. Oct 4, 2012

### uasaki

Thanks for the help. So for the short circuit current I placed a wire in-between terminals a and b. Then, I just used mesh analysis to solve for the circuit.

The KVL for mesh 1 is $$4I_{1} + 4(I_{1} - I_{3}) + 8(I_{1} - I_{2}) + 48 = 0 \Rightarrow 16I_{1} - 8I_{2} - 4I_{3} = -48$$

The KVL for mesh 2 is $$8(I_{2} - I_{1}) + 6(I_{2} - I_{3}) + V_{oc} - 48 = 0 \Rightarrow -8I_{1} + 14I_{2} = 28$$

And I_{3} = -4A.

I put these equations into wolfram alpha and I got $$I_{1} = -\frac{21}{5}, I_{2} = -\frac{2}{5}$$ The short-circuit current is just I2, but when I divide the open-circuit voltage, which is -4V (I calculated this with mesh analysis, and verified it by drawing the circuit with circuitlab) by I2, I get 10-ohms for the Thevenin resistance, which is different from the result I got by directly computing the equivalent resistance.

8. Oct 4, 2012

### Staff: Mentor

Recheck your KVL for loop 2. I suspect that you've erred on the voltage total.

9. Oct 4, 2012

### uasaki

Oh, wait. If we place a short from terminals a and b, then there is no voltage drop across that wire. So, we don't need to add $$V_{oc}$$ in mesh 2's KVL.

10. Oct 4, 2012

### Staff: Mentor

It's okay if you've set it to zero (it's shorted).