# Need help figuring this thevenin equivalent out

Discussion in 'Homework Help' started by Kevin2341, Mar 3, 2013.

1. ### Kevin2341 Thread Starter Member

Nov 1, 2009
19
0
Hey everyone,

I'm struggling pretty bad with my circuits homework and we are currently going over Thevenin Equivalents. I understand the basic theory of it all, a complicated circuit may be reduced into a simple voltage and series resistor, or into a current source and parallel resistor (norton equivalent).

What I really am struggling with though is all the different analysis methods. We've learned nodal, mesh, and source transformation. We've also covered supermeshes and superposition. I can never figure out when I need to use one method over the other, except source transforms, those are fairly easy.

Anyway, I'm dealing with this circuit...

I'm playing around with the circuit in Circuitlab, and I've pretty much decided I have to leave the circuit as is (I cannot combine any resistors or perform any source transformations)

and now I'm stuck. It goes back to the nodal analysis or mesh analysis. Only problem is, I have no friggen' idea what to do. I have come to the conclusion with nodal analysis, you typically are finding the voltage at each node, and with mesh analysis, you are finding the current in a mesh.

I have seen methods where people will short circuit their active elements (voltage and current sources), but I'm still pretty unsure on that also...

Anyway... I'm going to give that a shot, if I disconnect my sources I get...

So from here, if my thought process is correct, I can say my only valid resistors left are: R13, 14, and 15, which from there, it will be 1\16000 + 1\1000 = 16000\17 = 941.2 Ohms...

From that I could say that my Rth = 941.2 Ohms?

So continueing on, I need to find my Vth, which I can say the node where the current source, R13, and R14 meet up, is the same voltage that will be going into the load. I can calculate this voltage using my Rth, and the current that would be going into the load. My current is going to be some mess of values from the resistors, voltage and current source, and I'll be honest, the two resistors on the left hand side of the circuit are completely screwing me up because I don't know what the heck to do with them. I know their current is 5mA (I = 10\2000), and I proved that on my software. But I really do not understand where to proceed from here.

2. ### WBahn Moderator

Mar 31, 2012
23,384
7,096
You are confusing the notion of "turning off a source" with "removing a source". When you turn off a source, you reduce it's output to zero. For a voltage source, this means reducing its output to 0V. In order to maintain 0V across the terminals, the source has to let any amount of current flow through it. This is equivalent to replacing the source with a short circuit. For a current source, on the other hand, reducing the output to zero means setting the current to 0A, which means that it won't let any current through it and must, therefore, allow any voltage to appear across it. This is the same as replacing it with an open circuit.

In your circuit, you have set the current source to zero but you should have replaced the voltage source with a short. It turns out not to matter, though, because the current source servies to isolate everything to the right of it from everything to the left of it, as your diagram shows, since you need two connections between left and right to form a circuit. Thus the entire left-hand side of the circuit can be ignored (as far as its effect on a load resistor placed between the terminals on the far right side).

3. ### Kevin2341 Thread Starter Member

Nov 1, 2009
19
0
Thanks for the speedy reply! So then am I correct in saying that my Rth is going to be the "bottom resistor" (1k), be parallel with the "middle" and "top" resistors (15000, and 1000 in series)? (941.2Ohms was my calculated number off of that.)

4. ### WBahn Moderator

Mar 31, 2012
23,384
7,096
That would be correct if you can show that an electron, in going from one side of the load terminals to the other, can either go through the "bottom resistor" (much better to refer to it as R15, at first I though you were talking about R12) OR it can go through R13 and R14. But that's not how it looks to me. Hop on an electron and trace out the paths that are available.

5. ### Kevin2341 Thread Starter Member

Nov 1, 2009
19
0
Ok, I see what you're saying. Basically there are two paths one can take. The top terminal to the bottom, or vice versa. So my resistance would actually be 17k Ohms then?

6. ### WBahn Moderator

Mar 31, 2012
23,384
7,096
Correct, though generally you pick one node as the departure node and only look for paths to the other node. You don't worry about the other way around

7. ### Kevin2341 Thread Starter Member

Nov 1, 2009
19
0
Ok. Any pointers for me to begin with the next part now that I know my Rth to be 17k?

8. ### WBahn Moderator

Mar 31, 2012
23,384
7,096
Use any analysis method you want to find the open circuit voltage across the load terminals or the short circuit current through the load terminals. Either will do.

You've already started down the to using superposition, so why not stick with that? Turn on one source at a time and find the open circuit voltage (or, again, the short circuit current, your choice). And then add the results.

9. ### Kevin2341 Thread Starter Member

Nov 1, 2009
19
0
Ok! I'll try that, I've heard of that one before but totally forgot about it.

10. ### Kevin2341 Thread Starter Member

Nov 1, 2009
19
0
OK, so here's my further attempt at this using superposition to solve this.

I = I' + I"

I' will be my current source "off"
I" will be my voltage source short circuited

Because of where the voltage source is, no current will be flowing through the area of interest, so thus, I' = 0.

Now with I", from my SPICE program, I discovered that I can effectively eliminate ALL resistors that do not have an effect on the local current effecting my load. So therefore, I am left with only my Rth resistors (1k, 1k, 15k). Using current division I can succesfully derive the current through that branch, which is 88.24mA.

V = 0.08824*17000 = 1500

11. ### WBahn Moderator

Mar 31, 2012
23,384
7,096
Don't use a SPICE program to do your thinking for you.

By inspection you can see that the short circuit load current when only the voltage source is turned on will be zero.

By current division you can get that the short-circuit load current when only the current source is on will be:

Iss = 100mA(15kΩ)/(2kΩ+15kΩ) = 88.2mA

So you got the correct answer. However, don't be so sloppy with your units. Get in the habit of tracking your units through your work and checking that they work out correctly. Do NOT just tack them on at the end.

Now, notice that there is a much simpler way to get the answer.

You already have that Rth = 17kΩ.

What is the open circuit load voltage? The entire 100mA has to go down through the 15kΩ resistor giving 1500V across it. There is no current in either of the 1kΩ resistors and, hence, no voltage drop across either. So the Thevenin voltage is 1500V and the Thevenin resistance is 17kΩ, both obtainable by inspection from the original circuit.