# Homework Help: Thevenin Equivalent

1. Jan 30, 2016

### Marcin H

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

2. Relevant equations
V=IR

3. The attempt at a solution
I am not sure if I did this problem correctly. To find the thevenin equivalent do you have to get the circuit into series first? I combined the resistors in parallel and found the Vth to be 6V and Rth to be 8ohms. Is that correct? If so, the part I am stuck on is what the IV graph for this circuit will look like. Would it just be a straight line through the origin? y=x basically?

2. Jan 30, 2016

### Staff: Mentor

Your Thevenin values look good. But I suspect that you should have a better handle on the procedure for finding the Thevenin equivalent. When you ask, "To find the thevenin equivalent do you have to get the circuit into series first?", it sounds like you are close but maybe just missing the idea.

For the Thevenin voltage you remove any load and find the voltage that appears across the open output terminals.

For the Thevenin resistance you suppress any sources in the circuit and determine the net (or equivalent) resistance of the network from the point of view of the output terminals.

Suppressing the sources often creates new series or parallel combinations that you can exploit in the reduction of the circuit down to a single resistance value. In this case you correctly determined that all three resistors end up in parallel.

The equivalent circuit model is then comprised of the Thevenin voltage in series with the Thevenin resistance.

For your IV curve, have you drawn the Thevenin model yet? What's the maximum voltage you can ever see at its output? Under what conditions? What's the maximum current that it can provide? Under what conditions?

3. Jan 30, 2016

### Marcin H

I have not drawn a thevenin model yet. What is that exactly? I just learned about thevenin and it hasn't all sunk in yet. Also, wouldn't the thevenin voltage (6V) be the maximum voltage output? And for the current I=Vth/Rth, so I = 6V/8ohms = .75A. Is that right?

4. Jan 30, 2016

### Staff: Mentor

You should check your class notes or text, or do a web search on Thevenin Equivalent to find a description.
Yes that's right. What are the corresponding current and voltage at the output for those values? When the output voltage is maximum, what is the load current? When the maximum current is being drawn, what is the voltage across the load terminals? In other words, what are the (I,V) pairs at the two extremes? Plot them first.

5. Jan 30, 2016

### Marcin H

I'm not to sure about this one. What do you mean exactly. Am I looking for the readings across points a and b? So looking at the Thevenin circuit I drew and knowing that max voltage is 6V and max current is .75A... how can I find what is read across those terminals? Do I just follow the current? coming out the battery I go across the 8ohm battery and lose 4.5V, so the reading will be 6V-4.5V = 1.5V?

6. Jan 30, 2016

### Staff: Mentor

The output at the terminals depends upon the load that is placed across the output terminals.

What is the output voltage when there is no load (open terminals)? What is the load current in that case?

What is the output current when the load is a short circuit (as you've drawn)? What is the output voltage across the terminals in that case?

Those are the two voltage/current pairs that are the extremes of what is called the "load line" for the circuit.

7. Jan 30, 2016

### Marcin H

I don't really understand what you're saying. What do you mean by open terminals. I thought what I had drawn was open terminals. If there are open terminals, doesn't that mean that you will have max voltage (6V) and no current flowing through
.75A and 0V?? I am very confused about what I am looking for exactly.

8. Jan 30, 2016

### Staff: Mentor

Yes! Exactly. So that's one pair.
Yes again! That's the other pair.

Those are the loading extremes that the circuit can see: Open circuit and short circuit. So you have (V,I) pairs:

(0.75A, 0V) and (0A, 6V)

Scale your IV plot so that these two points will lie near the ends of the I and V axes. They represent the extreme ends of the load line. You can fill in other points by working out the current required to produce a given output voltage, or vice versa, or you can reflect on why the "load line" is called a line and not a curve