Help Calculating Thevenin Resistance

[maherelharake]

Homework Statement

I attached my question below.

Homework Equations

The Attempt at a Solution

I believe V^th is just 10 V, but I am having trouble calculating R_th. Can someone help me get started? Thank you.

 

[vela]

The Thevenin voltage isn't 10 V. Even with no load connected, there will be current flowing through the resistors so there will be voltage drops across them.

To find the resistance, just short the left end of the circuit, where the 10-V supply would have been, and find the equivalent resistance seen from A-B. It may help to redraw the circuit so it's easier to see what's in series and what's in parallel.

 

[maherelharake]

Hmmm I see. The nature of the drawing is a bit confusing to me. Would the 1 K and 2 K resistors be in series with each other, as the 4 K and 4 K would be as well? Or are they parallel. Normally, this is easy to identify, but I am struggling with this diagram for some reason. If I trace the current from +10 V, it would imply that 1 K and 2 K are in parallel, but I am unsure.

 

[vela]

Once the left side is shorted, those pairs are neither in series nor in parallel. To be in series, they need to be connected end to end with nothing branching off in the middle, which isn't the case because of the shorted source. To be in parallel, they have to be connected to the same two nodes in the circuit, which they aren't. The 1K and 2K resistors are both connected to the top node, but one's connected to node A and the other to node B.

Look at the 1K and 4K resistor below it. You should be able to see they're in parallel even though they don't look that way as drawn.

 

[maherelharake]

Oh ok so after it's shorted, then the 1 K and 4 K are in parallel because they both connect to the unlabeled node, as well as the other side due to the short. wouldn't that logic still apply to the 2 K and the 4 K below it as well?

 

[vela]

Yes, exactly.

 

[maherelharake]

Ok so that means I can replace the 1 K and 4 K by a 4/5 K equivalent resistor, and the 2 K and 4 K by a 4/3 K equivalent resistor. Now these two are neither in parallel or series, so how can a Thevenin circuit be constructed? Don't I need a Thevenin circuit to be able to figure out the V and R?

 

[vela]

But they are in series. If you have only two resistors, they have to be in series or in parallel.

 

[maherelharake]

Oh ok I think I see it now. So the ultimate equivalent resistance is 32/15 ohms. So I need to figure out how to use that to find I, right? I don't think V=IR applies here though.

 

[vela]

No, to find the Thevenin voltage, you have to go back and analyze the original circuit with the sources. You could short A and B and calculate the resulting current I_sc. Then since R_th=V_th/I_sc, you could solve for V_th. Alternately, you could just find the open-circuit voltage drop from A to B, and that will be equal to V_th.

 

[maherelharake]

I think see what you are saying, but I am still unsure how to carry it out.

 

[vela]

Calculate the voltage at A and the voltage at B. Then subtract the two to find the voltage drop across A to B.

 

[maherelharake]

Don't you need the current value to find that though? I don't have an current values.

 

[vela]

Try applying Kirchoff's voltage law to the circuit.

 

[maherelharake]

Ahhh ok. So 10V-(4/5)I=0 and 10V-(4/3)I'=0?

 

[vela]

No, you have to analyze the original circuit.

 

[maherelharake]

I think I just found that out. I think one current would be 2 mA and the other would be 5/3 mA. Hope that is right...

 

[maherelharake]

Can anyone help me with this one? Thanks in advance.

 

[vela]

What did you get for the voltage?

 

[maherelharake]

50/4 and 30/4 are the values I got for the voltage drops.

 

[vela]

The voltage drops across what? 50/4 is bigger than 10, so I doubt it's correct.

It would help if you'd show how you get your answers.

 

[maherelharake]

Oh no I was making the same mistake I made before. I suppose that I don't have an answer for the voltage.

 

[gneill]

Take the original circuit. Put your hand over everything to the right of terminal A and pretend it's gone. What's the voltage at terminal A with respect to the 0V line?
Hint: It's a simple voltage divider.

Now disappear terminal A and the 1K and 4K resistors. What's the voltage at terminal B?

 

[maherelharake]

Is it just V=(V_in)(R₂/R₁+R₂)? If so, isn't that just 8 V? And would it be 20/3 for the other one?

 

[gneill]

Is it just V=(V_in)(R₂/R₁+R₂)? If so, isn't that just 8 V? And would it be 20/3 for the other one?

Yes.

So what, then, is the voltage AB?

 

[maherelharake]

Subtract the two to get 4/3 V?

 

[gneill]

Yes.

 

[maherelharake]

Ok so how do I use that? Do I just subtract that value from 10?

 

[gneill]

Ok so how do I use that? Do I just subtract that value from 10?

Why would you do that? 4/3 V is the open circuit voltage between the terminals. You're done.

 

[maherelharake]

So that is the Thevanin voltage?

 

[gneill]

So that is the Thevanin voltage?

You should be in a position to answer that question yourself. Reread the definition and prescribed method of determining the Thevenin voltage.

 

[maherelharake]

Yes I believe it is based from previous responses. So now I need to find the shorted current, and use the Voltage I just found to find the Thevenin R?

 

[gneill]

Yes I believe it is based from previous responses. So now I need to find the shorted current, and use the Voltage I just found to find the Thevenin R?

Why? You found the R previously. Go back to post #9 and what led up to it.

 

[maherelharake]

That's what I thought, but my worksheet says that in order to find the R, I have to have an I first. Is there another way to find it using that procedure?

 

[gneill]

Well, perhaps they suggest that you do it that way, but it is not the only way. You can either find the resistance by shorting out the voltage sources, opening any current sources, and determining the resistance seen at the terminals, or you can determine the open circuit voltage and short circuit current and obtain R = V/I. You've already done the first method. If you want to do the second as well, fine. You'll get the same result.

 

[maherelharake]

I see. So just for my personal knowledge, how do you find the short circuit current? Is it by using loop rules?

 

[gneill]

I see. So just for my personal knowledge, how do you find the short circuit current? Is it by using loop rules?

That is one way. Any valid circuit analysis method that gets you where you want to go is fine. Use what you're comfortable with.

 

[maherelharake]

Ok. Thanks