Find Thevenin equivalent circuit

[leroyjenkens]

Homework Statement

Find Thevenin equivalent circuit as seen across terminals A and B for each circuit.

Homework Equations

$$V=IR$$

The Attempt at a Solution

Those resistors are in parallel because not all of the current going through the top one is going to go through the next one, right?
So I add the resistors to make an equivalent resistance, but I'm not sure what to do after that.

I need to find the open-circuit voltage, but I don't know how to do that mathematically.

And I need to find the short circuit current, which is $$I=\frac{V_T}{R_T}$$.

Is the $$V_T$$ just the voltage source? And is the $$R_T$$ just the equivalent resistance that I found?

Thanks.

 

[rude man]

Homework Statement

Find Thevenin equivalent circuit as seen across terminals A and B for each circuit.

Homework Equations

$$V=IR$$

The Attempt at a Solution

Those resistors are in parallel because not all of the current going through the top one is going to go through the next one, right?
So I add the resistors to make an equivalent resistance, but I'm not sure what to do after that.
Wrong argument.

I need to find the open-circuit voltage, but I don't know how to do that mathematically.
What's the voltage across terminals A and B? That's your open-ckt voltage.

And I need to find the short circuit current, which is $$I=\frac{V_T}{R_T}$$.
If you short those same terminals, what's the current? That's your short-ckt current.Is the $$V_T$$ just the voltage source? And is the $$R_T$$ just the equivalent resistance that I found?
No. The thevenin voltage is the open-ckt voltage and the thevenin series resistance is the open-ckt voltage divided by the short-ckt current.
Thanks.

See above.

 

[leroyjenkens]

What do you mean by wrong argument? That's not the reason they're in parallel? There's a different reason they're in parallel? Or are they not in parallel at all?

The open circuit voltage would be $$V_{Th}=\frac{(2)(100)}{100+100}=1$$ right?

And the short circuit current would just ignore the middle resistor and that would be $$I=\frac{2}{100}=0.04A$$ right?

Ok so $$R_{Th}=\frac{1}{0.04}=25$$

So to answer the question posed, I would make a circuit replacing the $$V_{in}$$ with $$1V$$ and replace the two resistors with a 25 Ohm resistor?

Thanks for the response.

 

[rude man]

What do you mean by wrong argument? That's not the reason they're in parallel? There's a different reason they're in parallel? Or are they not in parallel at all?

The open circuit voltage would be $$V_{Th}=\frac{(2)(100)}{100+100}=1$$ right?
Right. But call it 1V.

And the short circuit current would just ignore the middle resistor and that would be $$I=\frac{2}{100}=0.04A$$ right?
Wrong. 2/100 does not equal 0.04. And, you're not "ignoring the middle resistor", you are shorting out the resistor across the output terminals.

Ok so $$R_{Th}=\frac{1}{0.04}=25$$

So to answer the question posed, I would make a circuit replacing the $$V_{in}$$ with $$1V$$ and replace the two resistors with a 25 Ohm resistor?

Correct your mistake and you're on the right track.

There are two kinds of Thevenin equivalent circuits. One is a voltage source in series with a resistor. The other is a current source in parallel with the same resistor. If you do it right and derive both, you will see that both have the same open-circuit voltage and short-circuit current.


Thanks for the response.

See above.