 # Thevenin's Equivalent of only 1 Output Voltage

I'm a bit confused on how to find the thevenin voltage of this circuit. Usually I would deactivate the load resistor and go about finding the Rth and Vth, but in this case it doesnt specify which resistor is the load resistor. Does this mean that the load resistor is the 40 ohm since it is with respect to the 0v ground? Homework?

There is no load attached. That's all the help I'll give.

The equivalent of the circuit is an 80Ohm load. All the resistors are in series with a current of 0.5Amps and a Vout of 20V

-- Mark

Jiggy-Ninja: Homework?

There is no load attached. That's all the help I'll give.

:^) not homework. I have an exam in about 3 hours

mbferguson: :^) not homework. I have an exam in about 3 hours

Use a KVL loop to find Vo = Vth.

Short the dc voltage supply and combine resistances until you get a single resistor across the output (Rth). Things get more complicated when you have current sources (oh, and don't forget about dependent sources, too).

Hint: it's Vth = -20V and Rth = 20ohm

Power_Broker: Use a KVL loop to find Vo = Vth.

Short the dc voltage supply and combine resistances until you get a single resistor across the output (Rth). Things get more complicated when you have current sources (oh, and don't forget about dependent sources, too).

Hint: it's Vth = -20V and Rth = 20ohm

I see how you got -20V for Vth, simply add the 30 and 10 resistors and by inspection Vout is -20V. When I deactivate the 40V source and short circuit, the only step I can see to do is add the 3 resistors in series to get 80 ohms.

Ahh, I see now! To find Rth the two 40 ohms are in parallel with respect to Vout.

The source as seen by a load at Vout has :

a voltage of 40v * (40 / ( (30+10) +40 )) = 20v

an impedance of ( 30 + 10 ) || 40 = 1 / ( 1/ (30+10 ) + 1/40 ) ) = 20 ohms.

Allan

allanhurst: The source as seen by a load at Vout has :

a voltage of 40v * (40 / ( (30+10) +40 )) = 20v

an impedance of ( 30 + 10 ) || 40 = 1 / ( 1/ (30+10 ) + 1/40 ) ) = 20 ohms.

Allan

Close, the 40V is actually -40V, but everything else is correct.

Sorry - didn't notice that

But the sums are OK.

Allan

mbferguson: I'm a bit confused on how to find the thevenin voltage of this circuit. Usually I would deactivate the load resistor and go about finding the Rth and Vth, but in this case it doesnt specify which resistor is the load resistor. Does this mean that the load resistor is the 40 ohm since it is with respect to the 0v ground? Yes..... you are right in that - if they ask you to find a thevenin equivalent circuit, then they must also provide to you (us) a pair of terminals of interest. In this case, they did not specify a pair of terminals of interest. The all-important feature is - the pair of terminals of interest.

So, the only thing we could possibly do after that is what everybody did ..... which is to write down your assumption. The assumption can be - the terminals of interest are the ones on each side of the 40 Ohm resistor.

Setting the Thevenin circuit voltage to zero is equivalent to setting ALL independent sources in the original circuit to zero (and leaving all dependent sources alone, but there are no dependent sources in this particular circuit, so no need to worry about that). The only independent source in your circuit here is the voltage source. So that is set to 'zero'. And a zero voltage across voltage source terminals is essentially a short-circuit. Hence the "short-circuit" of voltage source when the voltage source is 'de-activated'. So, we go to the original circuit and 'short circuit' the independent voltage source.

This leaves [40 Ohm in parallel with 40 Ohm] across the terminals of interest, which amounts to 20 Ohm.

The significance of this 20 Ohm value: it is the resistance across the terminals of interest for the case where the Thevenin voltage source (from the Thevenin circuit) is set to zero. So when considering the Thevenin circuit by itself (with short circuited Thevenin voltage source), the resistance across the terminals of interest will be whatever we call it.... eg. Rth, the Thevenin resistance. Now, if we do the same thing with the original circuit, we found that the resistance across the terminals of interest is going to be 20 Ohm in unpowered state. So, the 20 Ohm value is the Thevenin resistance, as this is the resistance across the terminals of interest for BOTH circuits (original and Thevenin circuit) for unpowered condition.

One note: The Thevenin equivalent model by itself is open circuit across the terminals of interest. This is equivalent to the meaning of nothing "ELSE" connected across the terminals of interest in the ORIGINAL (not Thevenin) circuit. So, in the original circuit there is a 40 Ohm resistor across the terminals of interest. Yes there is. But there is no extra self-introduced load in parallel with that 40 Ohm (of the original circuit) - which is equivalent to the 'open circuit' condition of the Thevenin 'open-circuit' model. So 'open-circuit' condition just means nothing ELSE is connected across the terminals of interest (other than whatever is already connected in the original circuit).

Now, as for Vth. If you leave the voltage source as-is in the 'open-circuit' Thevenin equivalent model, then Vth will ALSO appear across the terminals of interest because current in the Thevenin model is zero, so no voltage drop across Rth in the Thevenin model. So, to find Vth in the 'original' circuit..... it is nothing more than just finding the voltage across the terminals of interest (in the original circuit.... using KVL, voltage divider, or whatever we like to use). Voltage divider will give us [40 Ohm /{40+10+30}] * (-40V source voltage), which amounts to -20 Volt. So Vth will be -20 Volt.

I got a 92 on my exam guys, thanks for the help.

Southpark: Yes..... you are right in that - if they ask you to find a thevenin equivalent circuit, then they must also provide to you (us) a pair of terminals of interest. In this case, they did not specify a pair of terminals of interest. The all-important feature is - the pair of terminals of interest.

So, the only thing we could possibly do after that is what everybody did ..... which is to write down your assumption. The assumption can be - the terminals of interest are the ones on each side of the 40 Ohm resistor.

Even to the most pedantic of pedants, I think the two dead-ended lines drawn out from the circuit clearly and unambiguously indicate the terminals of interest. It's not really an assumption when it's a universal convention.

And not to mention the obvious label "Vo" at the output terminals.

Just to tie it all up…

Allan

thev.pdf (12.3 KB)

Jiggy-Ninja:
Even to the most pedantic of pedants, I think the two dead-ended lines drawn out from the circuit clearly and unambiguously indicate the terminals of interest. It’s not really an assumption when it’s a universal convention.

It doesn’t necessarily mean that those two terminals with the lines coming out are the terminals of interest for a particular problem. It’s necessary to be clear and specify the terminals of interest, which can be a terminal pair of any component or across lumped components.

I disagree about it being a ‘universal convention’ associated with Thevenin Equivalent circuits.

Thevenin equivalent circuits allow you to study voltage and current relations at the terminals of interest (and hence the power in whatever extra component is connected across those terminals) as you already know. Best to be clear and put labels like “A” and “B” at the terminals of interest, and say that they are the terminals of interest. Write down our assumptions and everything should be fine.