# Help Understanding Thevenin Resistance

Discussion in 'Homework Help' started by djwinger, Jan 3, 2012.

1. ### djwinger Thread Starter New Member

Jan 3, 2012
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Hi there, hope this is in the correct forum. It's not so much homework but I'm studying for an exam and I've hit a brick wall.

My circuit looks similar to this:

When I calculate $V_{TH}$ in this circuit, there is no current flowing through $R_2$ so it is taken out of the equation. However when I calculate $R_{TH}$ , I am supposed to use it in the equation.

I don't understand this. If there is no voltage drop across $R_2$ and no current flowing through it, why is its existence even taken into account when calculating total resistance?

Thanks,
Winger

2. ### Adjuster Late Member

Dec 26, 2010
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The Thevenin model is supposed to predict the circuit behaviour when it is loaded, as well as when it is open circuit, so clearly RTH has to take all of the resistances into account.

As soon as any current flows in or out of the circuit shown R2 will matter. An extreme example is the short-circuit current, which must be equal in the real circuit and in the Thevenin equivalent.

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3. ### djwinger Thread Starter New Member

Jan 3, 2012
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Thanks for the reply Adjuster The penny still isn't dropping yet though. I'm sure what you're saying will make total sense once I have a better grasp on this stuff.

I guess what I'm trying to find out is; if the circuit is open at points A and B, is R2 still applying resistance to the circuit? If so how does this happen when only one side of it is connected to the circuit?

4. ### w2aew Active Member

Jan 3, 2012
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You're right. If terminals A and B remain open, then R2 doesn't affect anything. The whole concept with a Thevenin equivalent is to simplify the circuit to make it easy to predict what will happen at terminals A and B when a load is placed across them. I'm sure you'll agree if a load (like a resistor) is placed across terminals A and B, then resistor R2 will certainly have an affect - right? By calculating the Thevenin equivalent of the circuit, then calculating the A-B terminal voltage when a load is placed across them becomes a simple voltage divider problem.

Try replacing the resistors and voltage source with some real numbers, and calculate "e" and "r". Then place a load resistor and and calculate the the resulting V(A-B). Better yet, if you have access to a lab - verify your results with measurements.

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5. ### djwinger Thread Starter New Member

Jan 3, 2012
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Thanks w2aew! So if I have this right...in the following image, when the voltage source is replaced with a short circuit to calculate $\small{R_{TH}}$, for my own benefit I should visualise this circuit as "closed" between points A and B. In other words I should visually simulate a load being applied between points A and B. Is that correct?

6. ### djwinger Thread Starter New Member

Jan 3, 2012
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Actually, now I'm confused again. If a load is applied at points A and B, wouldn't R2 receive a voltage drop? In which case it should be included in the calculation for $V_{TH}$ (but it's not).

I must clarify, I already have my answers, so I know they're right. I just need to understand why the equations differ between $V_{TH}$ and $R_{TH}$ in terms of R2

7. ### Jony130 AAC Fanatic!

Feb 17, 2009
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The effect of a voltage drop across R2 when load is applied is included in Rth resistance not in Vth.

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8. ### crutschow Expert

Mar 14, 2008
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Back to basics:

You want the Thevenin equivalent to look exactly like the real circuit.

So if you were to measure the resistance across the output terminal of the real circuit with an ohmmeter (with the supply shorted) it would include R2. Thus the Thevenin equivalent also has to include R2.

But when you measure the V$_{th}$ voltage, there is no (theoretical) current flowing through R2 so it is not included for that measurement.

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9. ### djwinger Thread Starter New Member

Jan 3, 2012
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Cheers Jony, I get that, but the question is more to do with why?

I guess I'm really looking for a theory-based answer. There's a little nugget of information I'm missing which is preventing me from fully grasping the concept of how Thevenin came up with his resistance theorem.

I get the voltage theorem. There's no voltage drop on R2 so it's not included in the equation. But is there still resistance even if R2 is only partially connected? Until I fully understand this I don't think I'll be able to remember how to calculate this each time I come across the problem.

Thanks for everybody's help so far

10. ### djwinger Thread Starter New Member

Jan 3, 2012
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Aha, cool I think we're getting somewhere now! But in the same respect, if I measure the voltage output with a voltmeter, would it not close the circuit between points A and B; causing current to flow through R2?

Or is the voltage measurement not taken across points A and B?

11. ### Adjuster Late Member

Dec 26, 2010
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VTH should be equal to the open-circuit voltage from A to B. Any real-world meter connected there would draw some current so that the reading obtained would be different, though with reasonably low resistances and a sensitive meter the difference may be very small.

Actually, wherever you connect a real meter there would be some alteration in potential because of the meter current. This would be smaller for instance if the meter were connected between (the junction of R1, R2 and R3) and B, but it would not be zero.

In general however,
VTH is calculated, as is RTH , and having done so we can predict what a meter would read connected between A and B, at least within the limits set by the tolerances of the components used.

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12. ### djwinger Thread Starter New Member

Jan 3, 2012
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Thanks Adjuster, that's good to know. I'm only beginning with this stuff so all this info is really helpful (if a little overwhelming). If you or somebody else can clear up one last thing for me, I think I'll have this in the bag.

If I test with a voltmeter between points A and B, is the circuit bridged between these points? Or does this only happen when taking current measurements?

13. ### Adjuster Late Member

Dec 26, 2010
2,147
302
I'm afraid that I do not understand your question. What do you mean by "bridged"? Try to put that question in a different way.

I have to say though that I can't see what there is not to understand here. The Thevenin equivalent circuit is supposed to mimic the circuit it is calculated from - including the full source resistance. The fact that some resistor may not be carrying any current during the voltage calculation does not allow us to ignore it when calculating the equivalent resistance.

If you were to calculate the circuit's Norton equivalent instead, which consists of a current source in parallel with a resistance, you would evaluate the short-circuit current instead of the open-circuit voltage - and in this case R2 would count.

14. ### djwinger Thread Starter New Member

Jan 3, 2012
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Bridging is a term used when making current measurements with a multimeter. You break the circuit and then use the multimeter probes to "bridge" (close or join) the broken circuit...allowing current to flow up through the multimeter and back down through the circuit.

However my thinking is, this must not happen when doing voltage measurements. If points A to B are bridged when the probes are placed on them (making it a closed circuit) then R2 would draw the circuit current though it and it would have a voltage drop...in which case it would have been calculated in $V_{TH}$.

15. ### Adjuster Late Member

Dec 26, 2010
2,147
302
In these terms, a voltmeter should not "bridge" the two connections any more than is avoidable, but unfortunately nothing in this world is perfect.

An ideal voltmeter would have infinite input resistance: it would take no current. If such a meter could be made, it would measure the voltage without any current flowing and therefore the resistance of the circuit would not affect the reading.

Any practical meter will however take some current. Modern electronic instruments may take very little current, but some more basic types of analogue meter take rather more. This can load a circuit so that the reading is different from the open-circuit value when the meter is not connected. The size of the voltage shift depends on the meter current multiplied by the equivalent resistance of the circuit it is connected to.

See this article: http://www.allaboutcircuits.com/vol_1/chpt_8/3.html

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16. ### djwinger Thread Starter New Member

Jan 3, 2012
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Excellent, penny has dropped!

Thanks very much for your help. After all the fuss I've made, I don't think I'll forget this in a hurry!

17. ### thatoneguy AAC Fanatic!

Feb 19, 2009
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The theorem states that a system can be modeled as a black box having a current source and a resistor.

The Thev Equivalent ALSO has open terminals between A and B.

When you find the Equivalent, the voltage dropped across a load resistor should be the same for both the original circuit and the equivalent circuit, no matter the value of the resistor.

This is useful to simplify a more complex circuit into a current source and resistor, so that the rest of a circuit may be analyzed, such as when looking at the internal workings of an operational amplifier.