# Thevenin's Theorem Help

## Main Question or Discussion Point

Hello,

As part of a module of a HNC I'm looking at Thevenin's & Norton's Theorems. As it's distance learning and I work full-time it's easier for me to ask on here rather than contact the course lecturer - they're finished by the time I get home and I'm hoping I might get a response to my query faster on here rather than getting involved in an email exchange.

I am OK at reducing the various series and parallel resistors to a single resistance but I struggle then to use that to calculate the idea voltage source.

As a starting point let's look at this:  .

Total resistance = 7+5+((4x6)/(4+6)) = 14.4Ω

Current = V/R = 12/14.4

Voltage = IR = 12/14.4 x ((4x6)/(4+6)) = 2V

Now I can get that because it's similar to a worked example in the course notes, however I want to know why it's ... x ((4x6)/(4+6)) because when I come to the next question I can calculate the resistance but not to ideal voltage.

The next one being: The resistance is 4.5Ω, so the current is (10/4.5)A so the voltage is (10/4.5) x ????? V

Thanks.

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psparky
Gold Member
After skimming, It appears you are way off on your execution.

It is discussed in many wonderful ways in those 5 threads.

Study up and get back to us.

As a hint, assuming you are using thevenin, you want to find the open circuit voltage accross points a and b. In other words, when this circuit is running untouched, what would a volt meter read across points a and b.

AS far as thevinin resistance, you want to see what the total resistance is looking from the points "a and b." In other words, what would a ohmeter read at points a and b with the voltage source shorted.
YOu would take (5+7) which is 12 ohms. Then find what 12 in parralel with 4 and 6 and 2 equal. 4 parallel resistors of 12, 4, 6 and 2.

The open circuit voltage is what you call the "ideal voltage"......(hint: people call this the Thevenin Voltage....not "ideal voltage".

The Thevinin resistance I mention above is...well, the Thevinin resistance.

In reallity, its just a V=IR trick. Once you master this, you will automatically use thevinin to solve bigger, tougher problems. You will see.

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Cheers.

Can you elaborate on how I'm "way off" please? I had looked at the related topics earlier and the only one with anything useful (that I could see) had this:

1. Calculate the Open Circuit Voltage VOC - This is the voltage the appears between the terminals when no load is connected between them.

2. Calculate the Short Circuit Current ISC - This is the current that flows through the terminals when they are shorted (i.e. connected with no resistance in between).

3. The Thevenin equivalent resistance can then be calculated using

RTH=VOC/ISC
In the second example I have calculated RTH as being 4.5Ω

ISC will be 10V/6Ω (the 9Ω resistor being by-passed when you short A-B (Correct?))

4.5 x 10/6 = 7.5V which I know is incorrect.

However if I subtract 7.5V from 10V that gives me 2.5V which I know is the answer, but, other than that it gives me the correct answer(!), I don't know why I've just subtracted 7.5V from the 10V if I'm being honest.

psparky
Gold Member
In Thevenin terms........

In the second example, if you put an ohmeter on a and b....you will get 2.475 ohms.
This is also known as the RTh of the problem.

How did I get that? I looked from the open terminals a and b. So I start simplifiying resistors from the left.
A 6 ohm in paralel with a 6 ohm gives you 3 ohms. This is now in series with the 6 ohm resistor all the way on the right. Add these together and you get 9 ohms. You now have 9 ohms in parallel with 9 ohms which is 4.5 ohms.
You then have 4.5 ohms in parrallel with 5.5 ohms which equals 2.475.

Perhaps you stopped at the 4.5 ohms. That 5.5 ohm is definitely part of the Rth.

If you take the short circuit current you would bypass the 9 ohm resistor AND the 5.5 ohm resistor.
After simplifing, you will have 3.33 volts/6 ohms....or .555 amps.
I got that by bypassing 9 and 5.5 ohm resistors. YOu then have to parallel the 6 ohm resitor in the middle witht the 6 ohm resitor on the right.....that gives you 3 ohms. Using voltage dividier you then will get (3/(3+6)) X10 volts....or 3.3 volts across middle and branch on right. 3.3/6 = .555 amps.

So if you plug into your RTH=VOC/ISC above, you would get 2.475=VOC/.5555
So then VOC = 1.37 Volts.

Or you can solve the circuit finding the voltage across the 5.5 ohm resistor....this will also be your VOC.

I'll tell ya what, you sure picked a tough problem to learn on!!!!

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psparky
Gold Member
If you forget about theveinin for a second and just find the voltage across the 5.5 ohm resistor....you will find the voltage is 1.37 volts as I suggested above. I just worked it out on paper to make sure my answer was correct above.

That's a tough find tho, lots of parralel simplifing, voltage dividers, then you have to sweep back out. If you can do this, you definitely are a whiz a the basics of circuits.

Again, very tough prob for begginner.

psparky
Gold Member
Your first problem above is a cake walk compared to the second problem.

Again, looking from left....if you simplyify resitors in parrallel you get 12||4||6||2......those four resitors in parallel are 1 ohm.

Take short circuit current and you bypass the 4, 6 and 2 resistor...so you get 12 volts divided by 12 ohms.....or
1 amp.

So you VOC is obviously 1 volt as well.

Or you can solve the circuit for the VOC....and you will get 1 volt VOC as well. To do this, take 4||6||2 all in parallel....which simplifies down to 1.09 ohms. Then do voltage divider......1.09/(12 +1.09) X12......you get 1 volt.

Thanks for the refresher course on Thevenin!!! I was actually getting rusty.

Perhaps you stopped at the 4.5 ohms. That 5.5 ohm is definitely part of the Rth.
Yes, that's what I did as I thought when you were calculating the Thevenin's equivalent resistance you dropped the load (in this case the 5.5Ω)? Hence getting 4.5Ω.

If you take the short circuit current you would bypass the 9 ohm resistor AND the 5.5 ohm resistor.
Yes I get this.

After simplifing, you will have 3.33 volts/6 ohms....or .555 amps.

I got that by bypassing 9 and 5.5 ohm resistors.
Hmmm... OK.

You then have to parallel the 6 ohm resistor in the middle with the 6 ohm resistor on the right.....that gives you 3 ohms.
Yup

Using voltage dividier you then will get (3/(3+6)) X10 volts....or 3.3 volts across middle and branch on right. 3.3/6 = .555 amps.
What is (3/(3+6))? ((4Ω+2Ω)x6Ω)/(4Ω+2Ω+6)) / (((4Ω+2Ω)x6Ω)/(4Ω+2Ω+6)) + 6Ω (as it's now in series)).

What exactly does that equation give you?

The answer to the second question is this: psparky
Gold Member

Yes, that's what I did as I thought when you were calculating the Thevenin's equivalent resistance you dropped the load (in this case the 5.5Ω)? Hence getting 4.5Ω.

Yes I get this.

Hmmm... OK.

Yup

What is (3/(3+6))? ((4Ω+2Ω)x6Ω)/(4Ω+2Ω+6)) / (((4Ω+2Ω)x6Ω)/(4Ω+2Ω+6)) + 6Ω (as it's now in series)).

What exactly does that equation give you?

The answer to the second question is this: OH, sorry, my bad on that last part with the 5.5 ohm resistor. I was looking at a smart phone when I looked at it and missed that the 5.5 ohm was a removable load.

So adjust that into everything I said above.