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Electronics => Beginners => Topic started by: 256_bit on September 03, 2013, 01:32:05 pm

Title: Thevenin & Norton equivalent circuits
Post by: 256_bit on September 03, 2013, 01:32:05 pm

(http://media.tumblr.com/b2b0dbf09b036b3e9a7b6b05c68aa90f/tumblr_inline_msj1pjMTGZ1qz4rgp.jpg)

I've been having some issues with calculating the equivalent circuits for the problem above, I've attempted it and included my calculations where possible.
 
So far I have calculated that 50ohm//50ohm = 25ohm

Therefore for Norton we have a 2A source and a 25ohm resistor.

Then I’m using voltage = current x resistance to work out the voltage of the circuit so 2 x 25ohm = 50 volts.
So for Thevenin we have a 50 volt source and a 25ohm resistor.

The issue is when I use this site:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/thevenin.html (http://hyperphysics.phy-astr.gsu.edu/hbase/electric/thevenin.html)
to verify my answer.

With the input of 50ohm, 50ohm and 0.00001ohm (because it only calculates for a three resistor circuit) it returns an:
Open circuit voltage: 10v
Thevenin resistance of: 50ohm

So I'm not sure if this is correct because neither the book or my lecturer explained it very well. I managed to calculate a standard 3 resistor, 1 battery circuit but this one has stumped me!

Someone else insisted that these resistors were actually in series (not parallel as I assumed) and that they are used to calculate R_th but not V_th which I think has also caused me a little bit of confusion, so any help would be beneficial!

Title: Re: Thevenin & Norton equivalent circuits
Post by: Andy Watson on September 03, 2013, 01:41:06 pm

What do you want to know?  Your calculations suggest you are approaching this from the point of view of the current source, whereas (I'm guessing) what you want to know is what the circuit "looks like" from the terminals.

Title: Re: Thevenin & Norton equivalent circuits
Post by: Excavatoree on September 03, 2013, 01:43:01 pm

I think I see your problem.  For the Norton source, you want the short circuit current.  So, you imagine the terminals shorted, but then you are combining the parallel resistors.  You want the current through just one of them, not the parallel combination.  Because they are equal, that's 1 ampere.

For the resistance, you deactivate the source (consider the current source open) and see 100 ohms. (looking "in" from the terminals)  So the answer is a 1 ampere source, with a 100 ohm resistor in parallel.

I hope that's correct. It's been a while.

EDIT: 256_bit seemed to have gotten confused and didn't remember that for the resistance, the source is deactivated, resulting in the resistors in series.  For de-activation, voltage sources are shorted, current sources are opened.

Title: Re: Thevenin & Norton equivalent circuits
Post by: Excavatoree on September 03, 2013, 01:48:06 pm

For some strange reason, my thought was to first get the Thevenin equivalant of the Norton like part of the circuit on the left, then combine the resistors in series.

Doing that, the open circuit voltage for 2 amperes flowing through 50 ohms is 100V.  The resistance is 50 ohms.   Then we can add the other resistor, and we have a Thevenin circuit of a 100 volt voltage source with a 100 ohms resistor.

Converting to Norton, we have the the short circuit current of 1 ampere (100v across 100 ohms) and a 100 ohm resistor in parallel, which is what I got before.  So at least I'm consistent.

Title: Re: Thevenin & Norton equivalent circuits
Post by: Andy Watson on September 03, 2013, 01:53:57 pm

Quote from: Excavatoree on September 03, 2013, 01:48:06 pm

Converting to Norton, we have the the short circuit current of 1 ampere (100v across 100 ohms) and a 100 ohm resistor in parallel, which is what I got before.

I agree. It should be possible to calculate either the Thevenin or Norton and arrive at the same answer. But does 256_bit  understand why?

Title: Re: Thevenin & Norton equivalent circuits
Post by: Excavatoree on September 03, 2013, 01:58:00 pm

Quote from: Andy Watson on September 03, 2013, 01:53:57 pm

I agree. It should be possible to calculate either the Thevenin or Norton and arrive at the same answer. But does 256_bit  understand why?

I hope so.  I tried to explain where 256_bit went wrong.  I went ahead and worked out the whole thing because it seems as if 256_bit does know the theory and how to work the problem, but was just getting stuck.

Title: Re: Thevenin & Norton equivalent circuits
Post by: 256_bit on September 03, 2013, 02:14:22 pm

I think I understand it, it will take a few minutes for me to get my head around it - unfortunately we are only being taught how to use this sort of stuff and aren't given a very good explanation as to why or how to work out other problems except for the textbook example  :--

So obviously the 2amps x 50ohms gives us the 100volts.

Because nothing is connected across the terminals no current will pass through R₂ that's why they aren't in parallel.
(I'm a bit confused on why then we add the other resistor?)

EDIT: we add the second resistor because we are peering INTO the circuit so we are measuring between the two output terminals and thus closing the circuit and passing current through R₂.

And thus you have 100volts with a 100ohm resistor for Thevenin.

Therefore for Norton 100volts/50ohms = 2amps
which is split between the two resistors when the circuit is closed resulting in 1amp for each resistor.

Am I on the right track?