Norton/Thévenin

[Tiis]

I am practising for a electric circuits DC exam and I wondered if someone could check my way of solving a certain problem.

This is the circuit I start off with:

http://i.imgur.com/WFuTixF.jpg

I need to find the Norton equivalent if the source is 3 Volts. I add the 1 Ohm and 3 Ohm to make 4 Ohm, which gets me the following circuit:

http://i.imgur.com/40VAYpF.jpg

The line shows where I converted from Thévenin to Norton, this the circuit after I did that:

http://i.imgur.com/FKML5wd.jpg

I then simplify the two 4 Ohm resistors which gets me this:

http://i.imgur.com/xrk5AWQ.jpg

I then convert from Norton to Thévenin, the resulting circuit is:

http://i.imgur.com/8SGFkl2.jpg

Adding the 2 Ohm resistors together:

http://i.imgur.com/aThKWKq.jpg

I now have the simplified form of the circuit, but I needed the Norton equivalent so I am converting to Norton once more:

http://i.imgur.com/KH6UvHM.jpg

Is this the correct answer, did I do it correctly?

 

[Ratch]

I am practising for a electric circuits DC exam and I wondered if someone could check my way of solving a certain problem.

This is the circuit I start off with:

http://i.imgur.com/WFuTixF.jpg

I need to find the Norton equivalent if the source is 3 Volts. I add the 1 Ohm and 3 Ohm to make 4 Ohm, which gets me the following circuit:

http://i.imgur.com/40VAYpF.jpg

The line shows where I converted from Thévenin to Norton, this the circuit after I did that:

http://i.imgur.com/FKML5wd.jpg

I then simplify the two 4 Ohm resistors which gets me this:

http://i.imgur.com/xrk5AWQ.jpg

I then convert from Norton to Thévenin, the resulting circuit is:

http://i.imgur.com/8SGFkl2.jpg

Adding the 2 Ohm resistors together:

http://i.imgur.com/aThKWKq.jpg

I now have the simplified form of the circuit, but I needed the Norton equivalent so I am converting to Norton once more:

http://i.imgur.com/KH6UvHM.jpg

Is this the correct answer, did I do it correctly?

No, why do it in such a clunky clumsy way? First, find the open circuit voltage. That is 3*(9/13)=27/13 volts. Then find the short-circuit current. That is (3/(4+18/11))*(9/11) = 27/62 . Dividing the open circuit voltage by the short-circuit current gives 62/13 =4.76923 ohms for the Thevenin resistance. You can also calculate the Thevenin resistance directly by (36/13)+2 = 4.76923 .

Ratch

 

[Laplace]

Have never seen so many Norton/Thevenin conversions done in one problem, but there is no theoretical reason why it would not give the correct answer. Just seems like a lot of extra work. I got the same answer in the usual way. Are you sure you have the current going in the right direction?

It is much simpler to get this answer by the traditional method: zero the sources to find the equivalent resistance, find the open-circuit voltage, convert to the short-circuit current. Zeroing the 3V source gives 1Ω in series with 3Ω which is in parallel with 4Ω which is in series with 2Ω. ((1+3)║4)+2=4 Req=4Ω. For the open-circuit voltage there is 3V divided to the 4Ω resistor with 8Ω around the loop. 3∙(4/(1+3+4)=1.5 Voc=1.5V So the Norton current is Voc/Req=1.5/4=0.375A

 

[Tiis]

No, why do it in such a clunky clumsy way? First, find the open circuit voltage. That is 3*(9/13)=27/13 volts. Then find the short-circuit current. That is (3/(4+18/11))*(9/11) = 27/62 . Dividing the open circuit voltage by the short-circuit current gives 62/13 =4.76923 ohms for the Thevenin resistance. You can also calculate the Thevenin resistance directly by (36/13)+2 = 4.76923 .

Ratch

That's how it was explained to us. The goal here is to show how to get the answer correctly, regardless of how long the specific method takes, I get zero points for doing it quicker and if I divide it into as small as possible steps I reduce the chance of me making a mistake.

But anyway, the second part of this question requires me to load the A and B with an additional 2 Ohm resistor. I placed the 2 Ohm resistor between points A and B. I started from http://i.imgur.com/aThKWKq.jpg again and there I added the 2 Ohm and 4 Ohm together creating a 4/3 Ohm resistor. I then converted to Norton and got I = 9/8 A.

I believe this is where I made an error.