Is Using Norton and Thevenin Equations to Find Voltage Correct?

[darwinharianto]

Homework Statement

find the voltage by using norton and thevenin equation

is this method correct?

 

[darwinharianto]

if I change the pararelled resistor and send the most right resistor to the left, the voltage changed
is it right if I am doing it like this?
thx for reply in advance
sory for bad english

 

[Rellek]

Hey there!

You might notice that the voltage over your last resistor (on the right) will be the Voltage across your two nodes. So, if you were to find the voltage over that resistor, that will be your Thevenin voltage.

For your Thevenin resistance, try to deactivate all of your independent devices and starting at point A on your node, calculate the equivalent resistance from that point. That will be your Thevenin resistance.

From there you know the Norton circuit is a source transformation. Good luck!

 

[darwinharianto]

thx for the advice
after I deactivate, I found the resistor from paralleled 10k, 6k and 8k
then the ampere source have the resistor (4k) below it paralleled with volt source and the resistor ( from paralleled 10k, 6k and 8k).
after that I'm stuck.
any idea how to go on?

 

[gneill]

Hi darwinharianto, Welcome to Physics Forums.

Some of the component values in your attached figure are hard to make out. Can you clarify their values?

 

[Rellek]

It looks like you might benefit from using source transformations to make both of your sources the same, if you want to use this method instead. I see a voltage source in series with a resistor right off the bat. Also, if you have a resistor in series with a current source, what can you say about the current in that section of wire?

From there, it seems like it would only be current division and Ohm's law to find the voltage over your resistor on the right.

 

[NascentOxygen]

Hi darwinharianto! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

You converted the voltage source to its Norton equivalent. Why not also convert the left-most source and resistors to its Norton equivalent? You'll then have two current sources in parallel ...

... the rest is easy!

 

[darwinharianto]

Hi darwinharianto! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

You converted the voltage source to its Norton equivalent. Why not also convert the left-most source and resistors to its Norton equivalent? You'll then have two current sources in parallel ...

... the rest is easy!

actually the left one is not a volt source, its an ampere source (1mA)
which is trouble me
hope this pic will be giving a better view

 

[NascentOxygen]

actually the left one is not a volt source, its an ampere source (1mA)
which is trouble me

I know it's a current source. You can still represent it by an ideal current source in parallel with a resistance.

 

[darwinharianto]

how is that?
so the current source is multiplied? I got confused

 

[NascentOxygen]

how is that?
so the current source is multiplied? I got confused

What is the procedure for representing any source by a Norton (or Thévenin) equivalent? Follow that.

 

[darwinharianto]

okiee
ill figure something out