# How do i find Thevenin and Norton equivalent? [updated]

I tried calculating Vth, Rth and In but i am getting stuck at the V2 source.

simulate this circuit – Schematic created using CircuitLab

Can anyone explain why Vth=45?

– Chu
Jun 4 '17 at 15:03
• @Chu I happened to see the earlier schematic that was posted and then deleted. (electronics.stackexchange.com/questions/309003/…) So there was a little work done, at least. Both on a transformation as well as the voltage divider. (Added recently, though.)
– jonk
Jun 4 '17 at 15:57
• You've worked out that the divider node voltage should be $9\:\textrm{V}$. There is no current in the $10\:\Omega$ resistor, though. So no voltage drop across it. This should make figuring out the $a$ node voltage, relative to $b$, rather easy. Do you see?
– jonk
Jun 4 '17 at 16:02
• Vth=Voc+(voltage of source)? Down on the comments we found Rth=9.6 . Which is right? Jun 4 '17 at 16:20
• @Andyaka this is how we write Ω in Greece (most of us). Τhe first is 4Ω Jun 4 '17 at 16:30

Well for this circuit:

simulate this circuit – Schematic created using CircuitLab

Finding $V_{TH}$ is straightforward

$V_X$ voltage is equal to:

$$V_X = 18V*\frac{4Ω}{4Ω+4Ω} = 9V$$

And from KVL we find $V_{TH}$ :

$$V_{TH} = V_X+(4*V_X) = 9V + 36V = 45V$$

To be able to find $R_{TH}$ you need to solve this circuit and find $I_{SC}$

simulate this circuit

And if you do just that, you will find $R_{TH} = \frac{V_{TH}}{I_{SC}} = \frac{45V}{2.25A} = 20Ω$

In this question since u have a dependent source so The Rth will equal Voc/Isc So u need to find the voltage from the thevenin theorem and the current from the norton Theorem to get the Rth And if u change the current source to voltage then use mesh analysis because only two loops But it would be easier if u show ur work then I'll understand where is ur problem

• Ur work has a small mistake the current source has to change to a voltage source and v=i*R so the voltage source will Equal to 40Vo then short circuit not open circuit if u r using mesh to find the current for norton
– Laya
Jun 4 '17 at 15:35
• This is basically what i am trying to do. The source is voltage (i did current by mistake). So V=I*R=0.4*10=4 right? Jun 4 '17 at 15:40
• Yes correct my mistake I thought it was 4 not 0.4 and then Vo= (4/4+4)18 and then use KVL to get the Voc but the KvL has to include the voltages for the 10 ohm and the dependent source don't avoid them
– Laya
Jun 4 '17 at 15:49
• So lets say 18=4*i1+4*(i1-i2) and 36=4*(i2-i1)+10*i2. So i1=3.3 and i2=2.1 . Is this true? Wouldn't i1 be 2.25 ? Jun 4 '17 at 16:02
• Yes 18=8i1-4i2 and 36=-4i1+14i2 can solve using the calculator or manually will get an answer of i1=4.125 and i2 3.75
– Laya
Jun 4 '17 at 16:07