# Finding Thevenins equivlant using his thereom and nortons

Joined Mar 5, 2007
45
Hi there, problem sorted many thanks

#### The Electrician

Joined Oct 9, 2007
2,778
To find the resistance in the equivalent, replace all the voltage sources in the circuit with shorts, and all the current sources with opens, and compute the resistance seen at terminals A and B.

Starting at the left, we have 10 ohms in series with 6 ohms, giving 16 ohms. Now we have that 16 ohms in parallel with 24 ohms and 1.777 ohms, giving 1.499 ohms as the resistance seen at terminals A and B.

Next, I converted the current sources into Thevenin equivalent voltage sources. So, the 8 amp current source becomes an 80 volt source in series with 10 ohms. The 2.5 amp current source becomes a 4.4425 volt source in series with 1.777 ohms. Now you've got a network with two loops. Assuming the long line in the voltage source symbols represents the positive end, and you assume a current of i1 in the left-hand loop, and a current of i2 in the right hand loop, both flowing clockwise, I get i1=5.3167 amps and i2 = 4.7778 amps. From this you can compute the output voltage of the circuit as 12.933 volts. So the Thevenin equivalent at terminals A and B is a 12.933 volt source in series with 1.499 ohms. This might even be correct if my assumption about the polarity of the voltage source symbols is right.

Joined Mar 5, 2007
45
... the long line in the voltage source symbols represents the positive end, and you assume a current of i1 in the left-hand loop, and a current of i2 in the right hand loop, both flowing clockwise, I get i1=5.3167 amps and i2 = 4.7778 amps. From this you can compute the output voltage of the circuit as 12.933 volts. So the Thevenin equivalent at terminals A and B is a 12.933 volt source in series with 1.499 ohms. This might even be correct if my assumption about the polarity of the voltage source symbols is right.
Upto here I've seen where I was wrong and understand. Although I dont understand how you managed to get those values for i1 and i2?

Many thanks for the help so far,

#### The Electrician

Joined Oct 9, 2007
2,778
Do you know how to do a loop analysis of a circuit? You need that to follow what I did. If you have, I'll give the details.

Joined Mar 5, 2007
45
Using kirchoffs laws correct?

But doesnt that require a 3rd resistor alongside the 10ohm and the 1.777ohm? Or is the third resistor, the resistor worked out previously?

Many thanks

#### The Electrician

Joined Oct 9, 2007
2,778
You just keep the changes you made to get the Thevenin equivalent resistance at terminals A and B. Replace the 8 amp current and its 10 ohm resistor with an 80 volt voltage source in series with a 10 ohm resistor. Replace the 2.5 amp current and its associated 1.777 ohm resistor with a 4.4425 volt voltage source in series with a 1.777 ohm resistor.

Now you have two loops. The left hand loop has an 80 volt source, a 10 ohm resistor, a 12 volt source, a 6 ohm resistor, a 30 volt source and a 24 ohm resistor.

The right hand loop has a 24 ohm resistor, a 30 volt source, a 4.4425 volt source and a 1.777 ohm resistor.

The two loops have a 30 volt source and a 24 ohm resistor in a common branch in the middle, between the two loops.

Solve for the two loop currents using KVL, and then compute the voltage drop across the 1.777 ohm resistor plus (or minus, depending on the direction i2 ends up with, i2 being the current in the right hand loop) the 4.4425 volt source. That will be the voltage at terminals A and B.

Sorry for relying on all this textual description instead of a schematic. I'm not in a position to do a schematic. But if you're still having trouble visualizing all this, I can do a schematic later. Let me know.