# Norton's equivalent - finding the short circuit current

#### window789456

Joined Oct 17, 2018
3
So I first find the Thevenin's resistance:

3kΩ ll 6kΩ = 2kΩ
2kΩ + 2kΩ = 4kΩ
4kΩ + 4kΩ = 8kΩ
8kΩ ll 8kΩ = 4kΩ

Rth = 4kΩ

Then I tried to find the short-circuit current:

( 8kΩ / ( 8kΩ + 4kΩ + 2kΩ + 6kΩ ll 3kΩ) ) 2mA = ( 8kΩ / 8kΩ ) 2mA = 2mA

isc = 2mA

But I still have a feeling that this is wrong some how, any idea?

#### ericgibbs

Joined Jan 29, 2010
8,876
hi window,
Welcome to AAC,
Have you considered the location of the 2mA current source within the resistive network.???
E

• window789456

#### WBahn

Joined Mar 31, 2012
24,854
So I first find the Thevenin's resistance:

3kΩ ll 6kΩ = 2kΩ
2kΩ + 2kΩ = 4kΩ
4kΩ + 4kΩ = 8kΩ
8kΩ ll 8kΩ = 4kΩ

Rth = 4kΩ

Then I tried to find the short-circuit current:

( 8kΩ / ( 8kΩ + 4kΩ + 2kΩ + 6kΩ ll 3kΩ) ) 2mA = ( 8kΩ / 8kΩ ) 2mA = 2mA

isc = 2mA

But I still have a feeling that this is wrong some how, any idea?
I'm not seeing where you are getting your equation from (I'm not saying that it's wrong, just that I don't see how you are arriving at it).

Please provide a more detailed explanation or setup.

Keep in mind that you can almost always verify the correctness of a solution from the solution itself.

Assume that the answer is 2 mA and then solve the circuit backwards and see if it is consistent with the circuit parameters.

• window789456

#### window789456

Joined Oct 17, 2018
3
At this point I'm almost clueless, so how do we find the short circuit current in this 3-loop circuit?

#### ericgibbs

Joined Jan 29, 2010
8,876
hi w,
Redraw and post the reduced version of the circuit ie: the simplified resistor network at the stage where the short circuit is connected.
E

#### window789456

Joined Oct 17, 2018
3
Is this right?

#### WBahn

Joined Mar 31, 2012
24,854
At this point I'm almost clueless, so how do we find the short circuit current in this 3-loop circuit?
It's just a circuit like many other circuits you've analyzed before you ever heard of Thevenin or Norton.

One way to do it is put a resistor between 'a' and 'b' (call it R) and analyze the circuit to find the current in resistor R. Use any analysis technique you want to do this.

Now set R = 0 and see what the current in it is. That's the short-circuit current. Set R = ∞ and you will get the open-circuit voltage. Find the value of R that makes the output voltage equal to half the open-circuit voltage and that is the Thevenin resistance.

#### MrAl

Joined Jun 17, 2014
6,638
I'm not seeing where you are getting your equation from (I'm not saying that it's wrong, just that I don't see how you are arriving at it).

Please provide a more detailed explanation or setup.

Keep in mind that you can almost always verify the correctness of a solution from the solution itself.

Assume that the answer is 2 mA and then solve the circuit backwards and see if it is consistent with the circuit parameters.
Hello,

For that step he open circuited the current source and short circuited the 12v source, then computed the total resistance.

#### MrAl

Joined Jun 17, 2014
6,638
Is this right?
Hi,

Something is wrong though i think.
The source you have there is 12v, why do you have that there?
Either you have a Vth or an Ith but not both.
You could compute Vth and do it that way but Vth is not 12v.

So your final circuit looks like either a voltage source, resistor, and short at the output, or a current source and short.
If you use Vth then having Rth makes sense.