# Homework Help: Norton Equivalent Circuit: Find i(sc) and R(t)

1. Mar 3, 2013

### TheCarl

1. The problem statement, all variables and given/known data

The (b) circuit shown in the attached image is the Norton equivalent circuit of the (a) circuit. Find the value of the short-circuit current, isc, and Thévenin resistance, Rt.

isc being the current that would flow through the two open nodes if they were shorted.

3. The attempt at a solution

Right now I'm stuck on the current. I used mesh current analysis to find the current. Here are my equations.

KVLa: -10v + 3ia - 2ia + 6(ia - isc) = 0

KVLsc: 6(isc - ia) + 5isc = 0

I reduced and ordered the equations as follows:

KVLa: 7ia - 6isc = 10
KVLsc: -6ia + 11isc = 0

Which I solved as:

ia = 2.68A
isc = 1.46A

However the book has the isc equaling 1.13A

Where have I messed up?

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2. Mar 3, 2013

### Staff: Mentor

ia is the net current through the 6Ω resistor, not the mesh current in the first mesh.

3. Mar 3, 2013

### TheCarl

Perfect!

i1 = 2.08
ia = 0.94
isc = 1.13

I wish I wasn't struggling with the second part but I am so any help there would be much appreciated as well.

So with ia being 0.94, that means that the dependent voltage source is 1.88. add that to the other voltage source in series and you get 11.88 volts.

To get the Voc you would do voltage divide: 11.88v*(6Ω/(6Ω+3Ω)) which gives you a Voc of 7.92v

Divide that by the isc should give you the Rt. Which for me came out to be 7.01Ω but the book says that it is 7.57Ω.

Any suggestions?

4. Mar 4, 2013

### Staff: Mentor

On open circuit, ia will no longer be 0.94A will it? There's a new loop current.

5. Mar 4, 2013

### TheCarl

That's it! Thank you so much for your responses!

6. Mar 5, 2013

### Staff: Mentor

We're just happy to be able to help!