Engineering Norton Equivalent Circuit: Find i(sc) and R(t)

[TheCarl]

Homework Statement

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, i_sc, and Thévenin resistance, R_t.

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

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.

KVL_a: -10v + 3i_a - 2i_a + 6(i_a - i_sc) = 0

KVL_sc: 6(i_sc - i_a) + 5i_sc = 0

I reduced and ordered the equations as follows:

KVL_a: 7i_a - 6i_sc = 10
KVL_sc: -6i_a + 11i_sc = 0

Which I solved as:

i_a = 2.68A
i_sc = 1.46A

However the book has the i_sc equaling 1.13A

Where have I messed up?

 

[gneill]

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

 

[TheCarl]

Perfect!

i₁ = 2.08
i_a = 0.94
i_sc = 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 i_a 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 V_oc you would do voltage divide: 11.88v*(6Ω/(6Ω+3Ω)) which gives you a V_oc of 7.92v

Divide that by the i_sc 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?

 

[NascentOxygen]

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

 

[TheCarl]

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

 

[NascentOxygen]

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

We're just happy to be able to help!