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I have this circuit that I am trying to solve for the short circuit current.

enter image description here

I am not extremely strong with analyzing circuits (this is finding the short circuit current for a Thevenin equivalent), but I tried solving for currents two different ways and am still wrong. I performed mesh analysis using the three squares as loops which yields these three equations (the left loop is 1, the middle is 2, the right is 3):

10*(I1-I2)=0

10*(I1-I2)+12+40*(I3-I2)=0

40*(I3-I2)-5*I3=0

which gives: I1=2.1, I2=2.1, I3=2.4

Then I used the middle and rights loops using KVL with the current labelled as in the picture:

12-40*Ia-10*Ib=0

-5Isc+40*Ia=0

Ia+Ib+Isc=4.5*Ia

which gives Ia=2.4, Ib=-10.8, Isc=19.2

I understand the solution that they give (and I can post it but someone who is familiar with this probably doesn't need it), but what I don't understand is why my methods did not work and why they picked their particular method/loops to evaluate (it could also stem from the fact that I have not analyzed many circuits so I am inexperienced at how to approach).

Thanks!

This is the solution as given: enter image description here

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You begin with an incorrect statement: $$10(I_1-I_2)=0$$

You're saying that the voltage across the current controlled current source is 0 which is very likely incorrect. You can say $$10(I_1-I_2) = V_\text{node_between_CCCS_and_12Vsource}$$

But you can't assume that that voltage equals 0. If you have some current flowing through that resistor, then the voltage can not be 0.

Other than that, your equations look good. These should be them:

$$ 4.5\times I_a = I_1 $$

$$ 10(I_1-I_2)+12+40*(I_3-I_2)=0 $$ $$ 40\times(I_3-I_2)-5I_3=0 $$ $$ I_a = I_2-I_3 $$

4 equations and 4 unknowns. You should be golden to solve it from here.

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  • \$\begingroup\$ So then you cannot use a mesh on the left circuit with the independent current source? \$\endgroup\$ – Prevost Sep 8 '14 at 19:50
  • \$\begingroup\$ @Prevost I would still call it mesh analysis. It's just that you already "know" your I1 current because the dependent current source tells you that it's 4.5*Ia. Maybe it's just a po-tae-toe/po-ta-toe thing in terms of what you call it. Also, just a clarification, it's a dependent current source rather than an independent one. \$\endgroup\$ – horta Sep 8 '14 at 20:21
  • \$\begingroup\$ @Prevost Yup, definitely still mesh analysis according to wikipedia: en.wikipedia.org/wiki/Mesh_analysis#Setting_up_the_equations Also, you should read up how to use super-meshes there so that you can handle a current source that actually shares two loops. \$\endgroup\$ – horta Sep 8 '14 at 20:29
  • \$\begingroup\$ My bad, it is a dependent current source. Okay, then I guess where I don't understand it is that if I used mesh analysis, I would not use the 4.5*Ia=I1 and instead would use 4.5*I1=I2...and then once I solved for I1,I2,I3 then I would find the actual current through the resistors (which would be equal to Ia, Ib, Isc). And also can you please explain this equation further 10*(I1-I2) = V_node_between_CCCS_and_12Vsource. I would not get that from my Mesh analysis (even though I agree that my original equation was wrong). \$\endgroup\$ – Prevost Sep 9 '14 at 12:52
  • \$\begingroup\$ Would you use a super mesh that combines the middle loop with the right loop (and doesn't include the 40 ohm resistor)? \$\endgroup\$ – Prevost Sep 9 '14 at 12:59

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