# Problem Finding Thevenin Resitance

Discussion in 'Homework Help' started by thedog, Oct 11, 2013.

1. ### thedog Thread Starter New Member

Oct 11, 2013
5
0
I set a closed circuit voltage of 1 between a and b. I tried using a mesh current method to solve it, but because the dependent voltage source is shared by two loops, I'm not sure of that way is possible.

http://imgur.com/iJMSJfv

ix=i1-i2
i3=-4Vx
Vx=2i1
3(i1-i2)-0.5ix+2i1=0 ?
3(i2-i1)+3(i2-i1)=1

Am I even on the right track? Any help is appreciated. Thanks

2. ### WBahn Moderator

Mar 31, 2012
17,446
4,699
You are doing just fine.

As a general note, I strongly recommend that you start tracking units throughout your work. Most mistakes you make will mess up your units and let you catch and correct them almost immediately. But it only works if you become religious about it.

Since most authors are lousy about using units properly, that means that you will have to fill in the gaping blowholes they leave. In this case, you have a dependent voltage source whose output is given as 0.5ix. Well, 0.5 multiplied by a current is a current, not a voltage. The transfer coefficient for this supply is properly 0.5V/A. Similarly, the transfer coefficient on the dependent current source is 4A/V.

With this in mind, you equations become

ix = i1 - i2
i3 = -(4A/V)Vx
Vx= (2Ω)i1
(3Ω)(i1-i2) - (0.5V/A)ix + (2Ω)i1 = 0
(3Ω)(i2-i1) + (3Ω)(i2-i1) = 1V

Notice that this allows you to do a quick sanity check to see if there is even any point going any further. In order to add two terms, the units must be compatible -- you can add voltage to voltage but you can't add voltage to current. So scan each equation and see if all of the terms in that equation are compatible. In this case they are, so you can have some confidence that you didn't put a voltage source were you needed a current or vice-versa, which are very common mistakes.

As for the issue of a dependent source being shared between two meshes, so what? It's not a problem for an independent voltage source to be shared between two meshes and both are simply devices that produce a specific voltage between the terminals while sourcing/sinking whatever current is needed to do so. Remember, mesh current analysis is nothing more than a formalized way of applying KVL around a set of loops.

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