Helping setting up Nodal Equations

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To find the node voltages at N1 and N2 in the given circuit, KCL equations are established for each node. At Node 1, the equation is set up as (v2-v1)/12 + (v3-v2)/20 + (v2-0)/40 = 0, while at Node 2, it is (v2-v3)/20 - (0-v3)/60 = 0. The solution involves substituting v2 from the first equation into the second to solve for v3, and then back-substituting to find v2. The discussion emphasizes the importance of sign conventions in setting up equations. Two equations with two unknowns can be solved through elimination or using determinant methods.
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Helping setting up Nodal Equations...

Homework Statement



I need to find the node voltages at N1 and N2. See attachment for circuit.


The Attempt at a Solution




I think I need to setup KCL at each node. At Node 1, I got this:

(v1-v2)/12 - (v3-v2)/20 - (v2-0)/40 =0

v1 is the source voltage, and v3 is the voltage at node 2. Is this correct thus far?
 

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Yes, you are correct so far. However, I personally like to write the KCL equations from the perspective of the node in question, showing that the sum of the currents out of the node is zero. So I would have written your first equation like this instead.

(v2-v1)/12 + (v3-v2)/20 + (v2-0)/40 =0

It's a matter of personal preference, I guess, but thinking about the KCL equations in this way helps me to keep them straight in my head.
 
Cool.. thanks. And the equation at Node 2 should look something like:

(v2-v3)/20 - (0-v3)/60 = 0

Plug in v2 from previous equation, solve for v3, and then plug v3 back into first equation?
 
bengaltiger14 said:
Cool.. thanks. And the equation at Node 2 should look something like:

(v2-v3)/20 - (0-v3)/60 = 0

Plug in v2 from previous equation, solve for v3, and then plug v3 back into first equation?

I still have trouble with your sign conventions, but I think that is correct.
 
Any suggestions on solving these equations?
 
bengaltiger14 said:
Any suggestions on solving these equations?

You have two unknowns and two equations. You can either set up the equations so that you can add them to eliminate one of the unknowns, or you can use the determinant math trick to solve them.
 
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