How Do I Correctly Apply Node Voltage Analysis?

AI Thread Summary
To correctly apply Node Voltage Analysis, it is crucial to maintain consistency in the assumed directions of currents at each node. The equations derived for node voltages must reflect whether currents are flowing into or out of the node, which can affect the signs in the equations. For example, when analyzing Node a, the equation should account for the current flowing out, while Node b should reflect the current flowing into it. Adjusting the equations to ensure that all currents are treated as outgoing simplifies the analysis and leads to accurate results. Properly applying these principles will help in determining the correct node voltages.
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Homework Statement


Determine the node voltages, Va and Vb of this circuit:

http://img198.imageshack.us/i/circuitx.jpg/


Homework Equations





The Attempt at a Solution



Node a: 3 V - Va/4 + (Va - Vb)/2 = 0

3 - Va/4 + 2Va/4 - 2Vb/4 = 0

3 + Va/4 - 2Vb/4 = 0

--> 12 + Va - 2Vb = 0

Node b: Vb/3 - 4 - (Va - Vb)/2 = 0

2Vb/6 - 4 - 3Va/6 + 3Vb/6 = 0

---> -24 - 3Va + 5Vb = 0

Combining two equations yields:

-24 - 3Va + 5Vb
3(12 + Va - 2Vb)
12 - Vb = 0
Vb = 12 V

- 24 - 3Va + 5(12) = 0
Va = 12 V

According to the book, my answer is wrong. What am I doing wrong??
 
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Your node equations are not being consistent with the assumed directions of currents (what sign to assign to incoming versus outgoing currents from a node).

I find it's easier to always assume that currents are flowing out of a node unless it is a current source that leaves one no choice. The mathematics takes care of sorting out the actual directions via the node voltages it determines. So, for example, I would write:

Node a: -3A - Va/4Ω - (Va - Vb)/2Ω = 0 ... No choice for the -3A flowing out of Node a
Node b: +4A - Vb/3Ω - (Vb - Va)/2Ω = 0 ... No choice for the +4A flowing into Node b

Note how the (Va - Vb) term changes "direction" when looking from Node b towards Node a, versus looking towards Node b from Node a.

So, at any given node, to write the "outgoing" current for a branch, simply take the node's voltage and subtract the voltage of the next node over, and divide by the intervening resistance.
 
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