Node Voltage Analysis

  • Thread starter Quincy
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  • #1
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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??
 

Answers and Replies

  • #2
gneill
Mentor
20,925
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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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