How Do I Correctly Apply Node Voltage Analysis?

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SUMMARY

The discussion focuses on applying Node Voltage Analysis (NVA) to determine node voltages Va and Vb in a given circuit. The user initially derived equations for both nodes but received incorrect results compared to the textbook answers. The key issue identified was the inconsistency in the assumed directions of currents. The correct approach involves consistently assuming currents flow out of a node, which simplifies the analysis and ensures accurate voltage calculations.

PREREQUISITES
  • Understanding of Node Voltage Analysis (NVA)
  • Familiarity with Ohm's Law and Kirchhoff's Current Law (KCL)
  • Basic circuit analysis techniques
  • Knowledge of circuit components such as resistors and current sources
NEXT STEPS
  • Review Node Voltage Analysis techniques in electrical engineering textbooks
  • Practice solving circuit problems using consistent current direction assumptions
  • Explore advanced circuit analysis methods such as Mesh Analysis
  • Learn about simulation tools like LTspice for circuit verification
USEFUL FOR

Electrical engineering students, circuit designers, and anyone interested in mastering circuit analysis techniques, particularly Node Voltage Analysis.

Quincy
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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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