Use node-voltage method to find current?

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SUMMARY

The discussion focuses on using the node-voltage method for circuit analysis to determine branch currents Ia, Ib, and Ic, as well as the power associated with each source. The user successfully applies nodal analysis, deriving equations based on Ohm's Law (V = IR) and power calculations (P = IV). The final results indicate Ia = 2A, Ib = 4A, and Ic = 1A, with power calculations showing 100W and 120W supplied by the respective sources. The user also notes the redundancy in labeling Vb and Vc as the same node.

PREREQUISITES
  • Understanding of the node-voltage method in circuit analysis
  • Familiarity with Ohm's Law (V = IR)
  • Knowledge of power calculations (P = IV)
  • Basic circuit theory and nodal analysis techniques
NEXT STEPS
  • Study advanced nodal analysis techniques for complex circuits
  • Learn about mesh analysis as an alternative to node-voltage methods
  • Explore circuit simulation tools like LTspice for validating calculations
  • Investigate power factor correction in AC circuits
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Electrical engineering students, circuit designers, and professionals involved in circuit analysis and design will benefit from this discussion.

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use "node-voltage" method to find current?

Homework Statement




http://imageshack.us/a/img213/8582/homeworktest2prob1.jpg

a. Use the node-voltage method of circuit analysis to find the branch currents Ia, Ib, and Ic in the circuit

b. Find power associated with each source and state whether the source is delivering or absorbing power.


Homework Equations



V = IR

Nodal analysis ?

P = IV


The Attempt at a Solution



I assume it means nodal analysis, so:

http://imageshack.us/a/img594/8576/homeworktest2prob1edit.jpg


my equations starting are:

(Va - Vb)/5Ω = Ia

Vb/10Ω = Ib

Vc/40Ω = Ic

3A + Ia = Ib + Ic

Va = 50V and Vb = Vc right at the beginning


So:

3A + (Va - Vb)/5Ω = Vb/10Ω + Vb/40Ω

3A + (Va - Vb)/5Ω = (4Vb + Vb)/40Ω;

(then added (-Va + Vb)/5Ω to both sides and multiplied for common denominator:)

3A = (-8Va + 8Vb + 4Vb + Vb) / 40Ω

(substitute Va = 50V and combine like terms)

3A = (-400V + 13Vb)/40Ω

120V = -400V + 13Vb;

Vb = 520V/13

Vb = Vc = 40V


Ib = (40/10)A

Ib = 4A

Ic = 40/40A

Ic = 1A

Ia = -3A + 4A + 1A

Ia = 2A


P = IV

P from 50V = 2A*50V

= 100W supplied

P from 3A = 3A * 40V

= 120W supplied


Looks good or no?

and the power absorbed by the resistors adds up to the same amount
 
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I did a quick check and got Ia = 2 so I figure you've got the rest right as well, but I didn't actually check any of them
 


It looks funny to have both Vb and Vc, since they are the same node. I'd just label it Vb.
 

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