Nodal analysis, setting up the equations

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Homework Help Overview

The discussion revolves around nodal analysis in circuit theory, specifically focusing on setting up Kirchhoff's Current Law (KCL) equations for a given circuit with a voltage source and resistors.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to set up KCL equations for different nodes but expresses confusion about incorporating the 30V voltage source. They question whether to treat the current through the resistor as a current source entering node A.
  • Some participants clarify the current through the 15-ohm resistor and suggest how to incorporate it into the equations.
  • Further inquiries are made about calculating power at the independent voltage and current sources after determining node voltages.

Discussion Status

The discussion is active, with participants providing clarifications on the setup of equations. There is a confirmation of the equation for node A, and guidance is offered on calculating power using voltage drops. Multiple interpretations of the problem setup are being explored.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the amount of direct assistance provided. There is an ongoing exploration of assumptions related to the circuit configuration and the roles of the voltage and current sources.

gomezfx
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Homework Statement



[PLAIN]http://img844.imageshack.us/img844/8508/circuit.png

Homework Equations


KCL set ups


The Attempt at a Solution


I am having trouble setting up the KCL equations

The 30V voltage source on the left with the resistor is messing me up, how do I incorporate that?

Do I just do 30/15 and assume a 2A current source is going into node A?

Here are my equations:

VD = reference node = 0 V

KCL @ Node B
(VB-VA)/25 + (VB-VC)/50 - 1 + 4 = 0

KCL @ Node C
(VC-VD)/50 + (VC-VB)/50 + 1 = 0

KCL @ Node A?
(VA-VB)/25 + (VA-VD)/31.25 - 4 + ?? = 0
 
Last edited by a moderator:
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You have 30V at the other end of the 15 ohm resistor. So the current through it is (VA-30)/15.

ehild
 
ehild said:
You have 30V at the other end of the 15 ohm resistor. So the current through it is (VA-30)/15.

ehild

Ok, then the entire equation for voltage at node A is:

(VA-30)/15 + (VA-VB)/25 + (VA-VD)/31.25 - 4 = 0?

After I solve for all the voltages at the nodes, how do I find power at the independent voltage a current sources?

Do I just use voltage drops?
IE, Power at 4A = (VA-VB)*4
 
gomezfx said:
Ok, then the entire equation for voltage at node A is:

(VA-30)/15 + (VA-VB)/25 + (VA-VD)/31.25 - 4 = 0?

This is correct.

gomezfx said:
After I solve for all the voltages at the nodes, how do I find power at the independent voltage a current sources?

Do I just use voltage drops?
IE, Power at 4A = (VA-VB)*4

Yes, knowing the potential difference U across the terminals of a source and the current I flowing through it, the power is U*I.ehild
 

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