# Apply Nodal Analysis to the Circuit

1. Sep 29, 2011

### VitaX

1. The problem statement, all variables and given/known data

Apply nodal analysis to the circuit to find the power (released or absorbed?) by the 3-mA source.

[PLAIN]http://img15.imageshack.us/img15/1546/circuity.png[/CENTER] [Broken]

2. Relevant equations

Ohm's Law V = IR

3. The attempt at a solution

OK, so firstly I have to label all the nodes in the circuit and then pick a reference node on the circuit and ground it. I chose V3 as my grounding node. Then I put the current's on the diagram as shown with the arrows indicating the direction.

What I'm having trouble with now is formulating the equations necessary to find the voltages. Ultimately it wants me to find the voltage that runs through the current source and whether it is released or absorbed.

From my reference node, V3 = 0 V. So I just need to find V1, V2, and V3.

Applying KCL at each node:

Node 1: (V1 - 0)/2000
V1 - V2 = 2

Node 2: (V2 - V4)/1000

Can you automatically say V4 = 0 V as well due to the positioning of the ground?

That is what I have so far, can someone help me brush this up a little bit to help me understand things a little more? I'm mainly confused about how to write KCL for the current going through the 3 k-Ohm resistor and the 10 V source, but I'm really jumbled on how to set this problem up as a whole in order to solve for the voltages. Any help appreciated.​

Last edited by a moderator: May 5, 2017
2. Sep 29, 2011

### Staff: Mentor

Nodes labelled V3 and V4 are the same node: any parts of a circuit connected by unbroken wiring belong to the same node.

The 3K resistor and 10V supply constitute a branch that contains a pseudo-node at the join of the resistor an voltage supply. It's voltage is fully constrained by the supply since the supply connects it directly to the reference node. Its node voltage is 10V, that's it, that's all!

So when writing the KCL equation for V2, you write a term such as (V2 - 10V)/3K for that branch.

3. Sep 29, 2011

### VitaX

I finally finished this problem. Ended with it delivers 0.006 W of Power.