Can anyone verify that the nodal equations for the attached circuit.

-5m - V1/200 + V1-V2/1k8 (1)
-V1-V2/1k8 - 16/8k + V2-V3/2k4 (2)
-V2-V3/2k4 -6/12k + V3/6K (3)

Im new to this and trying to get my head around it.

 

You forgot to attach the circuit...

 

Two more things...

First, you should use try to make the equations a little more clear...

Is -V2-V3/2k4 really -(V2-V3)/2k4??

And second, the equations should all have an equal sign in them. Your equations are not complete. Typically, the equations are derived from the current entering the node being equal to the current leaving the node. Or, saying it another way, adding all of the current entering the node should be equal to zero...

 

yeah sorry they should be

-5m - V1/200 + (V1-V2)/1k8 = 0 (1)
-(V1-V2)/1k8 - 16/8k + (V2-V3)/2k4 = 0 (2)
-(V2-V3)/2k4 -6/12k + V3/6K = 0 (3)

 

Again, you should make sure things are very clearly labeled or it can get very confusing... Call the nodes something different than the voltage sources in the circuit.

Let's look at your first equation... I will start with current into the node equals current out of the node.

Current into the node is 5mA.

Current out of the node is current through R1 and R2. This assignment is arbitrary. Just be sure to be meticulous in keeping track of all of the + and - signs.

So, 5mA = I(in R1) + I(in R2)

Now, what are the currents in R1 and R2?

You have it correct for R2... With the current defined as out of the node, the current is (V1 - V2)/1k8.

For the current through R1, I(in R1) = (V1 - 0)/200 = V1/200.

***Assigning the current in the opposite direction would result in (0 - V1)/200 or -V1/200. The math all leads to the same equation. Putting it on the other side of the equal sign in the equation essentially introduces yet another - sign***

Be careful how you assign currents in the resistors. Currents through R3 and R5 are not determined solely by the voltage sources connected to them. That would only be the case if V2 and V3 were at 0V. The current through R3 is the voltage across R3 divided by the resistance. The voltage on one side of R3 is 16V. The voltage on the other side is what?

 

If we assume and label the current direction in this way (see attached pic ).

We can write three equations:

I1 = (A - B) / R2 + A / R1 (1)

(A - B) / R2 = (B - C) / R4 + (B - V1) / R3 (2)

(B - C1) / R4 = (C - V2) / R5 + C / R6 (3)


0.005 = (A - B)/1800 + A/200

(A - B)/1800 = (B - C)/2400 + (B - 16)/8000

(B - C)/2400 = (C - 6)/12000 + C/6000

And after solving this I get

A = 1.26V
B = 3.6V
C = 3V

http://wims.unice.fr/wims/en_tool~linear~linsolver.en.html

 

so how would I solve these using cramers rule? I have multiplied out the brackets.

will it look like

1/180 -1/1800 0
-1/1800 41/36000 -1/2400
0 -1/2400 1/1500

 

You should show the equations that you are starting from and the steps that you have taken to get there.