EE, Node voltages, can't figure out how to reach end result

[rod1866]

hey I'm stumped with a question that deals with finding unknown node voltages as I only know how to complete part of the problem. here is the circuit, sorry for the crude MSPaint I do not have a working scanner here.

http://img7.imageshack.us/img7/8730/circuiti.png

the answers are

v1 = 16.97V
v2 = 23.85V
v3 = 16.06V

I am stuck trying to figure out how these have been reached

First I tried KCL to get three equations on each of the three unknowns

v1

(v1-v3) / 20 + (v1 - v2) / 20 + v1 / 5 = 0

v2

(v2-v1) / 2 + (v2 - v3) / 5 - 5 = 0

v3

(v3 - v2) / 5 + (v3 - v1) / 20+ v3 / 10 = 0

I was told it is easy to put these together if it is imagined that the arbitrary current directions are thought to be leaving the node, except with v2 where the 5 amps are entering the node. apparently the answer will still be fine at the end of all of this.

at this point I have no clue how they manipulated these equations into finding the node voltages, what I keep reading is "use the equations to solve for node voltages" which means I'm missing something really obvious as they never bother to explain the process used to find the voltages. can someone please tell me what that is, I would really appreciate it as this is definitely the unit of study i need the most help with. thanks for reading.

 

[tiny-tim]

Welcome to PF!

Hi rod1866! Welcome to PF!

You must write each current (I1 I2 I3) on the diagram together with an arrow, or you'll get terribly confused with your pluses and minuses …

which I think is what's happened

(btw, it doesn't matter which way round the arrow is … if you put it the "wrong" way round, you'll simply get an I that comes out negative )

 

[KingNothing]

It looks like he's written that for each node, the sum of all currents leaving the node is zero. But they aren't all correct. For starters, the first equation should have (v1 - v2)/2 not /20.

What you need to do here is solve the system of equations. There are several methods. The easiest to understand is algebraic substitution. You essentially want to solve each equation for one variable (either v1, v2 or v3) and then plug it into the next equation.

For example, if I gave you these equations:
x + y = 4
2x + 3y = 6

First, solve for x in the first equation:
x = 4 - y

Then plug that x into the second equation:
2(4 - y) + 3y = 6

And solve for that one variable:
8 - 2y + 3y = 6
y = -2

Plug it back into any of the others:
x + (-2) = 4

so, x = 6.

Essentially, EEs use calculators and computers for circuits with more than 3 non-ground nodes.