# Circuits Question Node Voltage Method

1. Dec 26, 2012

### GreenPrint

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

I'm having problems finding the power.

I should my work to get the node voltages. Apparently they are correct. I showed how I got to those answers as well just in case they were wrong but I don't think they are.

http://img547.imageshack.us/img547/8812/capturebu.png [Broken]

2. Relevant equations

3. The attempt at a solution

I see from the reference node, which is at a potential of zero, that node 1 is 8 volts higher so...

$V_{1} = 8$

I then write the equations for the rest of the nodes

$\frac{V_{2} - 8}{3} + \frac{V_{2} - 3 - V_{4}}{4} + \frac{V_{2} - V_{3}}{2} = 0$
$\frac{V_{3} - V_{2}}{2} + \frac{V_{3}}{1} + \frac{v_{3} - 6 - V{4}}{3} = 0$
$\frac{V_{4} + 3 - V(2)}{4} + \frac{V_{4} + 6 - V{3}}{3} + \frac{V_{4} - V{5}}{2} = 0$
$\frac{V_{5} - V{4}}{2} + 2 - 3 = 0$

I than simplify these equations

$(\frac{1}{3} + \frac{1}{4} + \frac{1}{2})V_{2} - \frac{1}{2}V_{3} - \frac{1}{4}V_{4} = -( -\frac{8}{3} - \frac{3}{4})$
$-\frac{1}{2}V_{2} + (\frac{1}{2} + 1 + \frac{1}{3})V_{3} - \frac{1}{3}V_{4} = -(-\frac{6}{3})$
$-\frac{1}{4}V_{2} - \frac{1}{3}V_{3} + (\frac{1}{4} + \frac{1}{3} + \frac{1}{2})V_{4} - \frac{1}{2}V_{5} = -(\frac{3}{4} + \frac{6}{3})$
$0V_{2} + 0V_{3} - \frac{1}{2}V_{4} + \frac{1}{2}V_{5} = -(2-3)$

I than simplify these equations further

$\frac{13}{12}V_{2} - \frac{1}{2}V_{3} - \frac{1}{4}V_{4} + 0V_{5} = \frac{41}{12}$
$-\frac{1}{2}V_{2} + \frac{11}{6}V_{3} - \frac{1}{3}V_{4} + 0V_{5} = 2$
$-\frac{1}{4}V_{2} - \frac{1}{3}V_{3} + \frac{13}{12}V_{4} - \frac{1}{2}V_{5} = -\frac{11}{4}$
$0V_{2} + 0V_{3} - \frac{1}{2}V_{4} + \frac{1}{2}V_{5} = 1$

I than define matrix A to be

$\frac{13}{12} -\frac{1}{2} -\frac{1}{4} 0$
$-\frac{1}{2} \frac{11}{6} -\frac{1}{3} 0$
$-\frac{1}{4} -\frac{1}{3} \frac{13}{12} -\frac{1}{2}$
$0 0 -\frac{1}{2} \frac{1}{2}$

I than define matrix B to be

$\frac{41}{12}$
$2$
$-\frac{11}{4}$
$1$

I than do

$C = A^{-1}B$
C = 4.22
2.26
.1
2.1

So

$V_{1} = 8$
$V_{2} = 4.22$
$V_{3} = 2.26$
$V_{4} = .1$
$V_{5} = 2.1$

Then

$I_{x} = \frac{V{2} - V_{1}}{3} = \frac{4.22 - 8}{3} = -1.26$
$I_{y} = \frac{V_{3}}{1} = 2.26$
$I_{z} = \frac{V_{4} - V_{5}}{2} = \frac{.1-2.1}{2} = -1$

$I_{x} = -1.26$
$I_{y} = 2.26$
$I_{z} = -1$

So I guess I'm close within round off error

However for

$P_{6V} = IV = \frac{V_{3}-V_{4}}{3}6 = \frac{2.26 - .1 - 6}{3}6 = -7.68$

$P_{6V} = (-1.28)(6) = -7.68$

The answer key used the mesh current method. I don't like the mesh current method and try to use the node voltage method always. Either way I should still get the same answer which I don't. I'm not exactly sure why I don't. Thanks for any help.

Last edited by a moderator: May 6, 2017
2. Dec 26, 2012

### Staff: Mentor

Recheck the coefficient for V4 in the above equation. Everything else looks okay to this point.

Last edited by a moderator: May 6, 2017
3. Dec 26, 2012

### GreenPrint

Never mind I figured it out

4. Dec 26, 2012

### GreenPrint

Oh. For the final answer for the power. I forgot to subtract 6

((2.236-6-.016)/3)6 = -7.56

let me check the coeffecent for V4

5. Dec 26, 2012

### GreenPrint

ok i got it thanks