Did i solve my matrix correctly? (source sub. on 3 window circuit)

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marstery
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I've got two equations that i want to put into a matrix and solve for Va and Vb:

1.7ma = (1/5k + 1/20k) -1/20k Va
-2ma = -1/20k (1/4k + 1/20k) Vb

The answers I got were Va= 5.655v and Vb= -5.724

then, using I= (Vb-Va)/R ----> (-11.379/20k) = 0.65ma

..-----Va----20k----Vb-----
|...|.....|...|
^...5k....4k...V
1.7ma.|.....|...2ma
.| ___|___________|_____|

(ignore the periods, i needed them for spacing)
 
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marstery said:
I've got two equations that i want to put into a matrix and solve for Va and Vb:

1.7ma = (1/5k + 1/20k) -1/20k Va
-2ma = -1/20k (1/4k + 1/20k) Vb

The answers I got were Va= 5.655v and Vb= -5.724

Your equations look wrong, but I agree with your answers.

then, using I= (Vb-Va)/R ----> (-11.379/20k) = 0.65ma

Look again. You've got a negative divided by a positive. That should be negative. However, you've got Vb and Va reversed. Resistors are passive, so to compute the current you subtract the lower potential from the higher one, not the other way around.
 
**I set up my equations based on another example from class, what is the correct matrix set up for an example like this?

..-------Va----R2--I>---Vb------
|...|.....|...|
^...R1......R3...V
Is1...|.....|...Is2
.| _____|_____________|______|



**so after switching Va and Vb and correcting the neg/pos mistake, the current should be the same, right?
 
Regarding your equations: Let me show you how I interpreted them, using LaTeX.

marstery said:
1.7ma = (1/5k + 1/20k) -1/20k Va

I read this as:

[tex]1.7mA=\left(\frac{1}{5k\Omega}+\frac{1}{20k\Omega}\right)-\frac{1}{20k\Omega}V_a[/tex]

The equation, as I read it, is dimensionally wrong. The LHS has units of current. The first term on the RHS has units of reciprocal resistance. Also, Vb doesn't appear at all! Here is how the equation should read.

[tex]1.7mA=\frac{1}{5k\Omega}V_a+\frac{1}{20k\Omega}\left(V_a-V_b\right)[/tex]

On to the second equation.

-2ma = -1/20k (1/4k + 1/20k) Vb

This equation should read as follows.

[tex]-2mA=\frac{1}{4k\Omega}V_b-\frac{1}{20k\Omega}\left(V_a-V_b\right)[/tex]

But I checked your answers using KCL at nodes a and b, and they worked.
 
oh i see. i had set up a matrix as below where brackets above one another are actually one, but i just can't type them that way here.

[1.7ma] = [ (1/5k + 1/20k), -1/20k ] [ Va ]
[-2ma ] = [ -1/20k, (1/4k + 1/20k) ] [ Vb ]

thanks you for your help!
 
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