# Finding nodal voltages using KCL.

1. Aug 18, 2010

### Xmo91

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

Where XX is 60V

I'm trying to find the node voltages at point A and B, however each time I've tried this and then checked using circuit wizard it has come up incorrect. Yes my circuit in circuit wizards is correct.

2. Relevant equations

3. The attempt at a solution

V1: I1=I2+I3+Iground

60-V1/25 = V1-V2/40 + V1-40/30 + V1-0/10

then i multiplied by 600 to remove the denominators

1440-24V1=60V1 +15V1-15V2+2-V1-800
2240=119V1-15V2 (Eqn 1)

V2: I2+I3=2.5+Iground2

V1-V2/40 +V1-40/30=2.5 + V2-0/20

then i multiplied by 120 to remove the denominators

3V1 - 3V2 + 4V1 - 160 = 300 + 6V2
7V1-9V2=460

I then put them into a matrix and tried using cramers rule to solve for v1 and v2

119 -15 l V1 l l 2240 l
7 -9 l V2 l l 460 l

119x-9 - -15 x 7
Doriginal= -966

2240 -15
460 -9

2240 x -9 - -15 x 460
D1= -13260

119 2240
7 460

119 x 460 - 2240 x 7
=39060

V1= -13260/-966 = 13.7V (Circuit wizard shows it as being 2.37V)
V2= 39060/-966 = -40.43V (Circuit wizard shows it as being -9.71V)

2. Aug 18, 2010

### The Electrician

I would put everything on the left side of an equation, and equate to zero, include parentheses and fix a sign error and an omission:

(V1-V2)/40 + (V1-V2+40)/30 + (V1-0)/10 + (V1-60)/25 = 0

Multiply through by 600, collect terms and get Eqn 1:

119V1 - 35V2 = 640

For the second equation:

(V2-V1-40)/30 + (V2-V1)/40 + (V2-0)/20 + 2.5 = 0

Multiply through by 120 and get Eqn 2:

-7V1 + 13 V2 = -140

Solving, I get:

V1 = 570/217 = 2.6267
V2 = -290/31 = -9.3548