Finding nodal voltages using KCL.

  • Thread starter Thread starter Xmo91
  • Start date Start date
  • Tags Tags
    Kcl Nodal
Xmo91
Messages
4
Reaction score
0

Homework Statement



jayn7s.jpg


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.

Homework Equations





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)

please help!
 
Physics news on Phys.org
Xmo91 said:

Homework Statement



jayn7s.jpg


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.

Homework Equations


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)

please help!

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
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
20
Views
5K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
3
Views
7K
  • · Replies 5 ·
Replies
5
Views
9K