Finding nodal voltages using KCL.

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SUMMARY

The discussion focuses on calculating nodal voltages V1 and V2 using Kirchhoff's Current Law (KCL) in a circuit with a 60V source. The initial attempts yielded incorrect results when verified with Circuit Wizard. The correct equations derived from the circuit analysis are 119V1 - 35V2 = 640 and -7V1 + 13V2 = -140. The final calculated values for the node voltages are V1 = 2.6267V and V2 = -9.3548V, which differ from the initial incorrect values of 13.7V and -40.43V.

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Xmo91
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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!
 
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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
 

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