Finding nodal voltages using KCL.

  • Thread starter Xmo91
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    Kcl Nodal
In summary: Circuit Wizard shows V1 = 2.63 V and V2 = -9.33 V.In summary, the conversation was about finding node voltages at points A and B with a circuit that had a voltage of 60V. The equations and attempts at solving for V1 and V2 were provided, including the use of Cramer's rule. However, after correcting a sign error and omission and solving the equations, the calculated values for V1 and V2 did not match those shown in Circuit Wizard.
  • #1
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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  • #2
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
 

FAQ: Finding nodal voltages using KCL.

1. What is KCL?

KCL stands for Kirchhoff's Current Law. It is a fundamental principle in circuit analysis that states that the sum of currents entering and exiting a node (or junction) in a circuit must equal zero.

2. How is KCL applied to find nodal voltages?

KCL can be used to find the unknown nodal voltages in a circuit by setting up a system of equations based on the currents flowing into and out of each node. By solving these equations, the nodal voltages can be determined.

3. What are the steps for using KCL to find nodal voltages?

The steps for using KCL to find nodal voltages are:1. Identify all nodes in the circuit.2. Apply KCL at each node by writing an equation that relates the currents entering and exiting the node.3. Substitute known values and solve the resulting system of equations to find the nodal voltages.

4. Can KCL be applied to any circuit?

Yes, KCL can be applied to any linear circuit, regardless of its complexity. It is a general principle that applies to all circuits, as long as the circuit is well-defined and has a finite number of nodes.

5. Are there any limitations to using KCL to find nodal voltages?

One limitation of using KCL to find nodal voltages is that it assumes the circuit is in a steady state, meaning that all voltages and currents are constant. It may not be accurate for circuits with time-varying voltages or currents. Additionally, if the circuit has dependent sources, the KCL equations may become more complex and may require additional techniques to solve.

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