Equivalent resistance, using node voltage analysis

• yaro99
In summary: The conversation discusses a problem involving setting up a system of node voltage equations to find the equivalent resistance in a circuit. The problem suggests using MATLAB and the matrix method to solve the equations. The conversation also includes some confusion about the correct syntax for the equations in MATLAB, but ultimately the correct G matrix is determined and the equivalent resistance is found to be 0.111 Ω for the total equivalent resistance and 0.333 Ω for the equivalent resistance at node 1.)
yaro99

Homework Statement

(I added in the R values for reference)

GV=I
KCL

The Attempt at a Solution

I am not concerned with the MATLAB portion of the problem, only the setup.

I think what the problem wants me to do is:
1) connect a 1A test current from the ground source to v1 (like shown in the picture), find v1 using node voltage analysis, then find the equivalent resistance by dividing the test voltage (v1) by the test current, which is 1A.

2) connect a 1A test current from the ground source to v2. Repeat steps, where now v2 is the test voltage, to get equivalent resistance.

3) same as above, except with v4.

When I set up my system of equations I get too many unknowns.
For example, trying KCL on case 1 (where 1A is connected to v1), I get this system of equations:

v1/R1 + (v1-v2)/R2 + (v1-v3)/R6 = itest

(v2-v1)/R2 + (v2-v4)/R7 + (v2-v7)/R3 = 0

(v4-v2)/R7 + (v4-v3)/R11 + (v4-v6)/R12 = 0

Plugging in 1 for i and for all R values still gives me 6 different voltages to solve for, and only 3 equations.

Using the same method for case 2 and 3 (where vtest is v2 and v4 respectively), I get nearly identical systems of equations, the only difference being which equation is set equal to i instead of 0.

There are seven essential nodes, plus the reference node. So you'll need to write seven separate node equations. That's a lot of equations to bash away at by hand which is why the problem suggests using a computer and the matrix method; The node equations can be written by inspection into matrix form.

gneill said:
There are seven essential nodes, plus the reference node. So you'll need to write seven separate node equations. That's a lot of equations to bash away at by hand which is why the problem suggests using a computer and the matrix method; The node equations can be written by inspection into matrix form.

Thanks, this helped a bit but I am still stuck.
I wrote GV=I in matrix form, where G is a 3x7 matrix with all values being 1, and then V must be a 7x3 matrix. This means that I must be a 3x3 matrix. However, when I did the node voltage equations it ended up being a 1x7 matrix: [1; 0; 0; 0; 0; 0; 0]

Even when I make I a 3x3 matrix with 1 for the top left value and all others 0, MATLAB tells me I have the wrong dimensions when I perform the calculation.

Also, the problem says to use this in MATLAB: V=G\I

since the formula is GV=I, shouldn't the command then be V=I\G ?

yaro99 said:
Thanks, this helped a bit but I am still stuck.
I wrote GV=I in matrix form, where G is a 3x7 matrix with all values being 1, and then V must be a 7x3 matrix. This means that I must be a 3x3 matrix. However, when I did the node voltage equations it ended up being a 1x7 matrix: [1; 0; 0; 0; 0; 0; 0]

Even when I make I a 3x3 matrix with 1 for the top left value and all others 0, MATLAB tells me I have the wrong dimensions when I perform the calculation.

Also, the problem says to use this in MATLAB: V=G\I

since the formula is GV=I, shouldn't the command then be V=I\G ?

You should have 7 equations in 7 unknowns, the unknowns being the node voltages. The G matrix will be a square 7 x 7, the V and I vectors 1 x 7.

I must admit that I am not familiar with the particulars of MATLAB equation syntax.

I should have mentioned... also note that not all the entries of the G matrix will be 1. The entries on the diagonal will be the sum of the conductances connected to the given node, while the off-diagonal entries will have negative values (or zeros).

Last edited:
gneill said:
You should have 7 equations in 7 unknowns, the unknowns being the node voltages. The G matrix will be a square 7 x 7, the V and I vectors 1 x 7.

I must admit that I am not familiar with the particulars of MATLAB equation syntax.

Yeah I was just figuring this out. I ended up with a 7x7 G matrix, with 3's across the diagonal and either -1 or 0 values for the off-diagonals. I set I=[1;0;0;0;0;0;0], and using MATLAB obtained these voltage values:

v1 = 0.5V
v2 = v3 = 0.25V
v4 = v5 = v6 = v7 = 0.125V

Do these values make sense?

EDIT: They don't seem to make sense to me. Since Req = vtest/itest, I get Req = (0.5)*(1) = 0.5 Ω
That can't be true because the equivalent resistance is R1 which is 1 Ω just like all the other resistors.

Last edited:
yaro99 said:
Yeah I was just figuring this out. I ended up with a 7x7 G matrix, with 3's across the diagonal and either -1 or 0 values for the off-diagonals. I set I=[1;0;0;0;0;0;0], and using MATLAB obtained these voltage values:

v1 = 0.5V
v2 = v3 = 0.25V
v4 = v5 = v6 = v7 = 0.125V

Do these values make sense?
They aren't the node voltages that I'm seeing.

EDIT: They don't seem to make sense to me. Since Req = vtest/itest, I get Req = (0.5)*(1) = 0.5 Ω
That can't be true because the equivalent resistance is R1 which is 1 Ω just like all the other resistors.

The equivalent resistance between node 1 and the reference node won't be R1 since there are many more paths between those nodes than just the one through R1.

Recheck your G matrix. If you can, post an image of it.

I tried to check the values by part C) 3) of this problem. I ended up getting 0.111Ω for the total equivalent resistance, and 0.333Ω for the equivalent resistance at node 1. (Not sure which one I am supposed to be looking for)

This is what I got for my G matrix:

EDIT: Just realized I made a mistake in row 4. Here is the new G matrix:

(The only difference is a -1 value on the 4th row, 2nd column)

These are my newly obtained values:

v1 = 0.5833V
v2 = v3 = 0.3750V
v4 = 0.3333V
v5 = v7 = 0.2083V
v6 = 0.2500V

Last edited:
yaro99 said:
These are my newly obtained values:

v1 = 0.5833V
v2 = v3 = 0.3750V
v4 = 0.3333V
v5 = v7 = 0.2083V
v6 = 0.2500V

Yup. Matches what I'm seeing now. That's for the current source driving node 1.

You'll have to move the current source to other nodes to find their resistances to ground.

1 person
gneill said:
Yup. Matches what I'm seeing now. That's for the current source driving node 1.

You'll have to move the current source to other nodes to find their resistances to ground.

So then would this mean that for the current source driving node 1, Req = v1/itest = 0.5833/1 = 0.5833Ω ?

These are my results for the other 2 nodes:

node 2:
v1 = v3 = v6 = v7 = 0.3750V
v2 = 0.7500V
v4 = 0.5000V
v5 = 0.2500V

Req = 0.7500/1 = 0.75Ω

node 4:
v1 = v5 = v7 = 0.3333V
v2 = v3 = v6 = 0.5000V
v4 = 0.8333V

Req = 0.8333/1 = 0.8333Ω

Yes that looks good for nodes 2 and 4

1 person
Great, much appreciated!

1. What is equivalent resistance?

Equivalent resistance refers to the total resistance of a circuit when all resistors are combined into one single resistor. It represents the overall resistance that a current experiences as it flows through the circuit.

2. How is equivalent resistance calculated?

Equivalent resistance can be calculated using Ohm's law, which states that resistance is equal to the voltage divided by the current. For series circuits, equivalent resistance is the sum of individual resistances. For parallel circuits, it is the reciprocal of the sum of the reciprocals of individual resistances.

3. What is node voltage analysis?

Node voltage analysis is a method used to analyze circuits and determine the voltage at each node (connection point) in the circuit. It involves defining a reference node and using Kirchhoff's current law and Ohm's law to write equations that can be solved to find the unknown voltages.

4. Why is node voltage analysis useful for finding equivalent resistance?

Node voltage analysis allows us to determine the voltage at each node in the circuit, which is necessary for calculating equivalent resistance. By knowing the voltage at each node, we can use Ohm's law to calculate the current through each resistor, and then use Kirchhoff's current law to determine the total current in the circuit. The equivalent resistance is then calculated using Ohm's law.

5. Are there any limitations to using node voltage analysis for finding equivalent resistance?

Yes, there are some limitations to using node voltage analysis. It can only be used for circuits that can be represented by a system of linear equations, and it is not suitable for circuits with dependent sources or non-linear elements such as diodes. Additionally, it can become more complex and time-consuming for larger and more complex circuits.

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