Double checking my mesh analysis equations

Thread Starter

Reese Bautista

Joined Mar 4, 2018
9

I turned meshes 1,4 and 5,3 into supermesh as they have a current source between them.

These are the equations I came up with. Can you guys point out any mistakes?


Extra questions:
1) Am I correct in assuming that there are 8 nodes in total? Or do the connections between the 1.5k ohms resistors and 10mV sources not count?
2) How can I determine Vo?
2) What does it mean when I'm told draw the Thevenin equivalent with respect to Vo?

Thank you.
 

WBahn

Joined Mar 31, 2012
25,545
1) Am I correct in assuming that there are 8 nodes in total? Or do the connections between the 1.5k ohms resistors and 10mV sources not count?
Yes, you have 8 nodes total. However, the two that you mention specifically are known as "nonessential" nodes because the solution to the KCL equation at each is trivial and merely tells you that whatever current flows in one of the connected components flows in the other, so we can ignore them from an analysis standpoint.

Since we get to arbitrarily declare what the voltage is at a node of our choice, that leaves us with five essential nodes having unknown voltages. Note that this matches the number of meshes you have. This should always be the case.

2) How can I determine Vo?
By analyzing the circuit.

If I were to tell you (just making up numbers) that i1 = 10 mA and that i3 = 12 mA, can you tell me what Vo is?

If so, then once you analyze the circuit to find all of the mesh currents, can't you find Vo?

2) What does it mean when I'm told draw the Thevenin equivalent with respect to Vo?
Just that.

What is a Thevenin equivalent circuit? Describe it in your own words (but feel free to read up on it as much as you need to in order to do that).
 

Thread Starter

Reese Bautista

Joined Mar 4, 2018
9
Yes, you have 8 nodes total. However, the two that you mention specifically are known as "nonessential" nodes because the solution to the KCL equation at each is trivial and merely tells you that whatever current flows in one of the connected components flows in the other, so we can ignore them from an analysis standpoint.

Since we get to arbitrarily declare what the voltage is at a node of our choice, that leaves us with five essential nodes having unknown voltages. Note that this matches the number of meshes you have. This should always be the case.



By analyzing the circuit.

If I were to tell you (just making up numbers) that i1 = 10 mA and that i3 = 12 mA, can you tell me what Vo is?

If so, then once you analyze the circuit to find all of the mesh currents, can't you find Vo?



Just that.

What is a Thevenin equivalent circuit? Describe it in your own words (but feel free to read up on it as much as you need to in order to do that).
Thanks. I double checked my equations and realized I made some mathematical errors. Was eventually able to solve Vo which was (not 100% sure on this one but hey) 19.99 mV which I realized also work as the thevenin voltage.
 

WBahn

Joined Mar 31, 2012
25,545
Thanks. I double checked my equations and realized I made some mathematical errors. Was eventually able to solve Vo which was (not 100% sure on this one but hey) 19.99 mV which I realized also work as the thevenin voltage.
Ask yourself is this makes sense. Consider the symmetry of the left and right halves of the circuit. Do you see ANY difference between them? If not, then is there any reason to think that the voltage one either node that defines Vo will be different than the other?
 

Thread Starter

Reese Bautista

Joined Mar 4, 2018
9
Ask yourself is this makes sense. Consider the symmetry of the left and right halves of the circuit. Do you see ANY difference between them? If not, then is there any reason to think that the voltage one either node that defines Vo will be different than the other?
It does. While there is no difference between the right and left halves of the circuit, the polarity of the resistors connected to Vo have different polarities once the currents are solved.
 

WBahn

Joined Mar 31, 2012
25,545
It does. While there is no difference between the right and left halves of the circuit, the polarity of the resistors connected to Vo have different polarities once the currents are solved.
Polarity of the resistors????

If there is no difference between the right half and the left half, then on what basis can it be argued that the voltage on the left is higher or lower than the voltage on the right.

It's like taking two tanks of water and putting a fitting at the same height in each and connecting them together with a hose and then filling both tanks to the same height with the same amount of water. If asked which direction the water is flowing through the hose, the symmetry says that any argument that says that it flows to the left is equally valid for an argument that says it flows to the right. Hence any flow at all would result in a contradiction and, thus, the conclusion is that no current flows at all.

Since you have chosen not to show your corrected work, there is no way for us to help you find where your mistake is.
 

Thread Starter

Reese Bautista

Joined Mar 4, 2018
9
Polarity of the resistors????

If there is no difference between the right half and the left half, then on what basis can it be argued that the voltage on the left is higher or lower than the voltage on the right.

It's like taking two tanks of water and putting a fitting at the same height in each and connecting them together with a hose and then filling both tanks to the same height with the same amount of water. If asked which direction the water is flowing through the hose, the symmetry says that any argument that says that it flows to the left is equally valid for an argument that says it flows to the right. Hence any flow at all would result in a contradiction and, thus, the conclusion is that no current flows at all.

Since you have chosen not to show your corrected work, there is no way for us to help you find where your mistake is.
I'm so sorry. It completely skipped my mind that I should have posted the corrected work. Here it is. I'm really confused.

 

WBahn

Joined Mar 31, 2012
25,545
Let's suppose that I1 = 0 and I2 = 10 mA. Will V1 be +15 V or -15 V ?

First determine the answer direction from the schematic. Then determine the answer from your set up equations in the upper right corner of your work.

Look at your equation for Vo at the end. Instead of trying to do both the set up and the substitution of values at the same time, do the set up first and check if you agree with it.

Your set up is

Vo - (50 kΩ)(I4) - (50 kΩ)(I5) = 0

NOW substitute in your values for I4 and I5.
 

Thread Starter

Reese Bautista

Joined Mar 4, 2018
9
Assuming I1=0 and I2=10 mA, V1= -15000 mV = -15V.

Vo-(50k)(1.999E-7)-(50k)(1.999E-7) = 0 => Vo = 0.0199 V

I'm assuming you wanted me to put in the negative current valye for I4 but aren't negative currents imply a wrong assumption about its direction?
 

WBahn

Joined Mar 31, 2012
25,545
Assuming I1=0 and I2=10 mA, V1= -15000 mV = -15V.

Vo-(50k)(1.999E-7)-(50k)(1.999E-7) = 0 => Vo = 0.0199 V

I'm assuming you wanted me to put in the negative current valye for I4 but aren't negative currents imply a wrong assumption about its direction?
So what if it does?

Your definition of I4 says that a positive value for I4 means that current is flowing down through the 50 kΩ resistor and the reason you have a subtraction operator after Vo in your equation is because IF I4 is positive then it results in a voltage drop across that resistor. But since I4 turns out to be negative, that means that it produces a voltage gain in the resistor. The negative sign associated with the value of I4 cancels out the subtraction operator and results in a voltage gain. If you want to use a positive value for the magnitude of the current in that resistor, you better change that subtraction operator after Vo to an addition operator to be consistent with it -- but it is FAR better to just let the math take care of itself.
 

The Electrician

Joined Oct 9, 2007
2,777
Your results for I1 and I3 have the wrong signs. In your image of your calculations you have V1 = 1500(I1-I2), but if you look at the indicated sign of V1 in the schematic you will see that actually V1 = 1500(I2-I1). The same problem exists for your calculation of V2.

WBahn has suggested a couple of times that you consider the extreme symmetry of this circuit. Plainly I2 must be exactly zero, not some number like 1.312*10^-23, which is obviously just residual round-off error.
 
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