Is it redundant to apply KVL to the outermost loop of a circuit if mesh eq. are already formed

Thread Starter

Reese Bautista

Joined Mar 4, 2018
9


The problem requires me to solve for the voltages across all the elements. I applied mesh analysis to the circuit above and came up with the following equations:
IA=-I2
Mesh A:
-2000Ia+1000Ib-V1+V2=0
Mesh B:
1000Ia-2000Ib+1000Ic=15
Mesh C:
1000Ib-2000Ic+V1=0
KCL on middle left node:
I1+I2=-IC

It can be easily seen that this is not enough to solve for everything as I have 6 unknowns and only 4 equations.
I was thinking of performing KVL on the entire outer loop but worried that it may be redundant as I've already performed mesh analysis. Is this possible?
If it is I'd still be left with one missing equation and at this point I'm not sure where I'm supposed to get it.
 

WBahn

Joined Mar 31, 2012
26,321
You surmised correctly. You only have three independent loop equations. All others are linear combinations of those three and hence are redundant.

What are those rectangular blocks? Based on what you've given here, they could be anything from unknown resistors to independent voltage or current sources to dependent voltage or current sources. Without additional information or constraints, the circuit has an indeterminate number of solutions.
 

Thread Starter

Reese Bautista

Joined Mar 4, 2018
9
You surmised correctly. You only have three independent loop equations. All others are linear combinations of those three and hence are redundant.

What are those rectangular blocks? Based on what you've given here, they could be anything from unknown resistors to independent voltage or current sources to dependent voltage or current sources. Without additional information or constraints, the circuit has an indeterminate number of solutions.
Oh sorry. I forgot to mention that they are nonlinear devices. Can I make use of the equation for I1? I wasn't really sure so I didn't include it.


Thank you by the way I've been asking a lot of questions as this is really confusing to me and I have an exam coming up soon it's been really helpful.
 

WBahn

Joined Mar 31, 2012
26,321
So first and foremost you have to keep in mind that mesh current analysis is fundamentally only valid for LINEAR circuits.

The reason is simple. At the heart of linear analysis is the assumption that the voltage across a device can be written as a function of the superposition of the currents through the device.

So, for instance, if

V = I·R

and I = I1 - I2

we can write the voltage as

V = (I1 - I2) = I1R - I2R

This allows us to look a the voltage as consisting of purely a voltage due to the presence of I1 added to a voltage due solely to I2.

But as soon as you add in a nonlinear component, this is no longer the case.

For instance, let's say that

V = I²·R (note that R here does NOT have units of ohms, it has units of volts/amps², or Ω/A)

Now we have

V = (I1 - I2)²·R = I1²·R - I1·I2·R + I2²·R

That cross term means that we can't superimpose a voltage due to I1 with a voltage due to I2.

It is seldom the case that you can get a closed form solution for a circuit having nonlinear elements.

Normally what you do is an iterative approach. You first solve the system assuming some initial value for the currents (such as zero) and then find out what the voltage across the elements would be. Then you use that to determine what the current would be and you solve the system again using those currents and find out what the voltages would be. Usually, you don't have to repeat this process too long before you converge on a solution to whatever level of accuracy you need.

Set up a spread sheet for something like this is VERY useful.
 

Thread Starter

Reese Bautista

Joined Mar 4, 2018
9
By
So first and foremost you have to keep in mind that mesh current analysis is fundamentally only valid for LINEAR circuits.

The reason is simple. At the heart of linear analysis is the assumption that the voltage across a device can be written as a function of the superposition of the currents through the device.

So, for instance, if

V = I·R

and I = I1 - I2

we can write the voltage as

V = (I1 - I2) = I1R - I2R

This allows us to look a the voltage as consisting of purely a voltage due to the presence of I1 added to a voltage due solely to I2.

But as soon as you add in a nonlinear component, this is no longer the case.

For instance, let's say that

V = I²·R (note that R here does NOT have units of ohms, it has units of volts/amps², or Ω/A)

Now we have

V = (I1 - I2)²·R = I1²·R - I1·I2·R + I2²·R

That cross term means that we can't superimpose a voltage due to I1 with a voltage due to I2.

It is seldom the case that you can get a closed form solution for a circuit having nonlinear elements.

Normally what you do is an iterative approach. You first solve the system assuming some initial value for the currents (such as zero) and then find out what the voltage across the elements would be. Then you use that to determine what the current would be and you solve the system again using those currents and find out what the voltages would be. Usually, you don't have to repeat this process too long before you converge on a solution to whatever level of accuracy you need.

Set up a spread sheet for something like this is VERY useful.
Are the currents that I'm supposed to assign a value for I1 and I2? Or all the currents? Am I still gonna use the mesh equations to determine the voltages across the elements?
 

WBahn

Joined Mar 31, 2012
26,321
That you have been assigned this problem indicates that you have been shown, at some point, how to work with nonlinear elements. You might want to go back and review that material.
 

Thread Starter

Reese Bautista

Joined Mar 4, 2018
9
That you have been assigned this problem indicates that you have been shown, at some point, how to work with nonlinear elements. You might want to go back and review that material.
We haven't been shown how to work with nonlinear materials which is why I'm really lost in what to do. I tried emailing my professor about this but he only said that node or mesh analysis would still be applicable but you said it wouldn't so I honestly don't know what to do now.
 

WBahn

Joined Mar 31, 2012
26,321
If you only have a single non-linear element in a circuit, then you can treat it as the load of a Thevenin/Norton equivalent and then use a load line analysis to find the operating point. I don't any obvious way to cleave off your two non-linear elements to create such a load.

Nodal analysis lends itself to setting up iterable equations for non-linear circuits much more so than mesh analysis does.

So first set up your node equations and then let's take it from there.
 
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