Hi !
Sorry about delayed response but when I'm on business trip I'm not always able to be online !
The two equations I come up with are :
4,2-82*I1-57(I2+0,6A)-76,9*I1=0
4,2-22(I1-I2)-150(I1-I2-0,6A)-76,9=0

Solving these I get I1=0,6 A and I2=0,05 A

Best regards

Hi,

Sorry to have to say but i dont get those two currents when i use your two equations, but more importantly i still dont know what you are trying to do in the analysis. I know you are trying to find the best resistor value for the max power, but i dont know what kind of circuit analysis you are trying to use when you write those two equations, and i also cant be sure what you are calling I1 and I2.

The schematic needs to be annotated such as that shown in this new attachement. Note it makes it clear where I1 and I2 are, and which way they are oriented. I1 is clearly in the upper left branch, and pointing downward, and I2 in the upper right branch, pointing downward also. Note the RED arrowheads show the assumed direction of currents.
These may or may not be your choices, but if not, then you have to show what your choices were otherwise i cant tell what you are doing wrong, if anything.

I can tell you though that the solutions you found using your two equations do not look correct. You can prove this by trying to calculate the node voltages and see if the currents add up properly at the nodes.

So really the first thing you should do next is either verify that this new drawing shows the currents correctly, the way you wanted them to be, or you'll have to change them to show the way you want them to be, or you could just accept them as a decent choice.
Note i also included a ground in a convenient place, but that's up to you too.

 

Hi,

Sorry to have to say but i dont get those two currents when i use your two equations, but more importantly i still dont know what you are trying to do in the analysis. I know you are trying to find the best resistor value for the max power, but i dont know what kind of circuit analysis you are trying to use when you write those two equations, and i also cant be sure what you are calling I1 and I2.

The schematic needs to be annotated such as that shown in this new attachement. Note it makes it clear where I1 and I2 are, and which way they are oriented. I1 is clearly in the upper left branch, and pointing downward, and I2 in the upper right branch, pointing downward also. Note the RED arrowheads show the assumed direction of currents.
These may or may not be your choices, but if not, then you have to show what your choices were otherwise i cant tell what you are doing wrong, if anything.

I can tell you though that the solutions you found using your two equations do not look correct. You can prove this by trying to calculate the node voltages and see if the currents add up properly at the nodes.

So really the first thing you should do next is either verify that this new drawing shows the currents correctly, the way you wanted them to be, or you'll have to change them to show the way you want them to be, or you could just accept them as a decent choice.
Note i also included a ground in a convenient place, but that's up to you too.

Hi !

Thank's for the patience !
The problem is as in the schematic !
It was one of the problems to solve on a exam ! We only did have this drawing and the question is what's the
resistance of the unknown resistor for maximum power dissipation in it and what will the power be ?
The only shown by the task was the resistor values in the bridge the current source 0,6 A and voltage source 4,2 V !
Nothing else such as current direction or...

It's why I ask this here because I only know how to calculate the eq resistance but I have no idea how to calculate current ,
direction of current or voltage in this setup !

 

Hi !

Thank's for the patience !
The problem is as in the schematic !
It was one of the problems to solve on a exam ! We only did have this drawing and the question is what's the
resistance of the unknown resistor for maximum power dissipation in it and what will the power be ?
The only shown by the task was the resistor values in the bridge the current source 0,6 A and voltage source 4,2 V !
Nothing else such as current direction or...

It's why I ask this here because I only know how to calculate the eq resistance but I have no idea how to calculate current ,
direction of current or voltage in this setup !

But YOU are the one that created I1 and I2 completely out of thin air. That's perfectly fine, but it is then up to YOU to define and communicate what I1 and I2 are!

Is I1 the current flowing out of the positive terminal in the voltage source?

Is I2 the current flowing downward in R4?

YOU have chosen to use I1 and I2 to mean SOMETHING. But WE have no idea WHAT you mean unless you TELL us!

If you don't know how to even make an attempt at analyzing the circuit once you have determined the needed value of the load resistance, then you need to drop the course and take it again from scratch, because you will only be digging yourself a deeper and deeper hole if you don't. If, instead, you want to draw on the stuff you are supposed to have already learned about analyzing simple circuits and at least make an honest attempt to communicate the work you have done, then we can help you find where you are going right and where you are going wrong and try to get you back on solid ground.

 

But YOU are the one that created I1 and I2 completely out of thin air. That's perfectly fine, but it is then up to YOU to define and communicate what I1 and I2 are!

Is I1 the current flowing out of the positive terminal in the voltage source?

Is I2 the current flowing downward in R4?

YOU have chosen to use I1 and I2 to mean SOMETHING. But WE have no idea WHAT you mean unless you TELL us!

If you don't know how to even make an attempt at analyzing the circuit once you have determined the needed value of the load resistance, then you need to drop the course and take it again from scratch, because you will only be digging yourself a deeper and deeper hole if you don't. If, instead, you want to draw on the stuff you are supposed to have already learned about analyzing simple circuits and at least make an honest attempt to communicate the work you have done, then we can help you find where you are going right and where you are going wrong and try to get you back on solid ground.

The idea is that current I1 is the current going out from the voltage source through the right branch and then I1 adding up with the 0,6 A from the current source out to the unknown resistor !
Then I2 is the part of the current from the voltage source going through the left branch adding up with the 0,6 A from the current source out
to the unknown resistor !
And I also understand a part of this current in both branches will also be opposed by the current from the 0.6 A current source !?
The 0.6 A current from the current source will absolutely go downwards in the two branches (57 & 82 ohm resistors)
The 0.6 A current source will generate 0.6 A no matter what but how will the adding (or subtracting) part from the voltage source act ?

 

The idea is that current I1 is the current going out from the voltage source through the right branch and then I1 adding up with the 0,6 A from the current source out to the unknown resistor !
Then I2 is the part of the current from the voltage source going through the left branch adding up with the 0,6 A from the current source out
to the unknown resistor !

THANK YOU!!!

Finally, we are making some progress. And it also starts to become clear why your equations weren't making much sense.

Try drawing the currents, as you have described them, on your diagram. You can't! You have I1 flowing out of the voltage source. Fine. But then you say that it is also the current flowing in the right branch. That is only possible if the current in the left branch is identically zero, otherwise you violate KCL. Leaving that aside, you say that it then adds up with the 0.6 A from the current source and then goes out to the resistor. But after it combines with the output of the current source, it does NOT go out to the unknown resistor. There is a junction with the unknown resistor, the 57 Ω resistor, and the 150 Ω resistor in the way and current in the 150 Ω resistor is either added or subtracted from it before the rest goes to the unknown resistor.

You then say that I2 goes through the left branch and adds up with the 0.6 A from the current source and then goes out to the unknown resistor. But this branch is going to the other end of the current source. Does it make sense that, in both cases, the 0.6 A from that source will add to the current in question?

So here's a novel idea. START with an annotated diagram:



All you need is I1 and I2. I added I3 and I4 to make the steps easier to follow.

We know that

I1 + I2 = I3 + I4
I3 = I1 - 0.6 A
I4 = I2 + 0.6 A

Now apply KVL around both the green and the blue loops by summing up the voltage gains (or drops) around each.

Green: 4.2 V - (I1 · 22 Ω) - (I3 · 150 Ω) - ((I3 + I4) · Rx) = 0

where Rx = 76.9 Ω as previously determined.

Now you write the one for blue.

Then, using the relationships above, solve for (I3 + I4) and you are done (well, ready to find the power in the unknown resistor.

And I also understand a part of this current in both branches will also be opposed by the current from the 0.6 A current source !?

That depends on what you mean by "opposed".

The 0.6 A current from the current source will absolutely go downwards in the two branches (57 & 82 ohm resistors)

On what basis do you make this claim? It could flow up in the 57 Ω and down in the 82 Ω. It could flow up in both. It could flow down in both. The only think you know is that it will not flow down in the 57 Ω and up in the 82 Ω.

The 0.6 A current source will generate 0.6 A no matter what but how will the adding (or subtracting) part from the voltage source act ?

That is what circuit analysis is for. But do you see how annotating your diagram, even with just the two red currents and nothing else, makes it painfully obvious how these current must interact with the current from the source?

 

FWIW -- I think the easiest way to tackle this problem is to find the Thevenin voltage via the open-circuit voltage when the unknown resistor is removed. This results in an extremely simple circuit to analyze since we know there is no current in the voltage source. At an appropriate time I'll show you how to do that solution. For now, it's probably better that you stick to classic plug-and-chug techniques.

 

THANK YOU!!!

Finally, we are making some progress. And it also starts to become clear why your equations weren't making much sense.

Try drawing the currents, as you have described them, on your diagram. You can't! You have I1 flowing out of the voltage source. Fine. But then you say that it is also the current flowing in the right branch. That is only possible if the current in the left branch is identically zero, otherwise you violate KCL. Leaving that aside, you say that it then adds up with the 0.6 A from the current source and then goes out to the resistor. But after it combines with the output of the current source, it does NOT go out to the unknown resistor. There is a junction with the unknown resistor, the 57 Ω resistor, and the 150 Ω resistor in the way and current in the 150 Ω resistor is either added or subtracted from it before the rest goes to the unknown resistor.

You then say that I2 goes through the left branch and adds up with the 0.6 A from the current source and then goes out to the unknown resistor. But this branch is going to the other end of the current source. Does it make sense that, in both cases, the 0.6 A from that source will add to the current in question?

So here's a novel idea. START with an annotated diagram:

View attachment 113841

All you need is I1 and I2. I added I3 and I4 to make the steps easier to follow.

We know that

I1 + I2 = I3 + I4
I3 = I1 - 0.6 A
I4 = I2 + 0.6 A

Now apply KVL around both the green and the blue loops by summing up the voltage gains (or drops) around each.

Green: 4.2 V - (I1 · 22 Ω) - (I3 · 150 Ω) - ((I3 + I4) · Rx) = 0

where Rx = 76.9 Ω as previously determined.

Now you write the one for blue.

Then, using the relationships above, solve for (I3 + I4) and you are done (well, ready to find the power in the unknown resistor.



That depends on what you mean by "opposed".



On what basis do you make this claim? It could flow up in the 57 Ω and down in the 82 Ω. It could flow up in both. It could flow down in both. The only think you know is that it will not flow down in the 57 Ω and up in the 82 Ω.



That is what circuit analysis is for. But do you see how annotating your diagram, even with just the two red currents and nothing else, makes it painfully obvious how these current must interact with the current from the source?

Thank's

What I did find difficult in this is when there was a current source involved in this instead of just a simple resistor !

Sorry that I have been unclear and this has been annoying for You but I am only a few weeks into my course !
I will try to understand your explanation now and try to solve this !

 

Hello again,

Another way to handle this is to analyze the entire circuit with Rx included and calculate either the current through or the voltage across Rx. The power is then either of:
1. P=P(Rx)=iRx^2*Rx
2. P=P(Rx)=vRx^2/Rx

Having the power now we can take the derivative and set it equal to zero to find the critical points:
dP/dRx=0

Solving this for Rx we can find the value of Rx that produces the highest power in Rx.
After using nodal and doing this i got Rx=23908/311 exactly. Approximately that is 76.874598 and rounded that is your 76.9 ohms.

Since you are approaching this from the standpoint of the current that would be the one to use (#1 above). That's if you dont want to use the maximum power transfer theorem.

 

Hello again,

Another way to handle this is to analyze the entire circuit with Rx included and calculate either the current through or the voltage across Rx. The power is then either of:
1. P=P(Rx)=iRx^2*Rx
2. P=P(Rx)=vRx^2/Rx

Having the power now we can take the derivative and set it equal to zero to find the critical points:
dP/dRx=0

Solving this for Rx we can find the value of Rx that produces the highest power in Rx.
After using nodal and doing this i got Rx=23908/311 exactly. Approximately that is 76.874598 and rounded that is your 76.9 ohms.

Since you are approaching this from the standpoint of the current that would be the one to use (#1 above). That's if you dont want to use the maximum power transfer theorem.

Which is the Rth value of the bridge network (76.9 ohms) - as we would expect. Buried in the TS equations is the value 76.9 so I assume he already calculated Rth - which is good. However, when he attempted to calculate Vth he did it looking into the bridge network from the terminals of the current source, instead of the same terminals that he looked into to get Rth. That is why his Vth is wrong. He needs to disconnect the bridge from the rest of the circuit and find the voltage at the top of the bridge and the bottom of the bridge and find the difference in those voltages to get Vth.


.

 

Hi,

Yes that is a good idea too. I'd like to see him do it at least two different ways now since he's having so much trouble with this.
Would be a very good exercise.

 

Hello again,

Taking another quick look at the first post, i see that this problem was in the D to Y and Y to D section, which means they probably wanted the student to convert those three resistors in the Y configuration into a delta, then solve.

 

Hello again,

Taking another quick look at the first post, i see that this problem was in the D to Y and Y to D section, which means they probably wanted the student to convert those three resistors in the Y configuration into a delta, then solve.

What three resistors? The bottom two resistors and the unknown load resistance? I guess you could, but I would think it would make the analysis more difficult since the unknown resistance would be reflected in all three of the new delta resistors. In general you would then need to transform back since your component of interest, Rx, is now embedded in a transformed equivalent, but in this case you could use the transformed circuit to find the current in the voltage supply and then leverage the fact that Rx is in series with that supply.

My take (which could well be wrong) is that he is saying that the circuit is "like" the configuration in that chapter, but that it is different. Note that the bridge resistor has been replaced by the current source and the unknown resistor has been added. So I don't think this is intended as a delta-wye transform problem.