# Value of the unknown resistor and Maximum power transfer inthe unknown resistor !

#### tmlekus

Joined Mar 11, 2016
10
If I have a configuration like the one in chapter 10 (D-Y and Y-D Conversions), Wheatstone image picture with the 10 V source and replace the 6 ohm in the middle with a current source for let say 0,6 A instead and add a resistor in series with the voltage source , what value should this resistor have for maximum power in that resistor ? How do I calculate the equivalent total resistance in the rest of the configuration ? Maximum power is transferred to this unknown resistor when this resistance is equivalent !?

Any ideas ??

Best regards
//Tommy

#### MrAl

Joined Jun 17, 2014
11,276
Hi,

What type of analysis did you learn in the past so far, such as Nodal, Loop, etc.?
There are other ways too but unless you say what you did already we wont know the best approach to help with this problem.

Also, did you start to analyze this yet, as that is the de facto standard in the homework section, unless you have absolutely no idea how to proceed.

For example, you could solve for the node voltages and then calculate the currents and the power in each element if necessary. You can then start to figure out what you need to do to solve your problem. First you probably want to choose a ground node, such as the bottom of the 10v battery.

#### tmlekus

Joined Mar 11, 2016
10
Hi,

What type of analysis did you learn in the past so far, such as Nodal, Loop, etc.?
There are other ways too but unless you say what you did already we wont know the best approach to help with this problem.

Also, did you start to analyze this yet, as that is the de facto standard in the homework section, unless you have absolutely no idea how to proceed.

For example, you could solve for the node voltages and then calculate the currents and the power in each element if necessary. You can then start to figure out what you need to do to solve your problem. First you probably want to choose a ground node, such as the bottom of the 10v battery.
Hi !
This was one of the questions on my latest exam !
I'm familiar with Branch , Nodal , Mesh , Superposition , Millman's and Norton Thevenin !

My idea is as follows !
4,2-12*I2-18(I2+0,6)-15*I2+I1=0 ( eq 1)
4,2-18*I1-12(I1-0,6)-15*I2+I1=0 ( eq 2)
Then solve these eq to find I1 and I2 !?

Best regards
//Tommy

#### MrAl

Joined Jun 17, 2014
11,276
Hi !
This was one of the questions on my latest exam !
I'm familiar with Branch , Nodal , Mesh , Superposition , Millman's and Norton Thevenin !

My idea is as follows !
4,2-12*I2-18(I2+0,6)-15*I2+I1=0 ( eq 1)
4,2-18*I1-12(I1-0,6)-15*I2+I1=0 ( eq 2)
Then solve these eq to find I1 and I2 !?

Best regards
//Tommy
Hi,

Sorry to say it is unclear what you are trying to do and where you are getting some of the values you are using in those equations. Maybe you can clarify a little what it is you are doing.
For example, in the circuit there are values of 10, 0.6, 12, and 16, so where are you getting 4.2 and 15?
Also, it appears that you have added a voltage and a current in both equations which means it cant be right.
If you state what you are doing and why you are doing it we can move on.
Remember your goal is to eventually solve for R3 based on the power in R3.

#### tmlekus

Joined Mar 11, 2016
10
Oh , sorry !
I mixed the values from the configuration in my paper with the ones from the copied picture I have pasted !
The 4,2 value is the voltage source I have and the 10 is the one in the picture !
But the 15 is the value in ohm if I disconnect the resistor and measure the resistance between the two points !

Here are all the values that I have to deal with
So in the two equations I have written it should be i.e replace the earlier ohmic values with :

4.2-82*I2-57(I2+0.6)-76,9*I1=0
4.2-72(I1-I2)-150(I1-I2-0,6)-76.9*I1=0

#### MrAl

Joined Jun 17, 2014
11,276
Hello again,

Well, perhaps you can describe carefully what your strategy is going to be for this network.
It does not seem clear because apparently you are calculating the equivalent resistance, so i am not sure what those equations are for then. If you do want to use those equations then show your currents also, I1 and I2.
I am assuming you still want to calculate the maximum power in the unknown resistor with choice of value of that resistor.

#### tmlekus

Joined Mar 11, 2016
10
Hi again !

Correct , and this is where I'm confused !
To get maximum power dissipated in the unknown resistor I assume the resistor value must
have the equivalent resistor value as the resistance in the bridge configuration... !?
And what will then the current and the power in this resistor be ?

Best regards

#### WBahn

Joined Mar 31, 2012
29,876
Hi !
This was one of the questions on my latest exam !
I'm familiar with Branch , Nodal , Mesh , Superposition , Millman's and Norton Thevenin !

My idea is as follows !
4,2-12*I2-18(I2+0,6)-15*I2+I1=0 ( eq 1)
4,2-18*I1-12(I1-0,6)-15*I2+I1=0 ( eq 2)
Then solve these eq to find I1 and I2 !?

Best regards
//Tommy
You use I1 and I2 and expect everyone to read your mind and know which current is which in the circuit. This forces us to guess -- and engineering is NOT about guessing. If you want to use I1 and I2 in your analysis, fine. But it is YOUR responsibility to annotate your diagram to communicate to your readers just what I1 and I2 refer to.

You have a '15' in there for no apparent reason, given that your circuit has two 12 Ω resistors and two 18 Ω resistors.

You give no hint as to which loop either of your two equations correspond to.

Where did this 4.2 come from? Is it a current? A voltage?

Your equations mix terms that are voltages and terms that are currents. You couldn't catch it because you couldn't be bothered to track your units. If you had, you would have spotted right away that your equations are guaranteed to be wrong and, therefore, there would be no need to go any further. Always, always, ALWAYS track your units!

#### WBahn

Joined Mar 31, 2012
29,876
Oh , sorry !
I mixed the values from the configuration in my paper with the ones from the copied picture I have pasted !
The 4,2 value is the voltage source I have and the 10 is the one in the picture !
But the 15 is the value in ohm if I disconnect the resistor and measure the resistance between the two points !

Here are all the values that I have to deal with
So in the two equations I have written it should be i.e replace the earlier ohmic values with :

4.2-82*I2-57(I2+0.6)-76,9*I1=0
4.2-72(I1-I2)-150(I1-I2-0,6)-76.9*I1=0

View attachment 113763
Where are some of these values coming from? Where does this magical 76.9 Ω (assuming it's a resistance) come from?

Again, what is I1 and I2?

Where does the 15 Ω come from?

"disconnect the resistor"? Which resistor -- there are five of them.

"resistance between the two points"? Which two points.

Learn to communicate, with precision, the information you are trying to convey.

#### MrAl

Joined Jun 17, 2014
11,276
Where are some of these values coming from? Where does this magical 76.9 Ω (assuming it's a resistance) come from?

Again, what is I1 and I2?

Where does the 15 Ω come from?

"disconnect the resistor"? Which resistor -- there are five of them.

"resistance between the two points"? Which two points.

Learn to communicate, with precision, the information you are trying to convey.

Hi,

He had mentioned that he wrote the wrong values on the original schematic, such as the "10v" when it should have been "4.2v" and the resistors all changed too

The magical 15 ohms is the bridge equivalent resistance of the first circuit, and the 76.9 ohms is a rounded version of the equivalent resistance of the second circuit.
Because of these two values i had to question what the equations were for then. He seems to be indicating now that he also wants to calculate the power in the resistor after finding the right value.

#### MrAl

Joined Jun 17, 2014
11,276
Hi again !

Correct , and this is where I'm confused !
To get maximum power dissipated in the unknown resistor I assume the resistor value must
have the equivalent resistor value as the resistance in the bridge configuration... !?
And what will then the current and the power in this resistor be ?

Best regards
Hi,

The maximum power transfer theory says that the resistances must be equal for max power.
So you are saying you want to now calculate the power in the resistor?
If so, then yes you have to calculate the current through it or the voltage across it.
It's a little easier to check if you provide some numerical values for the power, or the two currents or whatever it is you are calculating. If you calculate the current in the upper two bridge resistors, then you can add them and get the total current which is the current through the unknown resistor which will be known already because of your equivalent value.

#### DGElder

Joined Apr 3, 2016
351
Transform the bridge into the Thevenin equivalent. Rth is the resistance of the network looking into the top and the bottom of the bridge - with the current source gone. With the current source back in calculate the open circuit voltage at those two points to get Vth. You then have a simple series circuit with two resistors, each equal Rth for max power in R?, and two voltage sources. You can then easily calculate the current in R? to get its power dissipation.

Last edited:

#### tmlekus

Joined Mar 11, 2016
10
Hi !

Yes , if I transform the bridge into a Thevenin I will get 41,5 V in series vith 4,2 V but , will these two add
or oppose each other ?

Best regards

#### WBahn

Joined Mar 31, 2012
29,876
Hi !

Yes , if I transform the bridge into a Thevenin I will get 41,5 V in series vith 4,2 V but , will these two add
or oppose each other ?

Best regards
Analyze the circuit. If you remain uncertain, show your work and we should be able to show you how to determine the answer to that question unambiguously.

#### MrAl

Joined Jun 17, 2014
11,276
Hi !

Yes , if I transform the bridge into a Thevenin I will get 41,5 V in series vith 4,2 V but , will these two add
or oppose each other ?

Best regards
Hello again,

I assumed you wanted to do this with your two equations you wrote previously:
4.2-82*I2-57(I2+0.6)-76,9*I1=0
4.2-72(I1-I2)-150(I1-I2-0,6)-76.9*I1=0

however there are still problems with these equations, like where did the "72" come from (did you really mean "22").
Also, you have not identified where I1 and I2 are in the circuit, and it looks like they can not exist in this circuit anywhere, but i'll wait for you to clarify, if you still want to try it that way too.

It seems to me that if you want to sum voltages then you have three currents you have to work with. If you call the unknown resistance R3 and the current though the resistors as:
i1 current through R1
i2 current through R2
i3 current through R3

that will at least start to make some sense.

If you indicated your original strategy that would help understand how you wanted to solve this originally.

#### WBahn

Joined Mar 31, 2012
29,876
It seems to me that if you want to sum voltages then you have three currents you have to work with. If you call the unknown resistance R3 and the current though the resistors as:
i1 current through R1
i2 current through R2
i3 current through R3

that will at least start to make some sense.
It's a start, but still not sufficient. He also needs to indicate which way those currents are flowing.

#### DGElder

Joined Apr 3, 2016
351
Hi !

Yes , if I transform the bridge into a Thevenin I will get 41,5 V in series vith 4,2 V but , will these two add
or oppose each other ?

Best regards
Really? Is that the open circuit voltage seen at the two interfacing terminals of the Thevenin circuit? Which of the terminals is more positive?

How about drawing this out, notating your circuit and showing us your math as has been requested - repeatedly. Technical communication is a big part of engineering and good communication skills in general are important in business. You are dealing here with the valuable time of professional engineers who are showing more patience than you can expect on a job. This might be a good time to start learning how to communicate professionally if you want a career in the field.

#### tmlekus

Joined Mar 11, 2016
10
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

#### tmlekus

Joined Mar 11, 2016
10
Hi,

He had mentioned that he wrote the wrong values on the original schematic, such as the "10v" when it should have been "4.2v" and the resistors all changed too

The magical 15 ohms is the bridge equivalent resistance of the first circuit, and the 76.9 ohms is a rounded version of the equivalent resistance of the second circuit.
Because of these two values i had to question what the equations were for then. He seems to be indicating now that he also wants to calculate the power in the resistor after finding the right value.
Correct I want to calculate what the optimal resistance in the unknown resistor should be for maximum power in this ? If I can calculate the current through this then I will also know the power !

The resistance in the knows resitors are the written values in the schematic ( 22 ,82 ,150 and 57 ohm ) The current source in the bridge is 0.6 A !

#### WBahn

Joined Mar 31, 2012
29,876
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
How can we possibly know if these are the right results for I1 and I2 when you patently refuse to tell anyone what the hell I1 and I2 are in this circuit!?!?!?!?!

What is SO hard to comprehend about the notion of labeling your damn diagram with the currents you are using in your equations so that your readers can tell what the flip you are talking about?

Or, more to the point, why should anyone waste any time guessing at what you mean when you are either too lazy to communicate what you mean or you simply don't care enough to do so?

At this point, if you were working for me, I'd be telling you to clear out your desk. And that's before we even touched on the matter of units that you still refuse to use despite the fact that it's already been pointed out to you that your failure to track units has already resulted in your inability to catch an obvious mistake.

Unless you get with the program, no one is going to be able to afford to employ you.