I'm a mechanical engineering student, so taking this course which has an electrical component is very hard to understand.

We were given a "black box" with four terminals sticking out of the top, and told to take the voltmeter and do a set of readings for each combination of terminals. 6 combinations total, and they are listed below:

Segment Resistance(Ω)
AB 500
AC 500
AD 11.4
BC 11.4
BD 499
DC 499

Then, the instructions are to find 6 equations that will give you the internal resistors. I have found 4 of those:

1/Rab= (1/AB ) + (1/ BC+CD+AD)
1/Rbc= (1/BC) + (1/ CD+AD+AB)
1/Rcd= (1/CD ) + (1/AD+AB+BC)
1/Rad= (1/AD ) + (1/ AB+BC+CD)

Now, I can't seem to figure out the last two? A classmate said that the last two had to do with if the arrangement inside is somehow diagonal. I don't understand how to get it from there.

I fixed a typo in your post #1

I think I have just realized an assumption I have been making about your approach to the problem.

It appears in this post that the variables AB, AC, AD, BC, BD, DC are the resistances measured at the outside terminals of the black box, and the variables Rab, Rac, Rad, Rbc, Rbd, Rdc are the resistors internal to the box. Is this true?

If so, then you are following a wrong method in developing your equations. You have interchanged variables.

Your first equation should be 1/AB = (1/Rab) + (1/(Rbc+Rcd+Rad)) (still not quite right because you have ignored the diagonal resistors, but this shows the variables to use in which side of the equation)

You need to write equations showing the external measured resistances in terms of the internal resistors. Then you must solve those equations for the internal resistors in terms of the external measurements.

 

The problem is equally well satisfied by only four resistors in the "adjacent" positions (no "diagonal" resistors). Their values, starting at the A terminal and proceeding clockwise, are: 988.666 ohms, 11.4657 ohms, 988.666 ohms, 11.4657 ohms.

On closer examination I don't believe the circle approach is analytically correct; however, for this particular case with the wide separation between the high and low resistor values, it's close enough. Ideally, Req(AB)=Req(DC)=Req(AC)=Req(BD) but that is not the case. Nor does the offset explain the 499,500 difference as measured.

 

On closer examination I don't believe the circle approach is analytically correct; however, for this particular case with the wide separation between the high and low resistor values, it's close enough. Ideally, Req(AB)=Req(DC)=Req(AC)=Req(BD) but that is not the case. Nor does the offset explain the 499,500 difference as measured.
View attachment 92772

It wasn't represented as analytically correct, although 4 of the 6 measurements are.

As WBahn put it: "The other possibility is that there are still only four resistors. You can get very close to the measured data (easily within the tolerance of the meter) with just four"

Considering the high sensitivity of the 6 resistor solution, as a practical matter it isn't any better.

 

It wasn't represented as analytically correct, although 4 of the 6 measurements are.

Approximate solutions are fine providing the approximation errors are less than the possible measurement errors.
What is analytically correct is the square/diagonal combination of two resistor values giving two possible values of measured equivalent resistance. In that case the equations for finding the exact side (Rs) and diagonal (Rx) resistor values simplify to the following:

 

Hi everyone, I still haven't received the correct answers from the TA or professor, I'll go talk to one of them later today and find out. I know The Electrician wanted to know what they thought the correct solution was!! I'll let you know as soon as I find out

 

I made a spreadsheet to calculate solutions for the 4-resistor problem and I was able to make it fit the measured values (including the 499/500 discrepancy) using two resistors of value 987.6 ohms and two resistors of 11.5 ohms.
The calculated measurements using those resistor values are:
(AC) & (DB) (cross-corners) ...... 499.55 ohms (500 by a 2000-count DMM)
(AB) & (CD) (across hi-val resistors) ..... 499.4838 ohms (499 by a 2000-count DMM)
(BC) & (AD) (across lo-val resistors) ..... 11.43382 ohms

This is a real box, and real measurements were taken, so we should assume that real resistors are inside it. 987.6 ohms and 11.5 ohms are not standard values, but consider tolerance spec:
The closest standard resistor values to 987.6 are 910 ohms (-7.9%) and 1k ohms (+1.2%).
The closest standard resistor values to 11.5 are 11 ohms (-4.5%) and 12 ohms (+4.5%).
These values fall within a 5% tolerance spec for standard resistor values.
I think it is totally possible (and probable) that there are only 4 resistors in the box.

In this case, the formulas should be easy.
For example your box is like this:

A---B
|   |
C---D

For the left resistor (L):
Your measured resistance (AC) is equal to the value of the left resistor (L) in parallel with the sum of the top (T), right (R), and bottom (B) resistors. AC = 1/((1/L)+(1/(T+R+B)))
Now rearrange that for L:
L = 1/((1/AC) - (1/(T+R+B)))
Same for the other 3 sides.

Now plug some formulas into each other and chip away at it.

 

This is a real box, and real measurements were taken, so we should assume that real resistors are inside it. 987.6 ohms and 11.5 ohms are not standard values, but consider tolerance spec:
The closest standard resistor values to 987.6 are 910 ohms (-7.9%) and 1k ohms (+1.2%).
The closest standard resistor values to 11.5 are 11 ohms (-4.5%) and 12 ohms (+4.5%).
These values fall within a 5% tolerance spec for standard resistor values.
I think it is totally possible (and probable) that there are only 4 resistors in the box.

I was re-reading post #1, where the TS says "...the instructions are to find 6 equations that will give you the internal resistors."

Even though 4 resistors will do the job, one can't help but wonder why the requirement to find 6 equations.

 

I was re-reading post #1, where the TS says "...the instructions are to find 6 equations that will give you the internal resistors."

Even though 4 resistors will do the job, one can't help but wonder why the requirement to find 6 equations.

One can't help but wonder a few things ....

 

I was re-reading post #1, where the TS says "...the instructions are to find 6 equations that will give you the internal resistors."

Even though 4 resistors will do the job, one can't help but wonder why the requirement to find 6 equations.

Well if the exercise was as poorly conceived as it appears that it may have been, then who knows what the instructor was thinking?
You could have 6 possible equations like this [AC = 1/((1/L)+(1/(T+R+B)))], and maybe that's what he was after. But that doesn't seem like it's worked far enough to me. Seems to me they should at least be rearranged to something like this [L = 1/((1/AC) - (1/(T+R+B)))]. But then that leaves two of the resistors being solved for twice, with different equations.

And if you're going to boil them down to [L = 1/((1/AC) - (1/(T+R+B)))], then you really only need 2 equations; one for the top/bottom side and one for the left/right side. A 3rd (diagonal) equation could be used to check the first two.

 

I made a spreadsheet to calculate solutions for the 4-resistor problem and I was able to make it fit the measured values (including the 499/500 discrepancy) using two resistors of value 987.6 ohms and two resistors of 11.5 ohms.
The calculated measurements using those resistor values are:
(AC) & (DB) (cross-corners) ...... 499.55 ohms (500 by a 2000-count DMM)
(AB) & (CD) (across hi-val resistors) ..... 499.4838 ohms (499 by a 2000-count DMM)
(BC) & (AD) (across lo-val resistors) ..... 11.43382 ohms

This is a real box, and real measurements were taken, so we should assume that real resistors are inside it. 987.6 ohms and 11.5 ohms are not standard values, but consider tolerance spec:
The closest standard resistor values to 987.6 are 910 ohms (-7.9%) and 1k ohms (+1.2%).
The closest standard resistor values to 11.5 are 11 ohms (-4.5%) and 12 ohms (+4.5%).
These values fall within a 5% tolerance spec for standard resistor values.
I think it is totally possible (and probable) that there are only 4 resistors in the box.

In this case, the formulas should be easy.
For example your box is like this:

A---B
|   |
C---D

For the left resistor (L):
Your measured resistance (AC) is equal to the value of the left resistor (L) in parallel with the sum of the top (T), right (R), and bottom (B) resistors. AC = 1/((1/L)+(1/(T+R+B)))
Now rearrange that for L:
L = 1/((1/AC) - (1/(T+R+B)))
Same for the other 3 sides.

Now plug some formulas into each other and chip away at it.

Your diagram above doesn't match up with what the TS says is the case. As shown in an image from post #12: http://imgur.com/rUSjRZn, it should be like this:

A---B
|   |
D---C

Otherwise your description:

(AC) & (DB) (cross-corners) ...... 499.55 ohms (500 by a 2000-count DMM)
(AB) & (CD) (across hi-val resistors) ..... 499.4838 ohms (499 by a 2000-count DMM)
(BC) & (AD) (across lo-val resistors) ..... 11.43382 ohms

doesn't match up.

But besides that, your values don't match up with what the TS gave in post #1 (typo fixed):

Segment Resistance(Ω)
AB 500
AC 500
AD 11.4
BC 11.4
BD 499
DC 499

According to this, one cross corner value (AC) is 500, and the other (BD) is 499. Also, AB and CD are not the same. What the hey??

 

Your diagram above doesn't match up [...] your description [...] doesn't match up. [...] your values don't match [...] one cross corner value (AC) is 500, and the other (BD) is 499. Also, AB and CD are not the same. What the hey??

FUBAR.

Well , you guys have fun with this one. I've wasted enough time on it, not going back to the drawing board again.

 

It's amazing how so many people have so much trouble following very direct, very simple instructions.

If he asked for six equations to find the internal resistors, he's talking about six resistors. No question whatsoever.

The equations need a network calculation involving the Delta-Y transform as WBahn points out. They will likely be very ugly. But that is the assignment. Unless someone else can find a better way of calculating that kind of "tetrahedral" network.

Perhaps there is a simplification due to the apparent symmetry in the measurements as has been suggested multiple times? We just need to relate that to the equations and the derivations.

Ignoring the instructions is unproductive.

I've been off this site for a while but if I have some time later today I may take a stab at it. I'm a little weak in the math but I looked at the Delta-Y and it's hairy but usable. I'd like to see what the implied symmetry can do to simplify it.

 

It's amazing how so many people have so much trouble following very direct, very simple instructions.

If he asked for six equations to find the internal resistors, he's talking about six resistors. No question whatsoever [...] They will likely be very ugly. But that is the assignment.

did you see this post?

it seems like this assignment isn't as well thought out as they had planned, there seems to be a lot of underlying theory that the professor didn't think about. What you're talking about is LEVELS over what we're doing. To give you an idea, other experiments in this set of experiments included:
1. finding the value of resistors and measuring them to compare,
2. using a voltage divider and seeing if Ohm's law held, and
3. playing with an oscilloscope, for no real reason

I seriously doubt it was supposed to be this complicated, [...] I strongly believe that there are only TWO unique resistors inside this box, and there's a good chance there are no diagonals. I agree with the previous poster who said that four resistors would satisfy the equations, not six.

If my 4th grade daughter came home with "The 7's Multiplication Table" homework, containing problems like "What is 7 X 3" and "How many apples are in a basket containing 5 bags of 7 apples each," and there was one on her paper that said "What is 11 X 7^8," I'd have to ask if maybe there was a typo.

 

As an update, I went to ask the professor in charge of the class what the answers were. His exact reply: "I have them around here somewhere but they are such complicated [nonlinear] equations that you need a computer program to solve. As long as you put down something, you will get credit."

Ahh, well that still doesn't satisfy my curiosity. So I will ask the TA and report back.

 

As an update, I went to ask the professor in charge of the class what the answers were. His exact reply: "I have them around here somewhere but they are such complicated [nonlinear] equations that you need a computer program to solve. As long as you put down something, you will get credit."

Ahh, well that still doesn't satisfy my curiosity. So I will ask the TA and report back.

Frankly, I think that's an pretty asinine answer from the professor. It basically comes down to: "I gave you a problem that I knew was almost certainly beyond your capability to solve at this point, so as long as you put something down you will get credit for it regardless of whether it is correct or not."

I'm surprised he doesn't just give you credit based on how you feel about your answer.

 

I agree. I'm a little disappointed, but also mixed feelings because passing grades are good. I'll keep searching for the solution though.

 

Your problem was to find 6 equations, but it would also be of interest to know what the professor says the internal resistors are.

 

As an update, I went to ask the professor in charge of the class what the answers were. His exact reply: "I have them around here somewhere but they are such complicated [nonlinear] equations that you need a computer program to solve. As long as you put down something, you will get credit."

That pisses ME off, and I'm not even in the class. I hope you enjoy your participation trophy equally with your classmates who just took 5 minutes to scribble some crap down.

 

That pisses ME off, and I'm not even in the class. I hope you enjoy your participation trophy equally with your classmates who just took 5 minutes to scribble some crap down.

It definitely is a demotivator for many students -- those that are truly interested in learning, such as the TS apparently is, can probably shake it off better than most. But the only lesson that many students will walk away with is to not put in very much effort on any problem they can't solve immediately, since they can expect to get full credit anyway.