Subscripts?

[LarryFine]

I think we shared admirably; we each took half of the answers.

 

[steve66]

Yes, but who took the right half?

 

[LarryFine]

Yes, but who took the right half?

Now that you mention it, I think I got the left half.

 

[Smart $]

Now that you mention it, I think I got the left half.

But is (are?) half actually correct? There are 255 possible answer combinations and only one that is absolutely correct.

In evaluating the question, did you guys consider only the resistor portion of the circuit, or both the resistor portion and the battery portion? If both, does evaluating only the resistor portion change the answer?

View attachment 76

 

[steve66]

Can you show us the 255 combinations so Larry & I can try again?

But really, I don't see 255 combinations. I see two ways to define the voltage, and two ways to define the current. And you can have right answers or wrong answers. Plus there might be a couple of ways to write the same equation. So I am seeing only 8 or maybe 16 combinations.

I considered only the resistor. We don't need to consider the battery side. Kirchoffs voltage and currents laws tell us if it works for the resistor, it works for the source. In other words, looking at the battery only turns the equation around. Instead of VAB=IAB*R, you have VAB/IAB=R. Same equation, just moved IAB to the other side.

So if you say VAB is the voltage at B with respect to A, I think Larry got the right answers. If you say VAB is the voltage at A with respect to B, I think I got the right answers.

Steve

 

[Smart $]

Can you show us the 255 combinations so Larry & I can try again?

But really, I don't see 255 combinations. I see two ways to define the voltage, and two ways to define the current. And you can have right answers or wrong answers. Plus there might be a couple of ways to write the same equation. So I am seeing only 8 or maybe 16 combinations.

...

I was referring only to the possible combinations of choices for the answer. An answer can have one or more choices of the available eight, but if one choice is incorrect, or the answer does not contain all available correct choices, it is not the absolute correct answer. Showing you all 255 combinations is more typing...and brain cell usage ...than I care to do, as I know of no easy way to list them all. It is much easier to just calculate the number possible. However, I will at least elaborate enough so you are aware of how there are 255 combinations...

First, there are 8 combinations of any 1 of the 8 choices. They are:

1
2
3
4
5
6
7
8​

Then there are 28 combinations of any 2 of the 8 choices. They are:

1,2 ... 2,3 ... 3,4 ... 4,5 ... 5,6 ... 6,7 ... 7,8
1,3 ... 2,4 ... 3,5 ... 4,6 ... 5,7 ... 6,8
1,4 ... 2,5 ... 3,6 ... 4,7 ... 5,8
1,5 ... 2,6 ... 3,7 ... 4,8
1,6 ... 2,7 ... 3,8
1,7 ... 2,8
1,8

Then there are 56 when choosing any 3, 70 for 4, 56 for 5, 28 for 6, 8 for 7, and 1 combination of all 8. Add 'em all up and there's 255.

The following image is captured from Excel's Help on the Combination function [COMBIN(number,number_chosen)]...

View attachment 80

My Casio calculator makes it even easier yet, as it offers the function in the form of nCr. I just enter 8C1+8C2+8C3+...+8C8, where C represents pressing the nCr function, then press EXE. Perhaps there is an even easier way, but I'm not aware of it.

Additionally, all variations of the two-subscript notation are in the listed 8 equations. Actually there are only 4 true variations, but I doubled that number by allowing the negative valuation [of the current component]. Reformatting the equation is not a variation of the possible subscript notations.

...

I considered only the resistor. We don't need to consider the battery side. Kirchoffs voltage and currents laws tell us if it works for the resistor, it works for the source. In other words, looking at the battery only turns the equation around. Instead of VAB=IAB*R, you have VAB/IAB=R. Same equation, just moved IAB to the other side.

Reformatting the equation does not account for the battery side. Remember all circuits are a closed path (or loop), electrically. For a "simple" circuit, the current flowing into a point is equal to the current leaving the point. The current vector for one path between two points in a loop must be opposite that of the other path (or the same vector with a negative value). Otherwise, the loop cannot exist. Because a double subscript notation also serves as a direction-of-current indicator, the equation must also indicate the direction. Moving IAB from one side of the equation to the other does not do that.

...

So if you say VAB is the voltage at B with respect to A, I think Larry got the right answers. If you say VAB is the voltage at A with respect to B, I think I got the right answers.

Steve

The "Given" stipulates, "Vab would be the voltage at node a with respect to voltage at node b, and that Vab is positive."

PS: I would like to point out that it's quite hard to write this stuff and not provide by context what I perceive to be the one absolutely correct answer

 

[kingpb]

Smart$,

You have got way to much time on your hands!

 

[steve66]

Yea, I was only kidding about posting the 255 answers. :smile: But I thought you meant there were 255 possible equations like the 8 you posted.

Now I realize you meant 255 combinations of the 8 answers you gave. But you are making the math harder than it has to be, and you went to a lot of extra work. For any of the 8 answers, we can either pick it, or not pick it. So the number of combinations is really just:

2^8=256

This includes the posibility that none of the answers is right (i.e. we pick 0 answers). So if we assume at least one answer must be right, we get:

256-1=255

Which is the result you got.

 

[Smart $]

...

...But you are making the math harder than it has to be, and you went to a lot of extra work. For any of the 8 answers, we can either pick it, or not pick it. So the number of combinations is really just:

2^8=256

This includes the posibility that none of the answers is right (i.e. we pick 0 answers). So if we assume at least one answer must be right, we get:

256-1=255

Which is the result you got.

Yeh, that works for total combinations, but in my write up I felt I had to be more explicit... not knowing how much you are aware of in the math of probability (not that I'm all that versed, mind you). Now how do you do the math for the possible combinations if I tell you I only see 2 of the choices as being correct? That is, there are 28 possible combinations of picking 2 of the 8 choices. Anyway, it's not my intention to focus on probability math. Can we get back to the discussion of double subscript notation?

It's my contention your initial response is incorrect. Same for Larry's, btw ...but don't y'all get bent way out of shape on that remark for I am just as likely to be wrong in my assessment. Just thought I'd add that last part because there seems to be some awfully touchy partcipants on this forum.

 

[LarryFine]

Just thought I'd add that last part because there seems to be some awfully touchy partcipants on this forum.

Who are you calling touchy?@!

 

[Smart $]

Who are you calling touchy?@!

"Surely" not you

 

[LarryFine]

"Surely" not you

"Yes, me; and don't call me Shirley!"

 

[steve66]

kingpb is the one complaining we didn't leave him any answers. He must be the touchy one

I get 28 correct combinations of answers also. 8 choices for the first answer, and 7 choices for the 2nd answer. That gives 8*7 = 56 possible ways to list the answers. But that counts answer "1 and 5" and "5 and 1" as 2 different items. So there are 56 premutations, and only half that many combinations.

I still don't see how you think only two equations are correct. If #1 is correct, then I can prove #4 is the same equation since IBA = -IAB. In the same way, I can prove #6 is the same equation with VBA = -VAB and IBA = -IAB. Same for #7.

Steve

 

[kingpb]

Who Me?

Who Me?

kingpb is the one complaining we didn't leave him any answers. He must be the touchy one

I'm not touchy, I like to think of it as being "sensitive"

And don't call me Shirley either!

I did know a Shirley once, ahh, a story for a different forum.......

 

[Smart $]

kingpb is the one complaining we didn't leave him any answers. He must be the touchy one

Actually you did leave him answers. That's why I brought up possible combinations. The way I see it, you chose 4 of the 8 available choices with at least one of the 4 being incorrect. The result is an incorrect answer (Think "one bad apple..."). You would have been better off by chosing only one having a correctly notated formula. However, assuming more than one of the available choices is correctly notated, choosing only one correctly notated formula would still not be the one absolutely correct answer.

I get 28 correct combinations of answers also. 8 choices for the first answer, and 7 choices for the 2nd answer. That gives 8*7 = 56 possible ways to list the answers. But that counts answer "1 and 5" and "5 and 1" as 2 different items. So there are 56 premutations, and only half that many combinations.

That's exactly how I use to do it before I learned of the combination function, such as the one in Excel or on my calculator, where you don't have to reason it out. The reasoning gets a little more difficult as the numbers increase (at least it does for me, more so as the years go by, while not utilizing math of the sort in daily functions).

I still don't see how you think only two equations are correct. If #1 is correct, then I can prove #4 is the same equation since IBA = -IAB. In the same way, I can prove #6 is the same equation with VBA = -VAB and IBA = -IAB. Same for #7.

I wrote, "...if I tell you I only see 2 of the choices as being correct?" Does that mean I said there are only 2 correctly notated equations? Regardless, please reread the stipulations... carefully! Hint: There are no -VAB's or -VBA's in the equations. If you don't see it after providing this hint, I'll have not other choice than to provide a dead-giveaway hint...

 

[rattus]

A simple question:

A simple question:

Ask a simple question, and you know the rest.

 

[rattus]

What does it mean?

What does it mean?

I think we now agree that Vab means the voltage drop from point a to point b.

Now let Vab = 120V@-30

Now trigger a scope at t0 (time zero). What waveform do we see?

I have asked this question before, but didn't get a good answer.

 

[steve66]

Smart$:

No more clues or hints. Not even a dead-giveaway. Please just tell us why only two answers are correct:smile:

Steve

 

[steve66]

I think we now agree that Vab means the voltage drop from point a to point b.

I'm not sure we agree on Vab yet. Everything that has been said about it so far could apply only to DC. Now you are throwing in AC. I think we need to expand the Vab definition, and add a definition for 120@-30. We also need to define t=0, and define exactly what our scope will "trigger" on.

Is this coming around to the single phase - two phase debate again?

Steve

 

[rattus]

Surely we agree:

Surely we agree:

I'm not sure we agree on Vab yet. Everything that has been said about it so far could apply only to DC. Now you are throwing in AC. I think we need to expand the Vab definition, and add a definition for 120@-30. We also need to define t=0, and define exactly what our scope will "trigger" on.

Is this coming around to the single phase - two phase debate again?

Steve

Whether we agree or not, Vab means the voltage DROP from point A to point B. Iab means the flow of positive charge through the load from point A to point B. Furthermore, four of Smarts equations are correct.

Surely we agree that 120V@-30 means 120Vrms with a phase lag of 30 degrees.

Surely we agree that t0 (time zero) means that instant in time when wt (omega*time) is zero.

Surely we agree that a scope triggered at t0 starts the trace at that instant.

Surely we agree that the 120/240V service is single phase.

Surely we agree that virtually all utility power is AC.