Subscripts?

[Smart $]

Smart$:

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

Steve

View attachment 85

For current, the notation defines the direction of flow of current when the current is considered to be positive. For example, Iab would mean the current is flowing from point a to point b. Iab = -Iba.

In the diagram, current is flowing from B to A, therefore I_BA and -I_AB are positive.

For voltage, the double subscript denotes the nodes of the circuit between which voltage exists. As stated earlier, the convention is that the first subscript denotes the voltage of that node with respect to the node identified by the second subscript. Vab would be the voltage at node a with respect to voltage at node b, and that Vab is positive. Vab = Vba @ 180 deg = -Vab

In the diagram, voltage is positive when measured A with repect to B, therefore V_AB and -V_BA are positive.

There are no -V_BA's in the equations provided. Equations of 5 through 8 are eliminated simply because the notation V_BA is not positive, even though the equations of 5 and 8 would be mathematically correct. Equations 1 and 4 would yield a negative value for R, which we all know is not possible, and also do not meet the criteria for current notation.

Of course the above is only considering the resistor portion of the circuit. If the battery portion of the circuit was under consideration, the equations of 1 and 4 would be correct.

 

[Smart $]

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.

View attachment 86

Is this a good answer?

 

[rattus]

Half right:

Half right:

Smart's waveform is correct:

And, if you use electron current, eqns. 2, 3, 5, & 8 are correct. I am free to measure Vba which will be negative. Therefore I must insert a negative sign or swap subscripts to make things right.

If you use conventional current, eqns. 1, 4, 6, & 7 are correct.

Actually, the battery merely confuses things as do the polarity marks. In fact, you don't need the schematic. All you need to know is tied up in the symbols, signs, and subscripts.

 

[Smart $]

...

And, if you use electron current, eqns. 2, 3, 5, & 8 are correct. I am free to measure Vba which will be negative. Therefore I must insert a negative sign or swap subscripts to make things right.

...

So where are the negative signs for voltage in equations 5 and 8? And if you do put them in, you also have to put them in (or take them out, as the case may be) for current. Regardless, the question was "Which of the following is correct?", not "Which of the following is correct when modified to suit?"

 

[rattus]

Algebra:

Algebra:

So where are the negative signs for voltage in equations 5 and 8? And if you do put them in, you also have to put them in (or take them out, as the case may be) for current. Regardless, the question was "Which of the following is correct?", not "Which of the following is correct when modified to suit?"

By swapping subscripts on both sides of 2 & 3, you negate both sides which has no effect on the validity of the equations. Multiply both sides by -1 and you have equns. 5 & 8. Basic algebra. It matters not where the negative sign falls.

I prefer to think Iab means the flow of positive charge from point A to point B through the resistor. It matters not that the carriers are electrons flowing in the opposite direction. Then 1, 4, 6, & 7 are correct.

 

[Smart $]

By swapping subscripts on both sides of 2 & 3, you negate both sides which has no effect on the validity of the equations. Multiply both sides by -1 and you have equns. 5 & 8. Basic algebra. It matters not where the negative sign falls.

As I said, equations 5 and 8 are mathematically [or algebraically] correct, but they are invalid by the stipulations given (by William D Stevenson, not me) because the voltage value is negative. Stevenson stipulated the values are to be positive and implied if the value is negative, it must be preceded by a negative sign.

I prefer to think Iab means the flow of positive charge from point A to point B through the resistor. It matters not that the carriers are electrons flowing in the opposite direction. Then 1, 4, 6, & 7 are correct.

I believe the best format for double subscript notation is V_ab = I_ab x R [or Z]. Doesn't matter what you, I, or anyone else prefers or has to assume to make it work correctly, as long as it is applied consistently by oneself and his/her colleagues. The reason I posed the equations question was to point out how easily things can get turned around and messy when following someone else's rules. Steve, Larry, and yourself have proven it.

 

[rattus]

Smart,

Yes, the most direct way of writing the equation is,

Vab = Iab*R

We can manipulate this equation by the rules of algebra and it is still correct.

Bur you are misreading the definitions.

If we stick to DC, then

Vab = Va - Vb which can be positive, negative, or zero.
Iab is positive if positive charge flows from A to B outside the source. If positive charge flows from B to A, then Iab is negative.

Just forget the battery and look at this problem in its most general terms.

If we are working with AC, then the magnitudes will always be positive. We commonly use algebraic manipulation in all sorts of problems and we often find a negative sign in front of a phasor or DC value. We then deal with it.

 

[Smart $]

...

For voltage, the double subscript denotes the nodes of the circuit between which voltage exists. As stated earlier, the convention is that the first subscript denotes the voltage of that node with respect to the node identified by the second subscript. Vab would be the voltage at node a with respect to voltage at node b, and that Vab is positive. Vab = Vba @ 180 deg = -Vab

...

Note -Vab in the quote above. I believe that to be a typo, which should be "-Vba".

Bu[t] you are misreading the definitions.

If we stick to DC, then

Vab = Va - Vb which can be positive, negative, or zero.

Where in the "definitions" does it say V_ab can be positive or negative? It expressly states, "Vab would be the voltage at node a with respect to voltage at node b, and that Vab is positive." Additionally "...double subscript denotes the nodes of the circuit between which voltage exists", so therefore it cannot be zero.

I'm not misreading the "definitions" and only demonstrating strict application. I didn't write them and I don't completely agree with them.

We can manipulate this equation by the rules of algebra and it is still correct.

I don't disagree. But where in the definitions does it say you can? The closest it comes is where it states Iab = -Iba, and Vab = Vba@180 deg = [-Vba]. But in strict application that is a far cry from saying we can use "Vba" and "Iba" without a negative sign in equations.

The whole of your post says you put into practice exactly what I wrote in the latter part of my previous post. So why are you continuing to fuel this discussion?

 

[rattus]

Quote from Smart:

"I don't disagree. But where in the definitions does it say you can? The closest it comes is where it states Iab = -Iba, and Vab = Vba@180 deg = [-Vba]. But in strict application that is a far cry from saying we can use "Vba" and "Iba" without a negative sign in equations."

There is no logical reason to affix a negative sign to Vba or Iba. You do not need permission; neither do their DC values have to be positive. The magnitude portion of a phasor must be positive by definition.

For example, I might write,

Vba(t) = 170cos(wt)
Vab(t) = 170cos(wt + pi/2)

Vce = +5V for an NPN transistor
Vce = -5V for a PNP transistor

Clearly these functions are sometimes negative, sometimes positive. Your self imposed restrictions eliminate one half of the possible solutions.

I am trying to convince you that four of the answers to your question are correct and in proper form, not just two. No disclaimers are required.