re: formula wheel

[jasons]

re: formula wheel

I teach an electrical apprenticeship class.
I was teaching basic electrical math. We were on the section of the PIE and the EIR formulas. One of my students said that they had a customer ask them if the voltage was low to their business and the amperage went up would their electric bill be any higher? Of course he said yes!!!

Since the question came up I proceeded to show them that if the amperage went up and the voltage went down that there would be no change in KW on the service. Since the power company bills based on KW this would not effect the amount of their bill.

Since I like to do things a little different in class I gave them a question on the test to see if they were listening or just reading the lesson out of the book . The question goes as follows

An industrial service has a voltage of 465V and an amperage of 1350A. If the voltage is raised to 487V what is the new amperage on the service.

465V x 1350A = 627,750W
627,750W/ 487V = 1289.01A

One of my students gave the answer as follows

465V/ 1350A = 0.34 ohms
487/ 0.34 ohms = 1432.35A

Since he figured it with a formula I gave him, how do I explain that this method is wrong. Everyone else in the class got it right.

My next question was as follows;
If the voltage on a circuit to a piece of equipment is raised; what does this do to the amperage?

Everyone answered "lowers" including the one that missed the other question.

All opinions and reasons are greatly appreciated!

 

[Ed MacLaren]

Re: re: formula wheel

how do I explain that this method is wrong?

It is a difficult question to answer without knowing the characteristics of the load. Your solution assumes that the load watts remains unchanged with the change in voltage.

If the total load was pure resistance, such as electric heat, etc, the student would have been correct, because it would have been the resistance that remained constant.

Motors with a constant mechanical shaft load could actually draw less current at the higher voltage.

Ed

[ October 14, 2003, 11:40 PM: Message edited by: Ed MacLaren ]

 

[Guest]

Re: re: formula wheel

There has to be a simpler way to explain this.

First of all your student is rounding down the Ohm's to .34 so the final answer will be inaccurate.

More importantly he (in a roundabout way) is using Algebra. If you wanted to express his solution Algebraically you might say "465 is to 1350 as 487 is to X". The equation looks like this:
465/1350=487/x
(then cross-multiply to get)
1350*487=465x
1350*487/465=x
1413.87=x

If you do your students math without rounding down you get the same result (1413.87).

Ohm's law is a "closed" ratio. Your student used an "openended" ratio. In your student's solution X is openended. With Ohm's law the numerator and denominator must always be in "balance". If one goes up, the other goes down.

Look at this carefully:
465/1350=487/x
[Since 487 is higher than 465, X is going to be higher than 1350]. This contradicts Ohm's law. Ohm's law is a "closed" equation. If you raise one value, you have to lower another value to keep it all in harmony. With your students solution as long as you raise the numerator X will continue to rise.

I have to go to dinner & a movie now. I'll try to finish this up later. If somebody else wants to jump in with an easier (or more correct) answer please flog away.

../Wayne C.

 

[Ed MacLaren]

Re: re: formula wheel

There has to be a simpler way to explain this

Wayne, you did say "simpler", didn't you?

Ed

 

[Guest]

Re: re: formula wheel

Too funny. While I was generating my masterpiece you had already answered the question

I was typing while you posted. What I meant was somebody else will post a simpler explanation while I watched the movie. Your answer did not exist for me until I submitted my 1st attempt.

Movie's over now.

../Wayne C.

[ October 15, 2003, 01:30 AM: Message edited by: awwt ]

 

[scott thompson]

Re: re: formula wheel

If I was taking this test, that question would have been answered with many, many, many-many-many variable scenarios - since so many can apply!

Would also have a note to Instructor for a side bar discussion of said exam question.

But then again, I can make putting on a Shoe sound difficult!
That's really more accurate than the "Simple, Generally Thrown Around" idea of all load currents becoming lower when the impressed voltage is raised!!!
That certain term has to be the Number One most quoted mistake I hear in the field (and Office too!).
It's mistakenly referenced, due to Apparent Power figures, or the solution of increasing the "R" or "Z" of a given item - thereby making it suitable for use on higher voltage, for a decreased load current; yet result in a similar overall load KVA (or, if DC, load KW).

Hope this makes sense!

me.

 

[Guest]

Re: re: formula wheel

OK, I'm thinking a bit more before I head off to bed. I may be stopped midway again

An industrial service has a voltage of 465V and an amperage of 1350A. If the voltage is raised to 487V what is the new amperage on the service.

465V x 1350A = 627,750W
627,750W/ 487V = 1289.01A

Your calculation is correct for the first scenario:
If the voltage is 465 and the amps is 1350 then the watts equals 627,750 (I didn't check your math-- I'll just trust you for now).

You follow this up with a question:

If the voltage is raised to 487V what is the new amperage on the service.

Based on that question there is insufficient data to arrive at your desired answer. Your goal is to prove that if voltage goes up, then amps must come down. The only way for amps to come down is if the watts remain constant.

If you want your desired answer you must restate your questions to something like this:

Question #1:
An industrial service has a voltage of 465V and an amperage of 1350A. What is the wattage using Ohm's law? Volts x Amps = Watts

Please write your answer here: ________________
[Student does calculation and comes up with 627,750].

++++
Question #2:
If we increase voltage of the industrial service to 487 volts and the wattage stays the same as your answer above what is the new amperage? Use Ohm's Law: Wattage ? Volts = Amps.

Do you notice that if the load remains the same when the voltage is increased the amps is reduced?

If the load is open-ended when the voltage increases the amps also increase as Ohm's law dictates that they all be in proportion. That's Ed's point about lighting loads vs. motor loads-- and Scott's point about power factors, etc.

++++
Break time. Time for the Graveyard Shift to kick in Feel free to flog away ladies and gentlemen.

../Wayne C.

[ October 15, 2003, 01:32 AM: Message edited by: awwt ]

 

[iwire]

Re: re: formula wheel

Originally posted by scott thompson:
If I was taking this test, that question would have been answered with many, many, many-many-many variable scenarios - since so many can apply!

Would also have a note to Instructor for a side bar discussion of said exam question.

Scott as a someone that reads all your posts at ECN (and have learned a great deal from doing so) I would love to see your answers to this test question.

I imagine a couple of three ring binders with references and all the different possibility's. but that shoe will be on better than any shoe before it.

I have never lacked for length in my posts either, I just wish they always had the substance yours do.

Bob

 

[charlie b]

Re: re: formula wheel

Before you read this response, go back and read Ed?s first response.

Here, then, is ?Truth?:

How do I explain that this method is wrong?

You don?t. He wasn?t wrong. Your question was wrong (or more properly, it was incomplete). You described the load as an ?industrial service.? This does not necessarily mean that the load characteristics are ?constant power,? nor that they are ?constant resistance.? Without specifying the type of load (as Ed has already observed), the question can be answered (correctly) either your way or the student?s way.

Please note that for a constant power load (BEWARE, a blasphemy is about to be spoken), Ohm?s Law does not apply! A material (e.g., a wire) can be said to obey Ohm?s Law if its resistance does not depend on either the applied voltage or the current. In other words, Ohm?s Law only applies to constant resistance materials. You cannot have a device that displays both the characteristic of ?constant power? and the characteristic of ?constant resistance.? Run the algebra any way you want, and you will verify that those two characteristics are mutually exclusive.

 

[charlie b]

Re: re: formula wheel

Originally posted by awwt: With Ohm's law the numerator and denominator must always be in "balance". If one goes up, the other goes down.

Ohm's law is a "closed" equation. If you raise one value, you have to lower another value to keep it all in harmony.

I?m not sure what you are trying to say here, but these two phrases, as stated, are incorrect. First of all, as I say in my previous post, Ohm?s Law only applies if the resistance is constant. Secondly, since you are speaking of numerators and denominators, then you are discussing the versions of Ohm?s Law that are expressed with fractions. There are two such versions.

The first version is E/R = I. In this version, if the numerator goes up, the denominator does not have to go down to maintain ?balance.? Rather, the denominator stays constant, and the item on the right side of the equation goes up.

The second version is E/I = R. In this version, if the numerator goes up, the denominator does not have to go down to maintain ?balance.? Rather, the denominator goes up, in order to keep the ratio constant, which in turn keeps the item on the right side of the equation constant.

 

[Guest]

Re: re: formula wheel

Thank you for bridging the gaps in my answer, and dispeling any portions that were incorrectly stated.

I didn't want to come right out and say the question was wrong. It was just incomplete if a specific theory was to be the end result. There was not enough data or detail in the 2nd question to end up with instructor's anticipated result.

../Wayne C.

[ October 15, 2003, 12:30 PM: Message edited by: awwt ]

 

[roger willis]

Re: re: formula wheel

In my classes that I have taught I have always made the statement that the voltage and current are inversely proportional to the same load.
Now you guys have got me to thinking that I may not have completed my statement to the class.
Should I have expanded on my statement?
Roger

 

[charlie b]

Re: re: formula wheel

Originally posted by roger willis: voltage and current are inversely proportional to the same load.

I would word it as follows:

Voltage and current are directly proportional to each other, when the load is of the ?constant resistance? type. Voltage and current are inversely proportional to each other, when the load is of the ?constant power? type.

But you would have to know whether your audience was sufficiently familiar with algebra to understand the notions of direct and inverse proportionality.

 

[roger willis]

Re: re: formula wheel

Thanks, Charlie
And yes you would have to know what level your audience was on.

Roger

 

[jasons]

Re: re: formula wheel

I am starting to wonder if I need to be teaching this class afterall!!!

I appreciate all of the information that I have recieved. Theory is a weak area for me and that is why I am going to finish my college classes and get my EE degree.

I have found that a lot of electrical contractors in my area know absolutely nothing about theory of electricity. Since I am only 25 I have time to finish my degree and maybe will not be bringing topics up like this.

For those of you that do know this stuff real well, is EE school the route I need to take or do you have any other suggestions on learning theory?

By the way i am currently an electrical contractor myself.

Thanks
Jason

 

[roger]

Re: re: formula wheel

Jason, if the means to go to school is there, by all means go to school.

School, School, and more School.

Start with EE classes and you can decide from there.

If you don't you will wish you had.

Roger

 

[golfking]

Re: re: formula wheel

Originally posted by awwt:
Thank you for bridging the gaps in my answer, and dispeling any portions that were incorrectly stated.

I didn't want to come right out and say the question was wrong. It was just incomplete if a specific theory was to be the end result. There was not enough data or detail in the 2nd question to end up with instructor's anticipated result.

../Wayne C.

exactly

 

[dereckbc]

Re: re: formula wheel

Originally posted by jasons:
For those of you that do know this stuff real well, is EE school the route I need to take or do you have any other suggestions on learning theory?

If all you want is theory, than an ASEE will do it. It really depends on what you want to do with your career. If you like working with your hands and tools then do contracting. If you would rather work with design or prototype go for an BSEE then obtain licensure.

I started in high school working as an apprentice, obtained my residential contractor license in college, graduated with a BSEE and have never looked back. It taught me a great deal of respect for the trade and the profession. Go for it, you won't be sorry.