Exam question

[mwdkmpr]

The following question appears in mike holts journeyman simulated exam book.

Power loss in a circuit because of heat can be determined by the formula _______.
(a)P=IE (b)P=IR (c)P=I2R (d)P=E2/R

The given answer is (c) and the explanation is "Power is always = I2R"

Please explain why (a) and (d) are not also correct answers.

 

[Ed MacLaren]

Re: Exam question

Answers (A) and (D) are also correct.

Ed

 

[mkbuck]

Re: Exam question

Deleted

MKB

[ July 15, 2003, 07:26 PM: Message edited by: mkbuck ]

 

[mkbuck]

Re: Exam question

I beleive that the question includes the assumption that alternating current circuits are not always in phase and "power" means true power or watts.

That being the case, (a)and (d) would be apparent power or volt-amps when the circuit is not at unity.

An equation for watts or true power has to have resistance and current as the variables.

MKB

 

[Ed MacLaren]

Re: Exam question

After reading MKB's response above, I agree with him. The word "heat" in the question implies that the true power equation is being asked for.

Ed

 

[bennie]

Re: Exam question

Power loss is always I squared R.

Power consumed is P=IxE

 

[mwdkmpr]

Re: Exam question

Good point, I had not thought of the posibility of reactance. I agree that we E*I could be apparent power, and that I2R is true power because of no reactance but why is E2/R not also true power?

 

[charlie b]

Re: Exam question

Sorry guys, but I have to disagree with everything said so far (except for Ed?s first answer, which he later retracted). But before anyone takes exception (or worse, takes offense), please let me explain that we need to define our terms here. If you think that it is self-evident what ?E? is, or what ?I? is, or what ?P? is, or what ?R? is, then we are certain to have an error in communication.

Formulas (a), (c), and (d) ?CAN BE? equivalent (i.e., convey the identical information, but in different forms). I say ?can be,? because it depends on how we define the related terms. So let me restate each of the formulas, using words instead of symbols. I?ll do Formulas (a) and (d) twice each, and show why they can be interpreted two ways (a wrong way and a correct way).

Formula (a) ? Wrong Version (i.e., This version gives a wrong answer)
?Power loss in a wire is equal to the current in the wire times the voltage of the source.?

Formula (a) ? Correct Version (i.e., This version gives a correct answer)
?Power loss in a wire is equal to the current in the wire times the voltage drop along the wire.?

Formula (b) ? I?ll not bother, since this formula is total nonsense.

Formula (c) ? Correct Version (i.e., There is only one way to interpret this formula, and it gives a correct answer)
?Power loss in a wire is equal to the square of the current in the wire times the resistance of the wire.?

Formula (d) ? Wrong Version (i.e., This version gives a wrong answer)
?Power loss in a wire is equal to the square of the source voltage divided by the resistance of the wire.?

Formula (d) ? Correct Version (i.e., This version gives a correct answer)
?Power loss in a wire is equal to the square of the voltage drop along the wire divided by the resistance of the wire.?

What?s the bottom line? Well, have you ever tried to measure the voltage drop along a 200 foot branch circuit conductor? You would need a 100 foot or longer pair of leads for your voltmeter. In our business, we usually interpret the term ?E? as meaning the voltage of the source, and not the voltage drop along the wire. That is the reason that the Instructor is likely to consider answers (a) and (d) to be wrong answers.

 

[charlie b]

Re: Exam question

Originally posted by mwdkmpr:I agree that E*I could be apparent power, and that I2R is true power because of no reactance but why is E2/R not also true power?

It is. But only, as I said before, you use the voltage drop along the wire as the value of ?E.?

That being said, power is lost along a wire through the mechanism of heating. What heats the wire is the current passing through it. The current passing through the wire includes current associated with the wire?s resistive nature and current associated with its inductive and capacitive nature. To get the total value of power loss in a wire, you need to include total impedance (usually denoted with ?Z,? as opposed to ?R?). But this is not generally important to us, because the wires we use most often (or at least the ones we use in great lengths) are mostly resistive. Look in NEC Tables 8 and 9, and you will see that ?DC Resistances? of wire sizes 1/0 and smaller are 3 times or more higher than the ?AC Resistances.? If you are concerned with power loss along long runs of larger wires, then you need to include the AC Resistances in the formula.

 

[bennie]

Re: Exam question

I squared times R is the only correct formula. The value of E is not relevant for power loss calculation.

 

[charlie b]

Re: Exam question

Which is not to say that ?voltage is not relevant.? I agree that the variable ?E,? if it is interpreted in its most commonly used manner as meaning the system?s nominal voltage (e.g., 120, 480, 4160, etc), is not relevant.

 

[bennie]

Re: Exam question

A fuse melts when heat from, I squared times R, exceeds the melt point.

This can be on a 120 or 480 circuit.

 

[charlie b]

Re: Exam question

We?re not disagreeing, Bennie, in case that is not clear. And if you want to calculate the heat being released when the fuse melts, you won?t use either the 120 or the 480 in the E2/R version of the formula. Instead, you can use the voltage drop across the fuse (a small number, but not zero) and the resistance of the fuse (another small number, and also not zero). The answer you get will be the same as you would get from the I2R version of the formula.

 

[bennie]

Re: Exam question

Charlie B. I am only throwing this on the table for discussion purposes. No! I realize everyone is on the same page.

This calculation is the one used for working bare handed on live high voltage lines. When you clip on to the line, you have marked your area of work.
An old saying is "don't work outside of your clips". When you stake out four feet of line, you can determine the voltage across the four feet, by measuring the current.

You can clip on to a 500 KV line, and not have a high potential in four feet. Measuring the current is all that is possible for determining safety.

The hazard is... a nearby insulator flashover, and a high current flow, with both hands on your repair work.

[ July 16, 2003, 04:21 PM: Message edited by: bennie ]

 

[bennie]

Re: Exam question

We are accustomed to calculating voltage producing current. Nodal voltage is created by current.

 

[mwdkmpr]

Re: Exam question

I still believe that the answer could be E2/R as the questin is written. It says "power loss in a circuit", not power loss in a componet or wire in the circuit. Votage drop in a circuit = voltage applied to that circuit.

 

[brian john]

Re: Exam question

Charlie:

For a variey of reasons I have measured voltage drop up to at least 100 feet, utilized an extension cord, I have in my test lead case, male and female cord caps with banna plugs for just this type of testing.

 

[charlie b]

Re: Exam question

Originally posted by mwdkmpr: It says "power loss in a circuit", not power loss in a component or wire in the circuit.

Yes it does. But focus on the ?loss? word for a moment. Power is dissipated in the load, but that is the purpose for which we supply power to the load. That is not a ?loss,? it is a ?use.? For most branch circuits, the only place we lose power is in the branch circuit conductors. The question goes on to say that the power of interest is being lost ?because of heat.? That is different than power being used to generate light or to run a radio.

All issues of technical precision aside, if you don?t give the I2R answer to this question, your answer is going to be graded as being ?wrong.?