# Voltage drop

#### petersonra

##### Senior Member
Im new to this and studying could someone please explain this problem and how to achieve the answer please?

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My best guess is they are trying to tell you there is a total of five volts drop in the wiring. That would mean you have 5 volts at 20 Amps worth of losses in the wiring or about 100 Watts. I have no idea where they came up with 720 Watts.

#### LarryFine

##### Master Electrician Electric Contractor Richmond VA
That's about 1/3 of the total circuit power. Nothing makes sense with that number.

#### hbiss

##### EC, Westchester, New York NEC: 2014
My best guess is they are trying to tell you there is a total of five volts drop in the wiring. That would mean you have 5 volts at 20 Amps worth of losses in the wiring or about 100 Watts. I have no idea where they came up with 720 Watts.

That's the way I took it too.

-Hal

#### Darwin

##### New User
Ah the good old prov practice test. I can't figure out how they got 720 either

#### ggunn

##### PE (Electrical), NABCEP certified
20A @115V through the load is 2300W. 20A @120V through the circuit is 2400W. 2400W - 2300W = 100W.
Or the voltage drop is 5V and the current in the circuit is 20A. (5V)(20A) = 100W.

The question writer has a screw loose or the editor doesn't understand electricity. Or both.

I had an algebra "teacher" (A PE coach that the school called in to finish the term when the teacher quit) who gave this bonus question on a quiz: How many feet are in a square foot? When I called him on it he gave a heavy sigh and pulled out a pencil and paper. He drew a square, labeled it 1 foot on a side, and drew the perimeter while speaking very slowly, "one... two... three... four", and looked at me as if I were the idiot in the room.

#### Tulsa Electrician

##### Senior Member
It's an easy solve. 56% chance it's "C". I have been told if don't know the answer choose "C"

#### gar

##### Senior Member
211012-2337 EDT

Clearly there is no possible answer to the problem because the resistance of the distribution lines is not provided.

By the wording in the question there are two separate power sources connected in parallel by two wires. The resistance of the wire is not provided. The 20 A load at one end has nothing to do with the question.

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#### winnie

##### Senior Member
211012-2337 EDT

Clearly there is no possible answer to the problem because the resistance of the distribution lines is not provided.

Please note that this is a zombie thread several years old. OP is probably long gone.

I believe that the question was poorly edited and none of the answers are correct. But if we _assume_ that the intent of the question is roughly '120V provided at the supply end of the circuit, 115V measured at the load end of the circuit with 20A flowing, how much power is lost to the voltage drop in the wires? ' then the resistance can be calculated. Furthermore to calculate power lost to voltage drop it isn't necessary to calculate the wire resistance.

Jon

#### gar

##### Senior Member
211013-0833 EDT

winnie:

I agree that something is wrong with the question. But if you use the wording as presented, then there are two voltage sources separated by the distribution wires. I was only trying to point out the question was probably incorrectly written.

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#### ggunn

##### PE (Electrical), NABCEP certified
211013-0833 EDT

winnie:

I agree that something is wrong with the question. But if you use the wording as presented, then there are two voltage sources separated by the distribution wires. I was only trying to point out the question was probably incorrectly written.

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I agree that if that's what the question writer intended then the problem is non-deterministic, but this problem was in an elementary electrical course; I believe it is unlikely that two voltage sources in parallel was what was intended. Occam's Razor tells me that the question was poorly presented and the correct answer was not one of the choices.

#### gar

##### Senior Member
211013-0951 EDT

ggun:

I agree that I don't think the author of the question meant what I described. But the author of the question is ignorant and incumbent to be creating questions. There are many problems with the problem description, and I have no way to judge what was the original intent of the question.

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#### ggunn

##### PE (Electrical), NABCEP certified
211013-0951 EDT

ggun:

I agree that I don't think the author of the question meant what I described. But the author of the question is ignorant and incumbent to be creating questions. There are many problems with the problem description, and I have no way to judge what was the original intent of the question.

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I agree that the question as presented has no solution, but it was supposed to be a simple problem; it was in an entry level electrical course. Take away the word "supply" at the load and place 100W as an answer; then it makes sense and illustrates power loss due to voltage drop. Of course, leave it to us to explore all the ways to make it more complicated. I'm surprised no one has brought in differential equations and tensor math.

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#### gar

##### Senior Member
211013-1036 EDT

Had some additional symbols been used the digram would have been clearer.

One source of symbols is https://www.rapidtables.com/electric/electrical_symbols.html

In AC and DC circuit classes it is typical to have problems with several different voltage and current sources distributed in various ways in a single circuit to be analyzed.

What happens if you put a voltage and current source in series? And then add a resistor in series with the other two? Or make the resistor in parallel?

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#### ggunn

##### PE (Electrical), NABCEP certified
211013-1036 EDT

Had some additional symbols been used the digram would have been clearer.

One source of symbols is https://www.rapidtables.com/electric/electrical_symbols.html

In AC and DC circuit classes it is typical to have problems with several different voltage and current sources distributed in various ways in a single circuit to be analyzed.

What happens if you put a voltage and current source in series? And then add a resistor in series with the other two? Or make the resistor in parallel?

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You illustrate my point well. I think this was supposed to have been a very simple problem solvable with basic algebra. I do not believe that anyone designing an engineering school level problem would have screwed it up so badly.