Voltage drop

Merry Christmas

newbie2020

Member
Location
wv
Occupation
electrician
Im new to this and studying could someone please explain this problem and how to achieve the answer please?
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LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
Welcome to the forum. You'll find that we prefer helping you answer the questions, rather than answering it for you.

So we have an idea of your understanding, what do you see as the procedure to find the answer?

In other words, what is the question actually asking for?

Added: In my opinion, the correct answer is not given.
 

newbie2020

Member
Location
wv
Occupation
electrician
It seems to be asking what the power loss is due to the voltage drop but i cant seem to plug in the correct formula.. the answer key tells me its 720watts but im not getting that

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synchro

Senior Member
Location
Chicago, IL
Occupation
EE
A 115V supply is shown across the load, and not just a voltage measured across the load. So we have a 120V supply and a 115V supply connected by two wires. So the total voltage drop is 5V, but we need either the resistance of the wires or the current through them to determine the power due to this voltage drop on the wires.
The 20 amp load is irrelevant to the question posed by the problem if there's a 115V supply across the load because this supply will provide the 20A by itself.
I suspect that they may have intended it to be a 115V measured voltage instead of a supply, but even in that case the answer is not listed as Larry has already stated.
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
To me, the question is asking how much power is lost in the five volts dropped in the conductors.

Technically, all of the voltage is being dropped across the circuit, making the "trick" answer 2400w.
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
The question is horrible _and_ the suggested answer is wrong.

newbie: I will give you a hint: _none_ of the 4 given answers make sense to me. Knowing that the answer is some number not listed, what do you think the answer should be, and how do you calculate that result?

-Jon
 

infinity

Moderator
Staff member
Location
New Jersey
Occupation
Journeyman Electrician
Poorly worded but I think that you can see what they were thinking. They started with 120 volts, when it got to the load it was 115 volts, the load is 20 amps @ 115 volts so they "lost" 5 volts due to voltage drop.
 

sameguy

Senior Member
Location
New York
Occupation
Master Elec./JW retired
I would find the ohms of #12 copper/ #10 alum. and use that number when I found where 5v loss was (length) then use that info. to compute the wattage lost to v drop.
Ok, I just wouldn't answer the question and make sure it didn't count.
 

charlie b

Moderator
Staff member
Location
Seattle, WA
Occupation
Electrical Engineer
You don't need to know anything about resistance. My answer is 100 watts. 20 amps flowing through the wires yields a drop of 5 volts. 20 x 5 = 100.
 

charlie b

Moderator
Staff member
Location
Seattle, WA
Occupation
Electrical Engineer
This depends on whether the load is 20a at 120v or 20a at 115v.
It's both. At the starting point, a 120V source is pushing 20 amps through both the load and the intervening wires. That 120V source is therefore supplying 120 x 20, or 2400 watts. At the other end, a 115V source is pushing 20 amps just through the load. That 115V source is therefore supplying 115 x 20, or 2300 watts. The energy lost in the wires is the difference, or 100 watts.
 

sameguy

Senior Member
Location
New York
Occupation
Master Elec./JW retired
I too came up with 100w, but being a bad question I figured the OP could show he went beyond to still not answer the question and perhaps gain an advantage. Like when management "fat fingers" and it's ok, but if we do all hell breaks.
 

newbie2020

Member
Location
wv
Occupation
electrician
Yeah at this point im not sure the answer guide says c-720watts but noone is coming up with that so im glad i aint the only one confused by this if i had a way to contact the publisher of the study guide i bought with this in it id love to know how they came up with it
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