| Ok, this is a practice question I came across on a PSI exam. Does anyone know how they came up with this number? I apologize for my ignorance but I'm here to learn something. An appliance draws 40 amperes when used on a circuit. A bad splice is discovered on the line at the appliance on one of the phase conductors. The splice has a measured resistance of 0.5 ohms. How much power is lost at the splice? Answer: 800 watts |
Voltage drop question?
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gar
Senior Member
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220301-0758 EST
I potentially disagree with the answer. It could be correct, if the load was a constant current load, but most loads are not constant current.
If the load is an invariant resistance, and assuming a 120 V source, then Rload = 120 / 40 = 3.0 ohms. Add an 0.5 ohm series resistance, then Rload = 3.5 ohms, and I = 120 / 3.5 = 34.3 A, and then power loss in the 0.5 resistance = 34.3 * 34.3 * 0.5 = 588 W.
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I potentially disagree with the answer. It could be correct, if the load was a constant current load, but most loads are not constant current.
If the load is an invariant resistance, and assuming a 120 V source, then Rload = 120 / 40 = 3.0 ohms. Add an 0.5 ohm series resistance, then Rload = 3.5 ohms, and I = 120 / 3.5 = 34.3 A, and then power loss in the 0.5 resistance = 34.3 * 34.3 * 0.5 = 588 W.
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ptonsparky
Tom
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He gave the answer they were looking for and with that, we must assume the current was 40 amps at that time.
LarryFine
Master Electrician Electric Contractor Richmond VA
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Welcome to the forum.
The question requires you to first figure out how much voltage is dropped across a half of an ohm when the current is 40 amps, and then how mush power (P) is lost with that voltage and current.
Your unknowns are the voltage (E) and power (P), and the knowns are current (I) and resistance (R).
E = I x R = 40 x 0.5 = 20v
P = E x I = 20 x 40 = 800w
Of course, these are theoretical numbers, but that what you use when discussing theory and answering tests.
4x4dually
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Engineering answer is 588, test answer is 800, when in reality, the $65 ACFI breaker has tripped therefore saving the planet and actual answer is ZERO! LMAO
My apologies, I know that wasn't helpful, but I couldn't resist myself.
My apologies, I know that wasn't helpful, but I couldn't resist myself.
gar
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If the question had been --- what is the power dissipation in an 0.5 ohm resistor with 40 A current flowing thru the resistance? --- then what is the power dissipation in said resistor having an answer of 800 W is valid. Otherwise we have an incompetent author of the question.
Because a potentially incorrect answer is given as the answer to the question does not make the answer correct.
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If the question had been --- what is the power dissipation in an 0.5 ohm resistor with 40 A current flowing thru the resistance? --- then what is the power dissipation in said resistor having an answer of 800 W is valid. Otherwise we have an incompetent author of the question.
Because a potentially incorrect answer is given as the answer to the question does not make the answer correct.
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ptonsparky
Tom
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- EC - retired
But it didn't trip. It wasn't arcing and there was no ground fault so the house burned down, killing the cat. (I don't like cats)
4x4dually
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If that cat had nine lives, Clark, .... he just spent 'em all. LOL
ptonsparky
Tom
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OP, please note that sometimes we get side tracked.
Welcome to the forum.
Welcome to the forum.
wwhitney
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This part of your analysis is in conflict with the information given in the question. The 40A measurement occurred while the appliance was already in series with the 0.5 ohm splice.
That is "An appliance draws 40 amperes when used on a circuit." is referring to the particular circuit further described, not some other idealized circuit. I.e. it means "There's a circuit. An appliance draws 40 amperes when used on that circuit. . . ."
Cheers, Wayne
That's it!
Thanks!
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And this points out why poor connections can result in fires...that 800 watts is producing heat at the location of the poor connection.
LarryFine
Master Electrician Electric Contractor Richmond VA
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- Henrico County, VA
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Ha ha I like it lol!
Sorry for the late reply , there has to be a way to reconcile those equations with I squared x R as they both provide the same number. Thanks for the help!
Keep in mind it's the TN PSI exam. Likely for LLE, possibly CE test. They are basically wanting to see if you can apply ohms law.
They asked for P and gave you I and R, so, as Roger states, don't let the "engineering" here confuse you. For the test, it's Keep It Simple.
They asked for P and gave you I and R, so, as Roger states, don't let the "engineering" here confuse you. For the test, it's Keep It Simple.
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