# Curious to everyone's answer?



## Big John (May 23, 2010)

We'll pretend it's a resistive load that maintains the same resistance regardless of power, and then it's ohm's law: Reducing voltage reduces current, so you know it can't be the same or more than the wattage it draws at 240V.


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## 99cents (Aug 20, 2012)

751.1


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## wildleg (Apr 12, 2009)

751 x2


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## Pharon (Jan 20, 2014)

Assuming that 208V is single phase, the correct answer is (d).


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## 3D Electric (Mar 24, 2013)

D 751.1


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## bkmichael65 (Mar 25, 2013)

For every 1% drop in voltage in a resistive load, you'll have a 2% drop in wattage, so that narrows your choices down to the 2 below 1000 watts. It works out like this

240^2/1000W=57.6 ohms, 208^2/57.6 ohms=751.1 Watts


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## Expediter (Mar 12, 2014)

Sounds like homework. Many journeymen get this incorrect so don't feel bad. It is simple Ohms law, remember that and you will be ok. Don't make the mistake of taking transformer rules and apply it here. The above posts give the details.


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## Jhellwig (Jun 18, 2014)

wildleg said:


> 751 x2


1502 isn't an option.


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## industrial951 (Jan 29, 2011)

Lol the answer is actually b!! ... This is not home work btw, this was a test for a recent Job that I applied for, I guessed C. Due to the fact of what I was taught in school a couple years back.. my second answer would of been D. By assuming that this was a pure resistance load, however the question does not state pure resistance but by being a heating appliance I guess 1 can assume it would be.. 

Sent from my SAMSUNG-SGH-I337 using electriciantalk.com mobile app


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## Jhellwig (Jun 18, 2014)

There would have had to have been more information in the question for b to be the answer.


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## Big John (May 23, 2010)

industrial951 said:


> Lol the answer is actually b...!


 The only way the answer would be "B" is if they assumed current didn't change with voltage and the load drew a constant 4.16A regardless.

In the real world, that's almost always gonna be false, so that's a really stupid test question.


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## Dennis Alwon (May 9, 2009)

Jhellwig said:


> 1502 isn't an option.


He wasn't saying to multiply by 2 just that he also agrees with 1502 as an answer-- I agree times 2 next guy may say x 3 etc


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## Dennis Alwon (May 9, 2009)

B is NOT the correct answer-- sorry. And then they have the nerve to put the correct answer and say it is incorrect-- wow. A general rule of thumb that works fairly well is to take 75% of the wattage when going from 240V to 208. 

For instance a 1500 watt element at 240V would be .75*1500= 1125-- close but not exact. 1126.7 is the actual wattage


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## Mshow1323 (Jun 9, 2012)

P/E=I
1000/240=4.166

E*I=P
208*4.166=866.666

Or 

(E^2)/P=R
240^2/1000=57.6


(E^2)/R=P
(208^2)/57.6=751.111


Two different formulas, two different answers, both available options. Which formula do we use and why?


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## 3D Electric (Mar 24, 2013)

Mshow1323 said:


> P/E=I
> 1000/240=4.166
> 
> E*I=P
> ...


The resistance is needed since it is constant in this equation. The correct answer is D.


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## Dennis Alwon (May 9, 2009)

No matter what the voltage is the resistance of the element is the same. The amperage changes depending on the voltage so using the first calc does not make sense


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## 3D Electric (Mar 24, 2013)

Mshow1323 said:


> P/E=I
> 1000/240=4.166
> 
> E*I=P
> ...


BTW thank you for posting the entire formula typed out.


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## Mshow1323 (Jun 9, 2012)

In the third formula the resistance is calculated by using the voltage as well. Calling it constant in one but not constant in the other doesn't compute. In fact in we change any one number in the formula, the answer changes. 

I'm not being difficult, just really want to learn


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## Pharon (Jan 20, 2014)

The equipment is rated for 1000 watts at 240 volts. That's why you use that voltage to calculate its resistance. When you lower the voltage, the wattage output is lower because the resistance does not change.


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## AllWIRES (Apr 10, 2014)

Current is directly proportional to voltage. In the first formula you used the intensity from 240 and used that value on the 208. Which would change your resistance.

The second formula you are using the same resistance across the different applied voltages. Which gives you a different intensity and power for each voltage.


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## Mshow1323 (Jun 9, 2012)

I guess I can accept that. I need to eliminate as many variables as possible. Voltage is obviously variable, and wattage is variable directly effected by voltage. While the resistance is the constant. I'll buy it


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## AllWIRES (Apr 10, 2014)

AllWIRES said:


> Current is directly proportional to voltage. In the first formula you used the intensity from 240 and used that value on the 208. Which would change your resistance. The second formula you are using the same resistance across the different applied voltages. Which gives you a different intensity and power for each voltage.


(E/I)R 

240/4.166 = 57.60 ohms 

208/4.166 = 49.92 ohms


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## DriveGuru (Jul 29, 2012)

Mshow1323 said:


> P/E=I 1000/240=4.166 E*I=P 208*4.166=866.666 Or (E^2)/P=R 240^2/1000=57.6 (E^2)/R=P (208^2)/57.6=751.111 Two different formulas, two different answers, both available options. Which formula do we use and why?


The second formula you used is correct, 751.1 is the answer. Assuming it is a resistive load, and the resistance has not changed, once you drop the voltage to 208 your current is no longer 4.166 it is 3.614.


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## Pharon (Jan 20, 2014)

Interestingly, if the load were an inductive motor instead of a resistive heating element, the answer would be closer to (c).


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## DriveGuru (Jul 29, 2012)

Pharon said:


> Interestingly, if the load were an inductive motor instead of a resistive heating element, the answer would be closer to (c).


With a motor...the reduced voltage would cause greater slippage, and subsequently higher current, therefore more heat


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## Pharon (Jan 20, 2014)

Yep, I see it all the time. I wish U.S. freezer manufacturers would stop using 240 volt motors. They have to know that standard voltage here is 208. I mean, are 200 volt motors really that much more expensive?


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## Jhellwig (Jun 18, 2014)

Dennis Alwon said:


> He wasn't saying to multiply by 2 just that he also agrees with 1502 as an answer-- I agree times 2 next guy may say x 3 etc


I was trying to be funny....


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## DriveGuru (Jul 29, 2012)

Pharon said:


> Yep, I see it all the time. I wish U.S. freezer manufacturers would stop using 240 volt motors. They have to know that standard voltage here is 208. I mean, are 200 volt motors really that much more expensive?


I think the issue lies not only in cost, but what works in the majority of applications, 230/460v being the most widely used in factories. Care has to be taken the other way as well, if you take a 208 and over excite it, it will increase its efficiency only until the motor winding/core iron reaches a point of saturation. Then there is an increase in reluctance which again causes more slippage and higher current


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## Jhellwig (Jun 18, 2014)

Big John said:


> The only way the answer would be "B" is if they assumed current didn't change with voltage and the load drew a constant 4.16A regardless.
> 
> In the real world, that's almost always gonna be false, so that's a really stupid test question.


Agreed. Unless there is some context missing in the ops post it is an unanswerable question without making an assumption. Or the guy that came up with the question is a dummy.


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## wdestar (Jul 19, 2008)

The correct answer is indeed 751.1 watts. The problem that I have with most qualifying tests is that the generator of the test is usually unqualified.


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