0 registered members (),
57
guests, and
7
spiders. 
Key:
Admin,
Global Mod,
Mod



#130453  05/30/06 08:05 AM
Power dissipation problem


From Kenbo: Help!
I set my students some home work the other day.
Image 1 is the question and his answer Image 2 is his working out sheet Image 3 is my solution
The answers are greatly different but I can not see where it has gone wrong.
I know the reason for the differences is staring me in the face but I just can not see it.



#130454  05/30/06 08:07 AM
Re: Power dissipation problem


Hint: Look at the voltage he's used throughout the three tables.



#130455  05/30/06 09:52 PM
Re: Power dissipation problem


Just working this out real quick, I come up with about 9.14 ohms total resistance and approx 6.3KW. Line current = 26.26 amps.
Radar
There are 10 types of people. Those who know binary, and those who don't.



#130456  05/30/06 09:59 PM
Re: Power dissipation problem


To answer the actual question, I think the problem is in the calculation of watts. Regardless of who's total resistance is most correct, they're all in the 9.something range. 9.something line amps operating in a 240V circuit will yield somewhere around 26 amps, and that will result in roughly 6,400 watts (ballpark answers).
Radar
There are 10 types of people. Those who know binary, and those who don't.



#130457  05/31/06 08:26 AM
Re: Power dissipation problem

Member

Joined: Apr 2006
Posts: 233
Scotland


Realise the problem now thanks Paul Radar you are correct. The calculations in the 3rd sheet is the correct answer. What I did not realise the student had done wrong is to treat the whole calculation as entirely parrallel. He correctly worked out the equivalent resistance for each of the three groups of resistors but then he should have treated the rest of the problem as a series circuit. Then it would have worked out. Some times you just can not see something obvious when you look to hard...
der Großvater



#130458  05/31/06 09:00 AM
Re: Power dissipation problem


He correctly worked out the equivalent resistance for each of the three groups of resistors but then he should have treated the rest of the problem as a series circuit. By my calculations, the lefthand group of four resistors will have 63.1V across them, the middle pair 49.7V, and the righthand group 127.2 volts. I'm a little puzzled as to why he decided to work out the total power the long way like this. He'd already worked out the overall resistance as the first part of the problem, so why not just use E^2/R, or do as you did and work out the current then get the power from I x E? [This message has been edited by pauluk (edited 05312006).]




