# # 19 of http://www.allaboutcircuits.com/worksheets/dc_s.html

Discussion in 'Homework Help' started by kconrad, Jan 6, 2011.

1. ### kconrad Thread Starter New Member

Oct 9, 2009
9
0
On #19 of http://www.allaboutcircuits.com/worksheets/dc_s.html I can't figure out how to get P =321.1W. Seems like I need to figure out the resistance value of "the water heater" using 110V and 500W. Using P=Esquared/R I come up with .041 and that's wrong. Please help.

If you know of anywhere there are expanded answers to the worksheets I'd appreciate the link.

2. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,400
1,218
In the second part of that question, you added a length of wire (3 ohms) on each side of that 500W heater.

How did you compute the resistance of that heating element?

3. ### kconrad Thread Starter New Member

Oct 9, 2009
9
0
Joe,
To answer the question do I have to figure out the resistance of the heater? In other words the resistance of 500W?

Apr 5, 2008
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5. ### Adjuster Well-Known Member

Dec 26, 2010
2,147
300
Yes, you need to find the resistance needed to dissipate 500W when connected to a 110V supply. You will have to manipulate P=Esquared/R to give R=???

Next find out what voltage the heater would have across it when fed via the resistive cable. Treat this as a potential division problem -don't forget that there is 3Ω resistance in BOTH wires!

Finally work out the power dissipated in the heater at this lower voltage. You should indeed get 321.1W.