# 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
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    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
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    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
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    Joe,
    To answer the question do I have to figure out the resistance of the heater? In other words the resistance of 500W?
     
  4. bertus

    Administrator

    Apr 5, 2008
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  5. Adjuster

    Well-Known Member

    Dec 26, 2010
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    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.
     
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