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Archived What is the voltage drop?

  1. Jan 30, 2013 #1
    1. The problem statement, all variables and given/known data
    The light bulb is 60 watts at 230 volts, how much does the voltage need to drop for the lightbulb to be 40 watts'?


    2. Relevant equations

    V=IA, P=IV, P=IR^2, V=IR

    3. The attempt at a solution

    I tried 60w/230v=0,26A..... 40w=0,26A*XV<->x=154V so voltage drop is 230-154=76.. I know the current drops with the voltage but this is all I can think of.


    nevermind, I found out.... 40w=v^2/885ohms->> v^2=35400(ACCIDENTALLY DEVIDED) so V=188.
     
    Last edited: Jan 30, 2013
  2. jcsd
  3. Feb 25, 2016 #2

    CWatters

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    Science Advisor
    Homework Helper

    The key to understanding this problem is to realise that the resistance is the same for both situations. After all its the same bulb.

    Using the relevant equations...

    P=I*V
    and
    V=I*R
    You can write..
    P=(V^2)/R
    Then rearrange to give...
    R=(V^2)/P

    Then plug in the numbers giving...
    R=(230^2)/60=882Ohms

    Then using...
    P=(V^2)/R
    Rearrange to give..
    V=SQRT(R*P)

    Substitute the new numbers...
    V=SQRT(882*40)
    =188V

    However the question asks how much does the voltage need to drop and that is....

    230-188=42V
     
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