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Power dissipated

  1. May 22, 2015 #1
    1. The problem statement, all variables and given/known data

    What is the power dissipated by a 60 Watt (rated at 120 V) light bulb if it is connected to a 20 V power source (assume the resistance of the bulb is constant)?

    2. Relevant equations


    3. The attempt at a solution

    If it
     
  2. jcsd
  3. May 22, 2015 #2

    BvU

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    Hello E, welcome to PF :smile: !

    You were saying ?
     
  4. May 23, 2015 #3
    I dont know where to start i know that the power dissipated is in Joules but i dont know what equation to use
     
  5. May 23, 2015 #4

    CWatters

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    Power is actually measured in Watts.

    You initially have a 60W bulb running on 120V. What equation relates Power, Voltage and Current?
     
  6. May 23, 2015 #5
    Ohm´s law V=RI
     
  7. May 23, 2015 #6
    W = VC right ?
     
  8. May 23, 2015 #7

    CWatters

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    Yes Ok but better to write it as P = I * V.

    Now using that and Ohms law write a new equation relating Power, Voltage and Resistance.
     
  9. May 23, 2015 #8
    if 60 = 120 * I = 0.5 A

    I use this current and know that is 120V*0.5A = W right ? and W is 60 right
     
  10. May 23, 2015 #9

    CWatters

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    Yes Ok the current is 0.5A.

    You actually need to calculate the resistance as well.
     
  11. May 23, 2015 #10
    hum ok i understand know so if V= RI since 220 = R *0,5A = so the resistence is 110 ohms and since is constant

    20 = 110*i = so I =0.182 A

    So P = I*V = 0.182 A*20 = 3.6 Watts ? Am i right ?
     
  12. May 23, 2015 #11

    CWatters

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    should be 120V.
     
  13. May 23, 2015 #12
    so is 120 = R * 0.5A = 60 ohms

    And then 20 = 60*I = 1/3 A

    so W = 1/3A*20 V = 6,66 watts right
     
  14. May 23, 2015 #13

    CWatters

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    Back later.
     
  15. May 23, 2015 #14
    but the answers are 60W , 1.67 W , 20W or 120W i really dont understand then :x
     
  16. May 23, 2015 #15
    i know what i did wrong it was 240 ohms i made a stupid error

    W = 1/12A*20V and then it is 1,67 W
     
  17. May 23, 2015 #16

    CWatters

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    Sorry I got called away..

    Ok you got to the right answer but there is an easier route that I was trying to lead you down in post #7. I think you're substituting values too soon. There is no need to work out the current in either circuit. Here is how I would do it...

    P = I * V
    V = I * R

    Substitute to eliminate I..

    P = V2/R (Aside: I memorised that one and also P = I2 * R)

    Rearrange to give R...

    R = V2/P

    Substitute values..

    = 14400/60
    = 240 Ohms

    For second circuit R is the same. V changes to 20V....

    P = V2/R
    = 202/240
    = 1.7W
     
  18. May 23, 2015 #17

    CWatters

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    You can also do it without working out R....

    P1 = V12/R
    P2 = V22/R

    P2 / P1 = V22 / V12

    P2 = P1 * V22 / V12

    = 60 * 202/1202
    = 1.7W
     
  19. May 24, 2015 #18
    @estike you are right, the value is 5/3 = 1.666...
     
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