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Power dissipated?

  1. May 18, 2009 #1
    1. The problem statement, all variables and given/known data

    What is the power dissipated by the 5.0Ω resistor in the following circuit?
    4000123.gif
    2. Relevant equations


    I thought maybe to simplify the resistance like.

    1/r=1/30 +1/30
    r=15
    +10+5
    R=30ohms
    then find the current
    v=ir
    25=Ix30
    I=.8333
    then find power with the 5 ohms?

    P=vi
    P=25x5
    P=75w?
    answer given is 6.2w?

    thanks.
     
  2. jcsd
  3. May 18, 2009 #2
    Your current calculations look ok.

    You have used the wrong value for voltage.

    Remember that only a proportion of the 25V is used in the 5 ohm resistor.

    I'll leave it up to you to figure out what the value should be.

    Oh, and you've substituded wrongly also....
     
    Last edited: May 18, 2009
  4. May 18, 2009 #3

    diazona

    User Avatar
    Homework Helper

    You can also find a formula for power in terms of current and resistance, instead of current and voltage as you have it.
     
  5. May 18, 2009 #4
    Im trying to think of what the voltage is at that point,
    do I use,
    EMF=Vab+IR ?
    24= vab+0.83*5 or something?

    I thought voltage was constant everywhere in a circuit?
     
    Last edited: May 19, 2009
  6. May 19, 2009 #5
    The voltage drop across loads connected in parallel is constant(equal to supplied voltage) whereas the voltage drop across loads connected in series is proportional to their resistance (V=IR)

    So, calculate the voltage drop across the 5Ohm resistor and use that in you're calculation for Power.

    Edit: I got 3.47W as the power dissipation though... I might have missed something.
     
  7. May 19, 2009 #6
    your right its 3.47, the question was split over 2 pages so I saw the wrong answer.

    I'm still kind of confused about the voltage when in parallel though.
    so if you have a 20v power source that splits 4 ways each is 20v then as far as resistance you treat each on seperatly as if it was a series?
     
  8. May 19, 2009 #7
    Thanks.

    so v=IR
    25=Ix30
    I=.833333

    V=IR
    V=.83x5
    V=4.15

    P=IV
    P=4.15x.8333333
    P=3.46


    Is that right?
     
  9. May 19, 2009 #8
    Yes, that's how I did it.
     
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