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Power Dissipated by Resistor

  1. Mar 9, 2009 #1
    1. The problem statement, all variables and given/known data

    How much power is dissipated by the 12[tex]\Omega[/tex] resistor in the figure?
    32.jpg
    PR1=______W
    How much power is dissipated by the 18 resistor in the figure?
    PR2=______W

    2. Relevant equations

    P=IV
    I=[tex]\frac{V}{R}[/tex]

    3. The attempt at a solution

    P=[tex]\frac{V}{R}[/tex]*V
    P=V2/R

    P=12*12/12=12V

    After solving for my equation I plugged in my numbers but it's not correct. All I need are a few hints to lead me in the right direction. Thanks!
     
    Last edited: Mar 9, 2009
  2. jcsd
  3. Mar 9, 2009 #2

    hage567

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    Homework Helper

    Please show the details of your calculations so we can see what you're doing. You need to find the voltage drop across the resistor you're interested in. Note this will NOT be 12V.
     
  4. Mar 9, 2009 #3
    Try finding either the current through the resistors or the voltage across the resistors.
     
  5. Mar 9, 2009 #4
    What is the current in the circuit?

    The power is

    P=RI^2

    Where R is the resistanse of the component and I is the current through the component
     
  6. Mar 9, 2009 #5
    the current is 2 amps. what is the equation to find the voltage drop?
     
  7. Mar 9, 2009 #6
    It's Ohm's law. But all you need is the current. Try arrangin the power equation so that you only need to use the current and resistor. Is that the current for the whole circuit? If so it's incorrect.
     
  8. Mar 9, 2009 #7
    wait sorry they dont give the current my bad. i was looking at another problem. sorry
    all the information up there is what they give me to solve the problem, nothing more.
     
  9. Mar 9, 2009 #8
    You can find the current using the information in the problem. Okay, you know Ohm's law and you probably can create an Req for the two resistor in the circuit. That's should be enough to find the current.
     
  10. Mar 29, 2010 #9
    So I found the power in each resistor using P=I^2R, plugging in the current going through that resistor and the resistance and I get the sum of the power being 204 watts. This is not what I got in part c. Is it supposed to be?
     
  11. Mar 29, 2010 #10

    collinsmark

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    Gold Member

    Hello sweetdion,

    I think you might have posted to the wrong thread. :biggrin:
     
  12. Mar 29, 2010 #11
    Haha, yes, thank you. :blushing:
     
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