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Power Resistor Immersed In Water

  1. Jul 29, 2016 #1
    1. The problem statement, all variables and given/known data
    For the circuit shown below the 20-ohm resistor is immersed in water surrounded by insulating foam; the equivalent resistance through the circuit is 40 ohms. The current through it is 0.75A. What is the power dissipated through the 20 ohm resistor?

    2. Relevant equations

    $$ P=I^2*R$$
    3. The attempt at a solution
    $$ P=0.75^2*20$$
    $$ P=11.25W$$

    The solution to this problem gives 22.5W, why is the power dissipated through the resistor doubled here? Does the water or insulating foam affect the power at all? or is power calculated for the entire circuit?
     
    Last edited: Jul 29, 2016
  2. jcsd
  3. Jul 29, 2016 #2
    The water and insulating foam do not affect the power dissipated, but they will certainly affect the temperature. The higher the insulation, the higher the steady-state temperature will be. Are there 2 resistors in the circuit? If 0.75 A is running through 40 ohms, then you can calculate the power from the equation you showed above. However, it should be P = I2 * R
     
  4. Jul 29, 2016 #3
    Sorry forgot to give the picture oEHxNTW.png
     
  5. Jul 29, 2016 #4
    It looks like the answer they gave is wrong.
     
  6. Jul 29, 2016 #5

    CWatters

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    Science Advisor
    Homework Helper

    That would be the power dissipated in the whole circuit (all the resistors). Perhaps check the wording of the question?
     
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