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Power dissipated by a resistor

  1. Dec 30, 2011 #1
    hi just needed to know if the power dissipated by a resistor is equal to the total voltage. I am having a hard time understanding a problem i was given, and i know the formulas to find power dissipated, but i can't use P = I squared * R or P = V squared / R for this question...

    " A 10 ohm resistor dissipates 1 W of power when connected to a dc voltage source. If the value of dc voltage is doubled, the resistor will dissipate?"

    My teacher said the answer is 4 watts, but he never actually worked it out for me, and i would like to know how he got that answer, rather than just believe him. How do i find the power dissipated if i'm not given the voltage? Is the 1W of power dissipated by the 10 ohm resistor equal to my total Voltage? As you can see this problem is driving me crazy, and i've been stuck and not sure how to figure it out. Someone PLEASE help me out!! It is probably not as hard as i am making it, but thanks in advance.
     
    Last edited: Dec 30, 2011
  2. jcsd
  3. Dec 30, 2011 #2

    Doc Al

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    Staff: Mentor

    Power and voltage are two different things with different units. They cannot be equal.
    Why can't you?
    If you wanted to calculate the voltage across the resistor, you could. But there's no need.

    Hint: The second formula for power is all you need.
     
  4. Dec 30, 2011 #3
    You can use these equations for this question.
    You know that power P = I^2 x R
    also power P = V^2/R

    In each of these cases if you double I power will be 4x greater (if R does not change)
    Also if you double V power will be 4x greater (if R does not change)
     
    Last edited: Dec 30, 2011
  5. Dec 30, 2011 #4
    Thank you so much for the quick responses. Although the answer is probably right in front of my face, i am still getting confused by the fact that my power dissipated is in watts. If i plug in the numbers, my equation reads " 1W^2 divided by 10 ohms which gives me 0.1 Volts? But the question is asking for power dissipated by the resistor if voltage is doubled and resistance stays the same. Am i plugging in the numbers wrong?
     
  6. Dec 30, 2011 #5

    Doc Al

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    Try this: Don't plug in any numbers. Looking at the power formula, just state what happens to the power if the voltage doubles.
     
  7. Dec 30, 2011 #6
    I think this is where i am most confused. What is my Voltage for the problem? It can't be the power dissipated, right?
    Trying it your way, if i double the V, my power will also double? I feel like I am not given enough information and i would have to calculate my current in order to get the answer. But before i can do that, i would have to know my total voltage. I'm sorry if i am giving too many questions at one time, i just don't understand power dissipation that well i guess lol
     
  8. Dec 30, 2011 #7
    Don't forget that Power = V x I and for a resistor if you double V then I will also double.
    So making V 2x greater, makes I also 2x greater which means power mis 4x greater.
    However you do it
    P = VxI
    P = V^2/R
    P = I^2R
    These are all versions of the same equation
     
  9. Dec 30, 2011 #8

    Doc Al

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    You don't need to know the voltage, only that it doubles.
    No, that makes no sense.
    Nope. That would be true if power were proportional to the voltage, but it's proportional to the voltage squared.

    P = V2/R. What happens if you replace V with 2V?
     
  10. Dec 30, 2011 #9
    Again thank you very much for all your help. I finally understand what you both are telling me. If the voltage across a resistor is doubled, the power dissipation in the resistance increases by a factor of 4.
    But when i use 2V^2 i get 4V. So now i divide by resistance? Given the information you both have provided, i try and work the problem, my answer is a decimal.
    I tried this - 1w = V^2 / 10 ohms. I'm sure this is wrong since my answer isn't anywhere near 4 watts. What am i doing wrong?
     
  11. Dec 30, 2011 #10

    Doc Al

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    Good. That's all you need to know to solve this problem.

    What's the original power? When you double the voltage, you multiply that power by 4. So what's the new power?
     
  12. Dec 30, 2011 #11
    my original power was 1W. my new power is 4W since my voltage was doubled. I really appreciate Doc Al and technician for the past hour!!! Great forum you guys!!!!!!!!!
     
  13. Dec 30, 2011 #12
    If you look at your original statement. a 10ohm resistor dissipated 1Watt then this tells you that the original voltage must have been
    V^2/R = 1Watt so V^2 = 1 x R = 1 x 10 so V =√10 = 3.16V
    If this voltage is doubled to 6.32V then power = V^2/R = 6.32^2/10 = 4 watts
    See.... your numbers confirm it
     
  14. Dec 30, 2011 #13
    Can't thank you both enough! Simple problem but man was it giving me a heck of a time :) Have a great new year fellas!!!!
     
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