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Power Increase In Rheostat Problem

Discussion in 'Electronics Homework Help' started by BlueScreen, Jul 22, 2012.

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  1. BlueScreen

    BlueScreen

    4
    0
    Jul 22, 2012
    Hello All,
    Im self studying Electronics at home, and I at the moment on Ohms Law.
    Now I have a problem:
    "What is the increase in current when 15v is applied to 10,000 ohm rheostat, which is adjusted to 1000 ohm value?"
    Now I have worked out that the power through the Rheostat is:
    a. 0.015A at 1000 ohms, and
    b. 0.0015A at 10,000 ohms.
    But Im stuck on the power increase. It looks like ten percent, but I could be wrong.

    Any clues?

    Cheers
    Glenn.
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,411
    2,779
    Jan 21, 2010
    I^2R or VI

    Work it out for both.
     
  3. BlueScreen

    BlueScreen

    4
    0
    Jul 22, 2012
    I will thanks Steve.
     
  4. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    Are you using the terms 'power' and 'current' interchangeably? They are not the same thing.

    If a value changes for example from 0.0002 to 0.002 then that is not a change of 10% but rather a factor of ten, i.e., 1000%
     
  5. BlueScreen

    BlueScreen

    4
    0
    Jul 22, 2012
    Thank you for the reply.
    No, im not. Im aware of the power formula and im just using Ohm's Law for the questions (cause thats what the questions are).
    I shouldnt have written 'power increase', my error. Im just wanting to know the increase in current. Its been a long long time since school and I have forgotten how to work out the increase.

    I guess the disadvantage of self study is that I don't have a teacher to ask the next day.
    But, in retrospect, when I was at school we didnt have the internet to turn to.
     
  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,411
    2,779
    Jan 21, 2010
    If you calculate two values and you want to now the % change from first to second, the correct formula is:

    (second - first * 100) / first
     
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