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2 Ohm's law/Power equations questions

  1. Apr 16, 2013 #1
    1. The problem statement, all variables and given/known data
    I've done both these questions with correct answers, but I just want sure that I chose the right choices for the right reasons. The questions are:
    PhysQ.png


    2. Relevant equations

    V = IR
    P = I2R
    P = V2/R

    3. The attempt at a solution

    The answers for both are D.

    The reason I chose D for the first one is because P/I2 = R And due to Ohm's law R is constant, so I2 = P So the graph should look like a y = x2 graph.

    The reason I chose D for the second one is because the brightness of the lamp depends on the power output so since P = I2R an increase of I (Due to Kirchovs first law) across L would mean a larger power output across L, and a decrease in I (because of the increase in R across the parallel part) across N would mean due to the same equation that P would decrease. For this particular question is it 100% necessarily to use the power equations or could you just say that a larger current means a greater brightness and then just deduce the answer from Ohm's law because the Examiner's report didn't mention anything about using the power equations...
     
  2. jcsd
  3. Apr 16, 2013 #2

    tiny-tim

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    Hi Normalization! :smile:
    But there's no I on those graphs, only V … you need an explanation that uses V instead. :redface:
    Technically, you're correct … brightness means power …

    but in practice we just use the fact that (for a particular component) that brightness increases with current, so no you should just use current.

    The current is not the same through each lamp, so you'll have to describe what happens to the currents more carefully.
     
  4. Apr 17, 2013 #3
    It should be clarified that the current going through each lamp is not the same before the filament of lamp M breaks. Once the filament of lamp M breaks, you are no longer dealing with a circuit containing parallel resistors (where the current varies, but the voltage remains constant), but with a pair of resistors in a series (where the current remains constant, but the voltage varies).

    Hope this helps :smile:
     
  5. Apr 17, 2013 #4
    Thanks a lot guys that really helps!

    I meant V^2/P sorry not P/I^2 (typo). And yes the current stays constant throughout the whole circuit (after removal of parallel light bulb) and increases relative to L and decreases relative to N.
     
  6. Apr 17, 2013 #5

    tiny-tim

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    Hi Normalization! :smile:
    Why?

    Describe what the currents are before and after (with numbers).
     
  7. Apr 17, 2013 #6
    Before:

    RTotal = 3R/2 Because of R + 1/(1/R + 1/R) = R + R/2 = 3R/2
    So ITotal = V/(3R/2) = 2V/3R
    IL = ITotal/2 (identical lamps) = 2V/6R = V/3R
    IN = ITotal = 2V/3R

    After:

    RTotal = 2R Because RTotal = R1+R2 = R+R = 2R
    So ITotal = V/2R
    IL = IN = ITotal = V/2R
    V/2R < 2V/3R So IN Decreases
    V/2R > V/3R So IL Increases
     
  8. Apr 17, 2013 #7

    tiny-tim

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    excellent! :smile:

    that's the way to do it! :wink:
     
  9. Apr 17, 2013 #8
  10. Apr 17, 2013 #9
    BTW: If you have 2 resistors in parallel (R1 and R2), do you just find the current going through one resistor R1 by doing R1/(R1 + R2) × ITotal?
     
  11. Apr 17, 2013 #10

    tiny-tim

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    yes :smile:

    (can you prove that? :wink:)​
     
  12. Apr 17, 2013 #11
    Are you sure it's not I1 = R2 / (R1 + R2) IT? Because:

    I1 = VP/R1 Where VP is the potential difference across the parallel part. And VP = IT × 1/(1/R1+1/R2) And 1/(1/R1+1/R2) = 1/(R1+R2/R1×R2) = (R1×R2/R1+R2).

    So VP = IT × (R1×R2/R1+R2)

    So I1 = IT × (R1×R2/R1+R2) × 1/R1 = IT × (R1×R2/R1(R1+R2)) = IT × R2/(R1+R2)
     
  13. Apr 17, 2013 #12

    tiny-tim

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    oh, I didn't see the R1 at the end of the line in your previous post …

    it was off the end of my screen :redface:

    yes, that was the correct formula for I2

    your new formula is the correct formula for I1
     
  14. Apr 17, 2013 #13
    Awesome, thanks :P
     
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