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Homework Help: Lightbulb Brightness in a circuit.

  1. Mar 12, 2010 #1
    1. The problem statement, all variables and given/known data

    Rank the lightbulbs in order of brighness.

    http://img641.imageshack.us/img641/5871/lightbulb.png [Broken]


    2. Relevant equations

    All lightbulbs are identical.



    3. The attempt at a solution

    A is brightest, then B=C, then D=E

    A>B=C>D=E

    B and C are the same brightness level because they are in parallel, D and E are the least bright because they are in series.

    I even downloaded a circuit creator to verify my solution but apprently I am still not correct and I am not sure why. (program from PhET Interactive Simulations Copyright © 2004-2010 University of Colorado)
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 12, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi 1988Greens! Welcome to PF! :wink:
    Sorry, but that's just waffle. :redface:

    Physics is equations.

    So what equation are you using to find the power through each bulb? :smile:
     
  4. Mar 12, 2010 #3
    Re: Welcome to PF!

    Heh Waffle? I like waffles they are delicious.


    I would use V= IxR They give no numbers for the problem, so I would just have R or 1/R for the resistances in any given bulb?

    B and C Resistance = 1/R + 1/R
    = 2/R

    D and E Resistance = R + R
    = 2R

    I don't really know I am pretty lost at this point.

    Say if they gave you some numbers i.e. If it was a 12V battery with 1 Amp current Volts for bulb A would be V= 1xRx2/R
    = 2R/R
    = 2 Volts for bulb A? I feel like I am way off.
     
  5. Mar 12, 2010 #4

    ideasrule

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    You don't need to use any numbers. Can you figure out the current through each bulb? For example, the potential difference across D and E combined is V (since the two bulbs straddle the battery), so the current through each is V/2R. The larger the current through a bulb, the brighter it glows.
     
  6. Mar 12, 2010 #5
    Yea, I can't seem to figure out the current for bulbs B and C. I mean I get the voltage for D and E V/2R, add the 2 resistances together so the current for bulb B would be VR? I can't figure out what it would be for A? would it be:

    V/((1/R + 1/R) + R)?

    I have difficulty understanding this.
     
  7. Mar 12, 2010 #6

    collinsmark

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    Not quite. Remember resistors is parallel are not Req = 1/R1 + 1/R2.

    You need to divide the whole thing by 1:

    [tex] \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} [/tex]

    which is

    [tex] R_{eq} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}} [/tex]

    which is

    [tex] R_{eq} = \frac{R_1 R_2}{R_1 + R_2} [/tex]

    Try that first for bulbs B and C, then add that to the resistance of bulb A in series to get the overall resistance of the path.
     
  8. Mar 13, 2010 #7

    tiny-tim

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    :tongue2: mmm … waffles! :tongue2:​
    ah, now that's why it's not working out for you.

    Do it logically …

    the question asked for the power, so you need an equation for power, which V = IR isn't. :wink:

    Moreover, you know that all the bulbs have the same R, but not the same I, so preferably you want an equation involving P V and R but not I.

    Which is … ? :smile:
     
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