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DC Circuits - How to Determine a Bulb's Brightness

  1. Feb 7, 2012 #1
    1. The problem statement, all variables and given/known data
    tPo7e.jpg
    The circuit has identical bulbs and an ideal battery. When bulb D is unscrewed from its socket (the socket remains in its place), what would happen to the brightness of bulb F?

    2. Relevant equations
    V = IR


    3. The attempt at a solution
    Since bulb D is unscrewed from it socket and the socket remains in its place, we can replace bulb D with a switch that is open. Assigning resistance of 1 for each bulb, the right side of circuit has 2R and the current for bulb F would be I = V/2R. When the bulb D is back in its place, the resistance of the right side of the circuit would be 5R/3 and the current for bulb F would be I = (3V/5R)/(2/3) = 9V/10R. This shows that when bulb D is unscrewed from its socket, bulb F would have less brightness.

    When I entered the above explanation it turned out to be wrong. Can anyone explain why the brightness of the bulb increases?
     
  2. jcsd
  3. Feb 7, 2012 #2

    ehild

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    Check the current through F. Can it be higher than the main current?

    ehild
     
  4. Feb 7, 2012 #3

    PeterO

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    When globe D is removed, current passes through just B and F on the RHS. As they are in series, the current through each is the same - and the PD across each is thus equal, being half the Emf of the battery. [remember they are equal resistance globes]

    When Globe D is/was in position, the current through B is shared through the two parallel arms, so more current flows through B compared to that through F - so the PD across B is greater than that over F, so the PD across B must be more than one half of the Emf of the cell, while the PD across F must be less than a half; so F must be duller when D is there - or if you like brighter when D is removed.
     
  5. Feb 7, 2012 #4
    Now that you mention it, I'm beginning to realize how stupid my answer sounds. If I multiply 3V/5R by 2/3, would that be the right answer?
     
  6. Feb 7, 2012 #5
    Thank you. I think I'm beginning to understand the problem now. Would I reach the same conclusion if I only discuss voltage differences?
     
  7. Feb 7, 2012 #6

    PeterO

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    The brightness of the globe depends on the power being transformed/dissipated.

    Power can be calculated from P = VI or I2R or V2/R

    That 3rd form would suggest that Voltage difference would be sufficient.
     
  8. Feb 7, 2012 #7

    ehild

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    Yes, of course. :smile:

    ehild
     
  9. Feb 7, 2012 #8
    Thanks everyone for helping me!
     
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