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Circuit brightness problem

  1. Aug 18, 2007 #1
    Can someone help me with 1 (e) at the following site:

    http://ocw.mit.edu/NR/rdonlyres/Phy...6C858-A92B-431F-A43E-725D58A768EF/0/exam1.pdf

    The solutions say that all on-bulbs in 1 (e) glow with equal brightness. But how is that possible because in e-c, the equivalence resistance is

    (2/R) so the total current is V * R/2 and the current through each branch is V * R. In e-b, the current is V/R through B, which is different, right?
     
  2. jcsd
  3. Aug 18, 2007 #2

    learningphysics

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    2/R as the equivalent resistance is wrong. It should be R/2. You don't need to find the equivalent resistance though. The current through a resistor is just the voltage across the resistor divided by resistance. For e-c, the voltage across bulb B is V... the resistance of bulb B is R, so the current through bulb B is V/R... exactly the same way, the current through bulb C is V/R.
     
  4. Aug 18, 2007 #3
    I see. But you can just keep adding more and more light bulbs in parallel and they will each glow as bright as a single light bulb? Doesn't that allow you to generate an infinite amount of light (energy) with a finite amount of voltage?

    Yes, we are assuming the wires have negligable resistance. But this still seems fishy to me...
     
  5. Aug 18, 2007 #4

    mgb_phys

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    They are in parallel so it's an infinite amount of light with an infinite amount of current. This isn't a problem as an infinite amount of current at a finitie voltage gives an infinitie amount of energy - of course paying the bill might be tricky!
     
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