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Ceiling lights

  1. Oct 11, 2011 #1
    1. The problem statement, all variables and given/known data


    ceiling light are connected in a parallel circuit, if one of the light bulbs burns out, will the other two bulbs be brighter?
    pls help me >.< my science exam is tomorrow.:(
     
  2. jcsd
  3. Oct 11, 2011 #2

    NascentOxygen

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    Draw a circuit diagram showing some light bulbs in parallel branches, all powered from the mains supply. Now, pretend one bulb burns out. What can you see is going to happen in the other branches?
     
  4. Oct 11, 2011 #3
    it shouldn't affect right?
     
  5. Oct 11, 2011 #4

    NascentOxygen

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    How would you justify claiming that?
     
  6. Oct 11, 2011 #5
    erm well the voltage supplied to each of both bulbs doesnt change and their resistance doesnt change either which is why the current and brightness of the bulbs do not change. is this correct? is it possible to explain only with resistance and current?
     
  7. Oct 11, 2011 #6
    why does a voltmeter have to have high resistance? shouldnt it not affect the brightness of the bulb in the circuit at all cos it is connected in parallel to the bulb?

    OMG IM SO DEAD FOR MY EXAM.
     
  8. Oct 11, 2011 #7

    NascentOxygen

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    That is true. Emphasise how, with parallel branches, the current through each branch is independent of that through the others, because the arrangement causes the full mains voltage to be applied across each branch.
     
  9. Oct 11, 2011 #8

    NascentOxygen

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    That's true, as far as it goes. But if the voltmeter did draw a lot of current, then, in general, it will affect the voltage that is powering the bulb. This won't happen with mains voltage, but is likely to happen where you are using batteries or lower-power sources. To ensure that you can connect a voltmeter to practically any circuit, and be confident that it will not upset that circuit, voltmeters are designed to have a high resistance.
     
  10. Oct 11, 2011 #9
    but why would it affect?
     
  11. Oct 11, 2011 #10

    NascentOxygen

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    Because no voltage source is perfect. It's as though they have a smallish resistor in series with their terminals. If you cause more current to be drawn from the source, the voltage it supplies decreases by a bit.
     
  12. Oct 11, 2011 #11
    oh so now your viewing it as overall resistance in a parallel circuit? by the way, how do we know when to view the total resistance in the circuit and when to view the individual reisistance of the sub circuits?
     
  13. Oct 11, 2011 #12

    NascentOxygen

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    Yes, the imperfection in a voltage source appears like a small series resistor; it's in series with whatever circuit you connect to the battery. Most of the time, you can ignore it because its effect is generally slight. But the designer needs to keep it in mind, to ensure it doesn't become a significant problem.

    Generally you only consider the resistors that you are told about, in school science questions.
     
  14. Oct 11, 2011 #13
    huh? the resistors i'm told about?
     
  15. Oct 11, 2011 #14

    NascentOxygen

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    Going back to your original question. If one of the ceiling bulbs burns out, the others will continue glowing unaffected.

    But if you were powering 3 small light bulbs off a couple of C cells, and one of the bulbs burnt out, you would probably notice the other two glow a tiny bit brighter, because with fewer bulbs to power, the battery would be supplying slightly more voltage. That's a characteristic of batteries: the more current you draw from them, the lower (slightly) the voltage the battery puts out.
     
  16. Oct 11, 2011 #15
    hmm ok but whats the proper answer i should give if this were an exam question?
     
  17. Oct 11, 2011 #16

    NascentOxygen

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    The one you gave earlier on. If one of the ceiling bulbs burns out, the others will continue glowing unaffected for the reasons you gave.
     
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