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Voltage drop across a battery, a capacitor and 2 light bulbs

  1. Mar 30, 2012 #1
    A 3v battery is connected to a capacitor and 2 light bulbs all in parallel set up

    1. Where is the total voltage drop at t = 0
    2. Where is the total voltage drop at t = infinity
    3. Which light bulb is brighter?

    From my understanding, since this is in parallel set up at t = 0 the voltage across the light bulb should be the same but would the capacitor have the same voltage as the two light bulbs?

    Very unsure of this one, at t = infinity, the capacitor is fully charge the current drops to 0 thus there will be voltage drop across the two light bulbs and eventually there will be no light on both of them.

    If we take wire resistance into account then the light bulb that is closer to the capacitor/battery is the brighter one while the farthest light bulb would the dimmer. If we don't take wire resistance into account then the two light bulbs would have the same amount of brightness.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 30, 2012 #2

    phinds

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    2016 Award

    In an ideal situation, the RC time constant to charge the cap would be zero because R=0 and all 3 elements have 3v at t=0 and at t=infinity.

    In a non-ideal situation, you'd have to give more specifics.

    Ideal situations don't exist in the real world but they are commonly used in homework problems.

    I don't understand your statement
    There WILL be light on both of them.
     
  4. Mar 30, 2012 #3
    This question is more conceptual rather than calculation so yes, assume ideal situation.

    I said that the light will go out because I was thinking that it would be the same as in series which is when the capacitor is fully charged (at t = infinity), the current drops to zero which causes the light to go out or perhaps my understanding is incorrect on this one?
     
  5. Mar 30, 2012 #4

    gneill

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    Staff: Mentor


    If the components are in parallel and one of them happens to end up not conducting current (such as would occur for a capacitor after some long time interval), that doesn't affect the other parallel components that do continue to conduct current. All components in parallel "see" the same potential difference across them. In this case that potential difference is determined by the battery.

    It seems odd that the problem statement would put a capacitor directly across a battery without an intervening resistance. For ideal components this would lead to an unrealistic situation. Are you certain about the configuration of components described in the problem statement?
     
  6. Mar 31, 2012 #5
    The problem stated as I have written down above, but I think you can rearrange the circuit which would mean you can put the capacitor at the farthest loop from the battery but I still don't see how that would change things or would it?
     
  7. Mar 31, 2012 #6

    gneill

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    Staff: Mentor

    No, it would not change anything. I think that the problem is poorly conceived.
     
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