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Light bulb brightness (series and parallel)

  1. Feb 26, 2006 #1
    Hi, my question relates to light bulb brightness in a circuit with light bulbs connected in series and in parallel. Ill give you the question and then my argument.

    Q1:In this diagram light bulbs A,B, and C are all equal. Rank in order the most bright to the least bright.
    A1: I would argue that B and C are equal in brightness and are both greater in brightness than A because the amount of current running through B and C is equal and greater than A, and since the amount of current running through a bulb determines brightness, A must be the least bright.

    Q2: Suppose a wire is connected between points 1 and 2 running through bulb C. What happens to the brightness in all the bulbs
    A2: Bulb C would be short circuited and thus go out, but I am not sure what happens to A and B. Please help

    Q3: What happens to the current drawn from the battery when the wire is connected
    A3: The current drawn from the battery is decreased because there is no current flowing through C?? Please help

    Sorry for the long post :smile: but please look over my arguments and tell me if im right or wrong. THANKS!!
  2. jcsd
  3. Feb 26, 2006 #2


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    Staff: Mentor

    If A, B and C are equal, and the total current through A is divided between B and C, then . . . ?

    Remove C from the circuit, and what do you have?

    The resistance of two parallel resistances is half of one, i.e. R and R are in parallel then the resulting resistance is R/2.

    Resistances add in series, i.e. R1 + R2, but in parallel, 1/R = 1/R1 + 1/R2.
  4. Feb 26, 2006 #3
    Ok, so for Q1, A > B=C because A has twice the current of B and C
    Q2= C does not glow, A=B because the current running through A is equal to the current running through B. Thus A stayed the same brightness, and B's brightness increased.
    Q3= if Q2 is correct, then the light bulbs are in series and the ammount of current running through the circuit hasnt changed since C was removed thus the amount of current drawn from the battery is the same?
  5. Feb 26, 2006 #4


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    Let me back up. If a wire bypasses C, i.e. C is shorted, the current will go through the wire, thus also by passing B, which still has resistance. Therefore only A has current through it.

    If C was burned out (open circuit), the current would go through A and B, which would have resistance 2R (R = resistance of A and B each). For interest, what is the resistance of B and C in parallel, if resistance = R for each case.

    Any time total resistance drops, the current increases. If the total resistance increases, current drops. I = V/R.
  6. Feb 26, 2006 #5
    first off thank you very much for your help.
    If a wire is connected between 1 and 2, then C is short circuited and there is no light, however current still runs through the wire. Thus A stays the same brightness and both B and C lack light.
    the resistance of B and C in parallel is 1/Req = 1/Rb + 1/Rc and B still has resistance as does A. When the wire is connected, C lacks resistance? so then total reistance dropped and the current drawn from the battery increases?
  7. Feb 27, 2006 #6


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    When C is short-circuited, so is B, because they are in parallel. Then only A provides resistance (R) and draws current.

    If two resistances R are connected in parallel, the combined resistance = R/2. So with A (also resistance R) connected in series to the parallel set of B and C, the total resistance would be R + R/2 (initial circuit).

    Now short-circuit R/2 and one has R, so current increases.

    If C were to break (open circuit), then the circuit would include A in series with B, with the same current since A and B are in series, but the resistance is now R + R = 2R, and the current would decrease.

    Its important to remember the significance of open and short circuit, and series vs parallel connections.
    Last edited: Feb 27, 2006
  8. Feb 27, 2006 #7

    Tom Mattson

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    One thread per topic, please.
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