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Three Bulbs in a Circuit, two parallel, one series.

  1. Mar 13, 2013 #1
    1. The problem statement, all variables and given/known data
    So I have a circuit diagram that looks close to 5F20.50-series-and-parallel-light-bulbs-600x372.jpg
    This picture.
    The first bulb in the series is called A, then inner bulb in the parallel is B and the outer is C

    The questions
    a) Which, if any, is brightest?
    b) What happens if you remove C?

    2. Relevant equations



    3. The attempt at a solution

    A) I would assume B is brightest because the electrons flow from the negative terminal of the battery. However I'm unsure if this is correct.

    B) I would assume A and B would increase in brightness and the resistance of the circuit would decrease. Again, I'm unsure if this is correct.

    Thanks to anyone who helps!
     
  2. jcsd
  3. Mar 13, 2013 #2

    PeterO

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    Homework Helper

    That picture exactly represents the situation you have - so the answer to part (a) should be obvious - though the explanation might evade you.

    The thing to remember about a light globe is that if you halve the voltage drop across it you more than halve its brightness - simply on the way a light globe works.

    Half the voltage may only be sufficient to make the filament to have a gentle red glow because it is not very hot. Indeed, if you took a standard 240V globe [standard household electricity is at 240V here in Australia] and applied a potential difference of only 120V, it may not glow at all.
     
  4. Mar 13, 2013 #3

    CWatters

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    Science Advisor
    Homework Helper

    a) The brightness will be roughly proportional to the power dissipated in each bulb.

    Power = Voltage * Current

    Try modelling the bulbs as 1 Ohm resistors. Two 1 Ohm resistors in parallel have an equivalent resistance of 0.5 ohms.

    Voltage

    Using the potential divider rule the voltage across bulbs B//C will be...

    Vbat * 0.5/(1+0.5) = Vbat/3

    The voltage across bulb A will be...

    Vbat * 1/(1+0.5) = Vbat * 2/3

    Compare the two voltages. The voltage across bulb A is twice that of B and C.

    Current

    The current through bulb A splits and half goes through each of B and C.

    So the current through A is twice that of B and C.

    Power

    Power = Voltage * Current

    If you double both the voltage and current then the power is 4 times greater.

    So the answer is that A will be about four times brighter than B or C.

    b) If you remove C you are left with two bulbs in series. The voltage and current will be the same for both remaining bulbs so both will have the same brightness. However...

    The overall resistance will go UP. If we pretend the bulbs are still 1 Ohm resistors then the total resistance changes from

    1 + 1//1 = 1 + 0.5 = 1.5 Ohms
    to
    1 + 1 = 2 Ohms

    The total current will therefore change from Vbat/1.5 to Vbat/2 which for bulb A is a reduction in current by a factor of 1.5/2 = 0.75

    The voltage on Bulb A also changes from Vbat * 2/3 to Vbat * 1/2 which is also a factor of (1/2)/(2/3) = 0.75

    If the current changes by 0.75 and the voltage changes by 0.75 then the power and brightness of A reduces by 0.75 * 0.75 = 0.56. So you could say that removing C will make A about half as bright.

    I'll let you work out what the magnitude of the change to B is.
     
    Last edited: Mar 13, 2013
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