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Homework Help: Working out the current though bulbs in a circuit by applying Kirchoff's laws

  1. Nov 1, 2012 #1
    1. The problem statement, all variables and given/known data
    I am trying to determine the birghtness of the lettered bulbs relative to the brightness of the bulb in the first and simplest circuit by giving a mathematical explanation using Kirchoff's laws. All cells and bulbs are identical.

    2. Relevant equations
    Kirchoff's current and voltage laws

    3. The attempt at a solution
    I define that the bulb in the first circuit will have power P=I^2*R and because each bulb is identical, the resistance of all the bulbs in every circuit will be the same. Therefore the brightness of the bulb only depends on the current so if I work out the current for the lettered bulbs and compare that to the bulb in the first circuit with current (I=V/R) then I can say if the lettered bulbs are brighter or dimmer then the original bulb.

    For the circuit with bulbs lettered d and c I get stuck after splitting the circuit vertically into two closed circuits:

    Applying Kirchoff's voltage law I get:
    V - (R*I1)) - R*(I1 - I2) = 0
    V - R*(I2 - I1) - (R*I2)

    I need to find I1 and I2 in terms of I but I am unsure where to go from this.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 1, 2012 #2


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    There's an obvious way to simplify that equation. Then divide through by R so that you get I in there. And you have another equation for the second circuit.
  4. Nov 2, 2012 #3
    Multiply both equations out. Then multiply one of them by 2 and add the equations together. Don't be afraid to try things out - you just may find that you can solve them!
  5. Nov 2, 2012 #4
    Okay I will try
  6. Nov 2, 2012 #5
    V - (R*I1) - R*(I1 - I2) = 0
    V - R*(I2 - I1) - (R*I2) = 0


    I - 2*I1 + I2 = 0

    I - 2*I2 + I1 = 0

    I solved these to get
    I1 = I

    therefore I2 = I

    [STRIKE]Therefore D and C have the same brightness as the original bulb. Will now try for the third circuit.[/STRIKE]

    EDIT: Actually noticed that I1 and I2 have the same magnitude but opposite direction so the total current through bulb D is 0 so it is off. Bulb C still has the same brightness though.
    Last edited: Nov 2, 2012
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