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Series and parallel circuits

  1. May 17, 2008 #1
    if we have 2 circuits. series and parallel. both of them have 3 light bulbs connected to them. 1 ohm 2ohms 3ohms. both circuits have these light bulbs. will the 3ohm light brighter in both circuits?why or why not? im thinking yes because the higher the resistance, the more collisions there will be so more electrical energy is converted into heat and light, but im not sure, so willl the 3ohm light brighter in both circuits? why or why not .
     
  2. jcsd
  3. May 17, 2008 #2
    No. The Three Ohm light bulb will be the brightest only if it is connected parallel to the other light bulbs. This is because the voltage drop on this light bulb will be equal to the voltage of the battery. However, in series the voltage drop on this light bulb will be smaller than the initial voltage. Thus in a series, the 3 ohm light bulb might actually be the dimmest one. A parallel circuit will guarantee it to be the brightest light bulb.
     
  4. May 17, 2008 #3

    nrqed

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    It's actually the other way around.

    In series the current is the same in all three so the best equation to use for the power is P = R I^2 which shows that the largest resistance will dissipate the most power.

    In parallel they all have the same voltage so it's better to use P = V^2/R which shows that the largest resistance will dissipate the smallest power.
     
  5. May 17, 2008 #4
    Call the bulbs B1, B2 and B3.

    In series, the total resistance is 6 ohms. So the current is V/6. The power in each bulb is the same: PB1,PB2,PB3 = 1/6 V^2, so they are equally bright. This is because each drops a different amount of voltage. VB1 = V/6, VB2 = V/3, VB3 = V/2.

    In parallel, the voltage drop for each bulb is the same: V. The current is different. IB1 = V/1, IB2 = V/2, IB3 = V/3. So the power in each bulb is PB1 = V^2, PB2 = 1/2 V^2, PB3 = 1/3 V^2.

    So, in series all the bulbs are the same brightness. In parallel, B3 is the dimmest of the 3. But its still brighter than in series.
     
  6. May 18, 2008 #5
    Corrected: Call the bulbs B1, B2 and B3.

    In series, the total resistance is 6 ohms. So the current is V/6. CORRECTION: each bulb drops a different amount of voltage. VB1 = V/6, VB2 = V/3, VB3 = V/2. So, the power in each bulb is PB1 = 1/36*V^2, PB2 = 1/18*V^2, PB3 = 1/12*V^2.

    In parallel, the voltage drop for each bulb is the same: V. The current is different. IB1 = V/1, IB2 = V/2, IB3 = V/3. So the power in each bulb is PB1 = V^2, PB2 = 1/2 V^2, PB3 = 1/3 V^2.

    So, in series B3 is brightest. In parallel, B3 is the dimmest of the 3. But its still brighter than in series.
     
  7. Jun 5, 2008 #6
    series the current is constant ,parallel he voltage is constant
     
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