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Lightbulbs in parallel & brightness > counterintuitive

  1. Feb 3, 2015 #1
    1. The problem statement, all variables and given/known data
    Hey guys, this is actually a question from my roommate's homework. She gave her response and I figured I'd check it over. I have some questions too!

    Consider the following statement made by a physics 480 student while pondering the fact that two bulbs connected in series are dimmer than the bulb in a single bulb circuit. "A battery always produces the same amount of current. It's the same battery, so the current has to be the same. The brightness of the bulbs in a two bulb circuit changes because the bulbs are sharing that current 50-50." Do you agree or disagree with this statement? Justify your response.

    2. Relevant equations

    3. The attempt at a solution

    "In a two bulb series circuit the two bulbs are less bright than a single circuit bulb because the two bulbs must share the energy, therefore creating less flow. In a double bulb circuit the current is less than that of a single bulb because it must be shared. In a parallel circuit the flow throughout the circuit increases or is greater because of the set up and the transfer throughout goes more smooth. The current is split amongst the two devices in different paths 50-50. The brightness of the two bulbs will remain the same." -Alexa

    "A battery always produces the same amount of current."

    Incorrect. Power is conserved, and even that changes. The load determines current drawn from a battery. This is why the remaining bulb in a two-bulb series circuit will get brighter if you've removed one of them. You just cut your load in half! Just think of a light bulb filament as a resistor. That's really what it is, right? Assuming source voltage is constant, a decrease in resistance = increase in coulombs/second through the load, and thus a brighter light bulb, hence E=IR.

    And in the case of a two-bulb parallel circuit? This is where I need your guys' help. It's counterintuitive because you'd think that removing a "resistor" would give you the same results as before. But the brightness of the remaining bulb doesn't increase. Heck, it stays the same. Is it easier to go the other way with it and start adding light bulbs in parallel? How would you explain that the total resistance of the system is decreasing and therefore the current drawn is increasing ...without giving the impression that a greater increase in current will always equate to a brighter light bulb? It's easy to see that the voltage across each light bulb is equi...


    Oh, and she mentioned they have yet to cover Ohm's Law, so I'm shooting for an explanation that doesn't require it:/
    Last edited by a moderator: Feb 3, 2015
  2. jcsd
  3. Feb 3, 2015 #2


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    Yes, the voltage across each bulb is completely unchanged, as you state. So why would you expect the behavior of the first bulb to change? It is getting the same voltage across it regardless of how many other bulbs are put in parallel with it.
  4. Feb 3, 2015 #3
    Her instructor posted the correct answers. It's actually "disagree". When you disconnect one of the light bulbs in a two-bulb series circuit, you now have a break in the circuit, and the brightness of the remaining bulb is 0 (no current can flow). In the parallel scenario, it won't make any difference. You know, I wonder how she was supposed to answer the question without having an understanding of Ohm's Law...


    Reread the first statement of the question. Make note of the underlined.

    "Consider the following statement made by a physics 480 student while pondering the fact that two bulbs connected in series are dimmer than the bulb in a single bulb circuit."

    There's a difference between asking what would happen if you disconnected a light bulb in a two-bulb series circuit & asking to compare the brightness of a two-bulb series circuit to a single-bulb circuit. The question is poorly written. The remaining bulb is defined as part of a COMPLETED circuit.
    Last edited: Feb 3, 2015
  5. Feb 3, 2015 #4


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    Your response is correct. A battery produces a voltage, not a current. The current is determined by the load resistance as you state.

    If the parallel bulbs are connected by wires of negligible resistance, and if the battery's output resistance is negligible, then each bulb will have the same brightness as a single bulb would have. But if those resistances are not negligible, what happens...?
  6. Feb 7, 2015 #5
    Incandescent bulbs are non-ohmic but we can ignore that for the purposes of a simplified model with bulbs as resistors but keep in mind that it isn't either voltage or current alone but the power dissipated that determines the brightness of a bulb.

    For the series circuit, all bulbs connected in a series have the same current passing through them. The resistances form a voltage divider. If they are all identical then they each have the same voltage drop of V/n Volts, where V is the total difference in potential from the start of the series to the end and n is the number of identical bulbs. They have the same current and voltage so therefore the same power (p=i*e or i*v if you prefer.) Removing a bulb from the series leaves an open circuit and all bulbs go out, unless you close the circuit with a jumper or some other manner. If you do close the circuit whenever removing a bulb they will get brighter as they get fewer since the total voltage is being divided into fewer portions.

    For bulbs in parallel the same voltage is seen across each bulb since each node can only be at one potential and parallel bulbs are all connected between the same two nodes. They will share the current however. The current into the nodes gets divided into branch currents, one for each parallel branch. Again, if the bulbs are identical, the current in each branch will be equal. The reason the brightness doesn't change when removing or adding bulbs is the total resistance changes also. The total current pulled from the voltage source is a function of the total resistance of the circuit. For 2 bulbs of equal resistance, the parallel combination has half the resistance of each one individually.Half the resistance means twice as much current but it gets divided into twice as many branches. For more than two bulbs the concept scales with the total conductivity (inverse of resistance) being equal to the sum of the conductivity of each bulb. For unequal resistances, the current in each branch is weighted linearly, branches with higher resistance having less current.

    When it really gets fun is when you have parallel and series combinations like several series connected in parallel bunches that are in series with other parallel groups.
    Hope this helped a little. It takes practice to grow your intuition.
  7. Feb 10, 2015 #6
    Are you asking me what would happen if I were to parallel two light bulbs of unequal resistance (let's say 100 ohms & 200 ohms)?

    Higher wattage bulb = brighter light bulb

    The 100 ohm light bulb will be consuming .09 Watts while the 200 ohm light bulb will only be consuming .045 Watts.

    The 100 ohm light bulb will be brighter.

    BUT! Will it be twice as bright? Is the brightness of a bulb curve linear?
  8. Feb 11, 2015 #7


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    No. I was asking about the case where the *wires* to the two bulbs are different lengths, and hence have different resistances. Also, I was asking about the case where the output resistance of the power supply is not negligible... :-)
  9. Mar 9, 2015 #8
    Assuming like wires i.e. composed of same material, gauge, etc. A conducting length of Cu may be shorter than another conducting length of Ag, but that doesn't mean that it is less resistive. Your point is clear. I'm just trying to refine mine.

    It shouldn't make a difference. If I parallel light bulb A and light bulb B to a power source with light bulb B having longer lengths of wires to it giving that path a higher resistance, then that light bulb (B) should be less bright than light bulb A.

    Is it wrong for me to take the resistance of the conducting path and add it to the light bulb's resistance to get R total for that load? Voltage across each load is the same. If I jump some old corroded wires across the terminals of the power source to power light bulb B, that should have zero effect on light bulb A. Easiest to think of it as two separate circuits.

    *If I have a two-bulb parallel circuit, is it wrong of me to call light bulb A load 1 and light bulb B load 2 or need I consider the whole circuit as a single load all together?

    Lastly, are output resistance and internal resistance used interchangeable? The only time I've seen internal resistance or "output resistance" come into play is when I tried wiring [too many] L.E.D.'s in parallel. A chemical battery can only supply so much current. I'm not sure what happens to internal resistance when dealing with power sources that supply continuous power (generator). Maybe it becomes irrelevant and it's impedance that matters...
  10. Mar 10, 2015 #9


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    Correct as long as the power supply behaves like an ideal voltage source (eg zero internal resistance).

    You can treat it as two circuits provided the power supply behaves like an ideal voltage source (eg zero internal resistance).

    In this context (batteries and light bulbs) they are more or less interchangeable. However the term "internal resistance" could be applied to an input as well as an output. For example if you had a rechargeable battery on charge it would make more sense to call it "internal resistance" than "output resistance". It might be more appropriate to use "output resistance" to describe the resistance of the output of the charger.

    The output resistance/impedance of a generator or the National Grid is very low and for many applications it can be ignored but not all. For example you can't just build 10 new houses and connect them to the grid without checking that the local transformer can cope without the voltage dropping.
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