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Electric Circuits

  1. May 20, 2009 #1
    Hey guys, I've just starting learning about electric circuits at school and there's something I just can't seem to get my head around.
    If you have two identical light bulbs in a series connected to a battery, the brightness of both the bulbs is the same. However, both bulbs are dimmer than if you simply had one light bulb in the series. I don't understand why both bulbs would be the same. It seems logical to me that the one closest to the battery would be brighter (electrons use all of their energy on that bulb) and the second bulb would be dimmer. However, this isn't the case. It is as if the electrons know that there is going to be another bulb up ahead, so they conserve half of their energy for it. How is this possible?

    Thanks :)
     
  2. jcsd
  3. May 20, 2009 #2

    MATLABdude

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    Welcome to Physics Forums! Both bulbs are equally bright (assuming they're the same make and model) because they both have the same resistance. Wiring up two of them together increases the resistance, well, by two times what you had before (of course).

    The electrons don't "know" that there's a second resistor (or lightbulb) in series, the total branch resistance determines how much current will flow, and, as a consequence of this, each resistor drops half the total voltage.

    To take your line of reasoning further, why should any power supply respect ohm's law at all (or care how much resistance it's trying to push electrons through)? They would merely pump out the same amount of current and let the chips fall where they will (possibly literally as your chips fry and/or blow up).
     
  4. May 20, 2009 #3
    But what's the difference between having one light bulb in one circuit and two light bulbs in another circuit so that the same of their resistances is equal to the resistance of the first. In the first example, there is the same resistance but all of the energy from the electrons goes to the light. In the second example, there is the same total resistance, however each bulb uses half the energy. What makes both the bulbs in the second example use equal amounts of energy? If you took the second one bulb out what would make the electrons suddenly use all of their energy on the first one?
     
  5. May 20, 2009 #4

    MATLABdude

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    The bulbs use the same energy because they both have the same resistance. If you took out the second bulb (and shorted the connections together there), the light bulb would use twice the power because the voltage is all across the single light bulb (which has half the resistance of that original light bulb).

    Again, no electrons know anything; the amount of current is determined by the total resistance in the circuit. This current (multiplied by the resistance of a single light bulb) just happens to work out such that half the total voltage is dissipated across the first light bulb, and the other half across the second light bulb. In the end, it's resistance which is the answer to most of your questions.
     
  6. May 20, 2009 #5
    I think what I'm trying to ask is why does it work, not just how. It helps me to get a better understanding of it.

    Thanks for the help though, greatly appreciated :)
     
  7. May 20, 2009 #6

    MATLABdude

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    Ultimately, I don't believe science comes up with answers to the big 'why', just whys that are probably more along the lines of by what mechanism, or how something happens. You can keep on exploring the issue by taking more electromagnetics classes (or, more likely, electromagnetics classes once / if you go to college and take hard sciences or engineering), or materials science and condensed matter classes, but all these do is to give you more insight into how it happens.

    You'll learn later on that electric potential travels at the speed of light (through a wire), and must be established before current starts flowing, but that the only current that flows is in proportion to the total resistance of the stuff in its way, and that voltage division is always what happens.

    EDIT: And that current flows at a speed significantly less than the speed of light.
     
  8. May 20, 2009 #7
    Let's start with an easier example: one resistor and one battery connected to the circuit by a switch. We turn on the switch. Now, we learn in class that there is a voltage increase across the battery and a voltage decrease across the circuit, and that in general, when you sum the voltage differences in a circuit, they should sum to zero. We ask, how does the resistor know to negate precisely all of the voltage increase provided by the battery?

    One answer is that if a circuit's voltage differences didn't sum to zero, the current would go faster and faster, because the battery's pushing the charges and giving them energy while nothing in the circuit takes that energy away. What stops this from happening? A resistor limits the current in the same way air limits the speed of a falling object to its terminal velocity. My guess is that when you flip on the switch, the battery's voltage forces the current to go faster and faster until the current is high enough for the resistor to exactly negate the voltage imparted on the current by the battery, in the same way a falling object will increase in speed until the air resistance precisely balances out the force of gravity and the object's gravity becomes constant.

    My point is that what you learn in class is what happens in the circuit after it's had time to have its voltage givers and takers-way balance things out, i.e., after it's attained the steady-state. I would bet that it's possible for one light bulb to be brighter for a split second. I don't know for sure, and it'd depend on how exactly the potential sets itself up around the circuit. Maybe the potential travels both ways around the wire from the battery, if the wire was initially at V=0 while the battery's terminals were at +V/2 and -V/2. In this case, if the bulbs were each placed the same distance from each end of the battery, then maybe they'd light up identically as the new wire potential met each bulb at the same time. I don't know. I hope I've made things clearer, though.
     
  9. May 21, 2009 #8
    Thanks, that clears it up a fair bit. I'm trying to think of the current flow as an equilibrium of some sort, just that that equilibrium is reached extremely quickly.

    Thanks for all the help guys.
     
  10. May 21, 2009 #9

    russ_watters

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    What you just described is mathematically impossible. Light bulbs have a fixed resistance. The only way to make a circuit with one light bulb have the same resistance as a circuit with two of that light bulb is to add another resistor to the first circuit.
     
  11. May 21, 2009 #10

    russ_watters

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    There is no "why" in a mathematical relationship, really. Why does 1+1=2? The problem here is simply a matter of not being able to visualize what is happening.

    Maybe if you think about something easier and more intuitive to visualize, it will help. Consider: water in a horizonatal pipe, with a fixed pressure at one end (say it is fed by an elevated tank) of 100 psi and open at the other end. The pipe is large enough that there is no meaningful restriction to water flow for our purposes. In this pipe, you place an object that provides a certain resistance. The pressure drop across this restriction will be 100 psi because that is all of the pressure available.

    But then you put two of the same object into the pipe. Does the water flowing past the first object know that it must save some pressure to get past the second object? No, the second object creates a restriction that simply doesn't allow the pressure between the two objects to drop to zero. The system achieves a stable equilibrium based on the same physical principles applying to both restrictions, causing the pressure drop across each to be 50 psi.

    If the system is full of water and there is a valve at the (otherwise open) end, the total pressure in the system is 100 psi and it is static. When you open the valve, the pressure at the open end drops to zero and that pressure change propagates backwards through the system. The water starts flowing past the second object first. That lowers the pressure between the second and first object, causing the water to then start flowing past the first object. The pressure variations even themselves out due to the math of a dynamic situation reaching a dynamic equilibrium.

    Actually building a mathematical model of the development of the equilibrium is pretty tough, but it shouldn't be that hard to visualize for water. The mathematical model for your lights and battery looks exactly identical.
     
    Last edited: May 21, 2009
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