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Question about voltage divider rule in series-parallel circuits.

  1. Feb 29, 2012 #1
    Say a circuit has a 9V DC supply, a 2 ohm resistor in series with it and also a parallel combination of two 1 ohm resistors in series with it. So a simple series-parallel circuit.

    What I don't understand is why the voltage drop across the 2 ohm resistor isn't 6V and the voltage drop across both parallel resistors isn't 3V from a physics perspective?

    I think about it from the perspective of an electron, the electron effectively must traverse through 2 ohms and select one path of 1 ohm, and so it should use 2/3 of it's energy across the 2 ohms and the other third across the 1 ohm from my thinking.

    However, I know this is not true because when we simplify the parallel combination to 0.5ohms in series with the 2 ohms the voltage division changes completely. I'm aware the voltage divider formula relies on their being equal currents in resistor branches, but from an energy perspective it doesn't make sense to me as to why it doesn't work.

    I thank anyone in advance.
     
  2. jcsd
  3. Mar 1, 2012 #2
    I think you are questioning Ohm's law because you are assuming the energy per electron to be independent of the number of electrons passing through the resistors at the same time (in fact the current).

    What you say seems correct for one electron passing through the resistor.But as the number of electrons passing through the 2-ohm resistor increases, each electron needs more energy for passing through it. This is true for 1-ohm resistors too but here the the number of electrons is half.
     
    Last edited: Mar 1, 2012
  4. Mar 1, 2012 #3
    Short answer: electron does not 'require' particular energy to traverse the resistance. The electron will use whatever potential energy is available and happily convert it all to heat.
    The energy wasted by 1 electron does not depend on resistance at all, it depends only on voltage (and measured in electron-volts).

    Long answer: this comes down to Ohm's law on microscopic level.

    When you think of 1amp current you imagine 1 coulomb worth of electrons going through the conductor per second. But the electrons don't go all the way through the conductor. Instead, there is this huge cloud of free electrons bouncing around atoms and the electric field acts on all of them equally. The whole cloud slowly drifts from (-) to (+) like a swarm of mosquitoes in a light breeze. The overall effect is still the same: 1 coulomb worth of electrons enters on one side and 1 coulomb worth of electrons exits on the other.

    The important bit is all free electrons in the conductor always participate equally. So twice the current does not mean twice the number of electrons, it means twice the average drift speed of the cloud.

    Drift speed turns out to be proportional to the force acting on electrons, see gory details here. Since drift speed is proportional to current and the force is proportional to voltage, you get Ohm's V=IR as a result. If you apply twice the voltage, each electron will feel twice the force and the average speed of the electrons will be twice as fast so you use 4 times the power P=V I = V^2 R = I^2 / R.
     
  5. Mar 1, 2012 #4

    cmb

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    'Volts' represents the amount of energy per unit charge. Imagine you're an element of charge driving through a single lane of dense traffic (2 ohms) then you get to a dual lane road (1 ohm, half the resistance of the single lane). The traffic gets through the dual lane section easier. If you were driving through rush-hour traffic, it'd take you to work at the controls of your car to get through it all.

    Now imagine that when you get past the single lane road, you are actually given two dual-lane roads to pick, either route. You pick one way, the next guy picks the other way, half the traffic again on each of the dual lane routes - easier to get throught than if all the traffic had only one dual lane to use.

    Does that help?
     
  6. Mar 2, 2012 #5
    First of all thanks for all your replies.

    @Hassan2 Yes you are right, in my model of electrons if their is more current passing through the same resistance then their will be more collisions with the ions leading to a higher voltage drop. I wasn't thinking that more I => more collisions => more energy lost.

    @Delta Kilo An electron has a particular energy to traverse a resistance depending on the relative resistances. For example one 1 ohm and one 2 ohm resistor in series means 2/3Vs is across 2ohm resistor because it is higher than the other resistor by a factor of 2, it is all relative.

    So your saying that the energy lost across each resistor is proportional to each electron's speed because with more speed they have more kinetic energy to deliver to the ions they collide with hence altering the thermal energy? And also that the amount of electrons in a cross-section area of a wire at any time is a constant and so more current can only be achieved by the drift speed increasing?

    I'm having a bit of trouble visualizing it with the model of the sea of electrons, your basically saying that in the parallel section the E field is weaker than in the series section?

    I just don't understand how the velocities add at the series junction, when I think of it as for e.g 2 electrons passing through each parallel section and then combine together at the series section it should double the current? Or in the other way say you have 4 electrons passing through the series section, then I thought 2 electrons would go through each parallel section. I just can't see how the velocities would just add in the sea of electron model?

    @cmb Yes I understand what you are saying, it all comes down to the lower current passing through the parallel section resulting in less voltage loss than if the same resistance appeared as the series resistor.
     
    Last edited: Mar 2, 2012
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