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Electronics Understanding

  1. Jan 12, 2016 #1
    Hi

    Im studying electronics off the web (btw Im looking for a good online course). My background is in Biochemistry and I've gotten into programming (6+ years) and recently Arduino projects (1 year). Im trying to get a better handle on electricity.

    As Im reading about Ohms law here (http://www.learn-about-electronics.com/series-circuit.html) I read that in a series circuit, the current through one of the components is the only measurement we need to know to determine the current through the whole circuit because since there is only one path in the circuit the same current must flow through every point. My first confusion comes here because I thought that a resistor was a component that 'resisted' electrical current flow. And since I=V/R, then current must definitely be reduced by an increase in Resistance.

    Why is this confusing me? An amp (current flow measure) is C per second. So a resistor reduces the rate of movement but not the total movement? Is that what is confusing me?
     
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  3. Jan 12, 2016 #2

    phinds

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    Electrical current in a series path is like a bicycle chain. The analogy is, more resistance slows down the speed of the chain but it has the same speed, whatever it is, throughout the circuit. The "speed of the chain" is analogous to "the amount of current", so more resistance means less current (for a fixed voltage) but the same current throughout the circuit.
     
  4. Jan 12, 2016 #3

    jim hardy

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    Current is defined as amount of charge moving past a point per unit time , whatever charge is ....
    1 coulomb per second ...... that's quite a few individual charges .
    Charge is conserved, too - what goes into a component at one end must come out at the other end ...
    sp
    think of it this way,, real simple physical terms
    pick your point alongside the wire
    those charges can't get out the side of the wire, can they ? There's insulation there .
    So,, as the saying goes , "There's no way out but through " - through the next component in line.

    It takes work to push charges through resistance - think of resistance as friction --- work done shows up as heat .

    any help ?
     
  5. Jan 12, 2016 #4

    meBigGuy

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    When you increase the resistance, you decrease the current thoughout the whole circuit. The same smaller current goes in and comes out of the resistor.
    If you have 5 series resistors, when you increase the resistance of 1, you decrease the current through them all. The amount of current that can flow is determined by the total resistance (and the applied voltage, of course).

    When you apply voltage, the supply voltage is split (propagated) across the components at the speed of light in that medium and the proper current flows. The potential starts "pushing" electrons and only the proper amount will actually flow.

    (I refuse to resort to a hydraulic analogy)
     
  6. Jan 13, 2016 #5
    Im almost there. Ok, so when I add a resistor, the current, C/s, is reduced, meaning that where as before 2 C/s were traveling through a point, now only 1 C/s is traveling. So yes, the current was reduced and so the the movement of those charges slows down. So the current at any point will have slowed down. So yeah, I think I got it.

    Btw, when the voltage is supplied, it is propagated at the speed of light?
     
  7. Jan 13, 2016 #6

    phinds

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    No, you haven't quite "got" it because you do not yet understand drift velocity, which is the answer to your last question. Google it and you'll see what I mean.

    EDIT: and by the way, don't feel remotely bad about not having got that. It is (I think anyway) completely counter-intuitive and your assumption of light speed is very reasonable until you see what's really happening.
     
  8. Jan 13, 2016 #7

    jim hardy

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    amen


    it is easy to get incorrect mental models in one's head
    i want to say 'yes' to your question even though a more typical number is 2/3c. Electrical properties of whatever the field is traveling through get into the picture,
    but do not think the charge carriers(be they electrons or ionized molecules) move anywhere near so fast
    upload_2016-1-13_11-29-25.png

    It's like people on an escalator. They move slowly but a rumor will propagate the entire length almost peed of sound .

    hang in there.

    it's a simple observed relationship
    Current(in amps) = Voltage(in volts) / Resistance(in ohms)
    amps is Coulombs per second
    volts is Joules per coulomb
    resistance is volts per amp

    the units are not intuitive at first being just people's names
    after you've worked drill problems your brain accepts the linear relationship.

    I've always pitied mechanical and chemical engineers - their Ohm's law involves square roots , viscosity, elevation, local gravity and Reynolds numbers. We EE's have it easy, really.
     
  9. Jan 13, 2016 #8
    I actually think I got it so Im good. I just wanted to clear up why I was under the impression that if I added a resistor the the circuit, the current would be different in different parts of the circuit whether if before or after the resistor. :-)
     
  10. Jan 13, 2016 #9

    meBigGuy

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    I take a slight exception with your use of "slow down" when you really mean "decrease". The number of electrons per second decreases, but the current flow does not "slow down" in a velocity sense. Fewer electrons at the same drift velocity. I assume that was what you meant, but I had to comment anyway.
    EDIT: Actually, Am I wrong about this? I thought the drift velocity in a wire with constant voltage across it would be the same regardless of the length of the wire. But the formulas seem to disagree.

    It propagates at the speed of light in the material, which, BTW, is different than speed of light in a vacuum. Google velocity of propagation. Some people use the "ping pong balls in a pipe" analogy to explain the difference between the velocity of propagation and the drift velocity. Push one ball in, one immediately pops out, but its a long time before the one you pushed in makes it to the output (but don't take it too literally as it has limited usefulness as an analogy).
     
    Last edited: Jan 13, 2016
  11. Jan 13, 2016 #10

    jim hardy

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  12. Jan 14, 2016 #11

    dlgoff

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    What hasn't been said, your current, at a given voltage in your series circuit, will be calculated as if there is only one "equivalent resistor". i.e. for series, Requivalent = R1 + R2 + R3 + ... , but you may already be aware of this. :)
     
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