So I'm taking my intro to electromagnetism course and we are learning about currents and resistance right now and I'm running into a little confusion when it comes to the conventional flow and electron flow of this voltage division circuit. So the issue is when I try to calculate the voltages between each resistor using conventional current I get the same values like the one in the diagram below. However when I try to do this considering electron flow I'm getting different values, specifically the voltage in the wire from R3 to R2 is 3 volts but the voltage in the wire from R2 to R1 is only 1 volt. I don't know how I'm not getting the same values as with conventional current. Shouldn't I get the same values?? Voltage Divider

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    \$\begingroup\$ But electron flow is conventional current. They're not two different things. Electrons have negative charge. Therefore, their amps in the forward direction will subtract from any amps of positive charges in the same direction. ( Perhaps you're not aware that "direction of particle motion" is a different concept than "electric current?") To actually use a different convention, we'd have to declare that the charge on the electron is positive, with the protons negative. Without that, we find that negative charges, moving forward, gives a Conventional Current! It has a negative value. \$\endgroup\$
    – wbeaty
    Mar 30 '17 at 8:41
  • \$\begingroup\$ @Wbeaty I think the confusion is coming from the fact he's basically swapping the poles of the battery and thinking it's "electron flow" and then wondering why there's not 5V just after R1 \$\endgroup\$
    – Doodle
    Mar 30 '17 at 8:46
  • \$\begingroup\$ What do you mean by forward direction? Is that from positive to negative? \$\endgroup\$
    – XSoloDolo
    Mar 30 '17 at 8:47
  • \$\begingroup\$ Is there a separate "electron flow" methodology? Ohm's law has no reference to electrons. Just use conventional current, it's the approach that makes sense. \$\endgroup\$
    – pjc50
    Mar 30 '17 at 8:52
  • \$\begingroup\$ You're right, "forward direction" isn't clear. Actually it means the direction you've decided to connect your ammeter: where is the positive ammeter terminal connected. Feeding negative charges into the positive terminal of an ammeter (the "forward" direction,) would measure a conventional current that has a negative value. PS conventional current is hiding the polarity of charges, and hiding the charge density and flow-velocity. Very useful for non-metal conductors which may have electrons and protons both flowing past each other. That conductor just has one number: the amperes. \$\endgroup\$
    – wbeaty
    Mar 30 '17 at 9:09

You haven't posted your calculations you used to get the answers you got so I'm just going to post my thoughts.

You seem to be trying to find the voltages between the resistors, what you actually want to do is find the voltage drop across the resistors.

Regardless of whether you use conventional or electron flow, the voltage drop across the resistor is still the same. Once you find all your voltage drops using either method and write them on the circuit, then you can find the voltage between two of the resistors. You seem to be getting yourself confused but as I can't see how you arrived at your answer I don't know how.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Ok so finding the voltage in between r1 and r2 I took V1=6V-1amp*1ohm=5V for voltage in between r1 and r2 then I took that voltage and subtracted it by the resistance of r2 multiplied by the current which was V2= V1-2ohm*1amp=3. I did the same process in the oppposite direction for the elctron flow and got different values \$\endgroup\$
    – XSoloDolo
    Mar 30 '17 at 8:32
  • \$\begingroup\$ Electron flow ≠ Voltage flow. What you're essentially doing in your calculations is Conventional flow, then you're reversing the battery and calling it electron flow, however it's still conventional flow just with the poles on the battery swapped. \$\endgroup\$
    – Doodle
    Mar 30 '17 at 8:43
  • \$\begingroup\$ Ok I must be completely missunderstanding how voltage changes through a circuit. How do I find the voltage in between each resistor then with each voltage drop of the resistors? \$\endgroup\$
    – XSoloDolo
    Mar 30 '17 at 8:44
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    \$\begingroup\$ The points on a circuit don't "have a voltage," since voltage depends on where you connect your voltmeter leads (where you connect your black lead, the reference point.) In other words, your 6V supply could be 0, 6V, or -6V, 0. Or it could be +3v, -3V, if your meter reference is at the point between R2 and R3. As @Hayman says, resistors can have voltage across them, but points on a circuit don't "have voltage" except in relation to some chosen reference point. \$\endgroup\$
    – wbeaty
    Mar 30 '17 at 8:49
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    \$\begingroup\$ If you want my advice, stick to conventional flow. I've done 6 years in electronics without even considering electron flow even once as the results still come out to the exact same thing at the end of it all \$\endgroup\$
    – Doodle
    Mar 30 '17 at 8:50

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