This is something I have been thinking for a while. At my level, I don't know much about electric fields. Now, I will be asking two questions: One for series circuit, and the other for parallel.

Series: Let us consider a simple series circuit with one resistor, which I put close to the negative terminal. Now, we say that $\Delta V_{battery}$ $ = \Delta V_{R1}$ + $\Delta V_{R2}$ $ .... \Delta V_{Rn}$, where n is the total number of resistors. Now, in this case we have only 1 resistor, so $\Delta V_{battery} = \Delta V_{R}$. My doubts:

  • Why are these equations true?
  • To me, it seems that the electric field is set up in a way which I cannot think of? Can someone explain me how the electric field is set up in a series circuit, maybe by using a diagram? Does the physical nature of the circuit effect how the field is set up?

Parallel: We say that $\Delta V_{battery} = \Delta V_{R1} = \Delta V_{R2}$ and so on, depending upon how many resistors are there. My questions:

  • Again, why is this equation true? How is the electric field set up in this type of circuit?

Electric circuits are easily understood, if they are considered one single step at a time. Let's try...

  1. You have a battery of voltage (potential difference across it's terminals) $\Delta V_{battery}$ and a resistor $R$. They are now connected with wires end to end in a circuit. Initially no electric field has been established and no electrons (no current) are moving, and there is no voltage across the resistor.

    • High potential at one point means that many electrons are gathered here. The repel. As soon as there is a path for them to move along, they start moving because they are pushed away from all the others.
  2. Electrons now start moving to the connected end of the resistor. As more electrons arrive the repelling force persists, so electrons start squeezing through the resistor.

  3. The electron flow is slower inside the resistor, so more electrons flow from the battery terminal than what moves through the resistor, and therefore they pile up at the resistor-end.

  4. Soon there are so many electrons present at this resistor end that they repel each other just as much as they are repelled from the battery terminal. Then an equilibrium is reached and they don't want to move anymore.

  5. As soon as an electron makes it through the resistor, there is room for a new one (since the repulsion here is a bit lower) and a new one enters. Then another one is moving from the battery terminal to fill up the missing space at the resistor-end. The repulsion is as good as equal at the resistor-end compared to the battery terminal at all times after the initial setup-time. In other words, they have as good as equal potential!

    • This now gives a steady / constant stream of electrons through the resistor. As if there is a constant "pressure" on the electrons from one end of the resistor.
  6. When electrons arrive on the other side of the resistor, they might want to stay there, but right after another electron arrives as well. Now they are again repelled and want to move towards the other battery terminal, where the potential (the repulsion) is the lowest in the whole circuit.

  7. When this steady electron flow is set up at the lower-potential side of the circuit as well, the potential is equal from the (negative) battery-terminal to resistor-end and likewise equal from the opposite resistor-end to the opposite (positive) battery-terminal.

Same flow-consideration works for parallelflow and for circuits of several components in series.

  • $\begingroup$ @Steevan, the explanation was really good.... But in #3, you said that flow is lesser in the resistor, so more electrons arrive at the end. Is this not violating the fact the current is constant? $\endgroup$ – codetalker Mar 1 '16 at 12:17
  • $\begingroup$ @Siddhantinf The flow is resisted in a resistor. So while the current builds up (before equilibrium and steady current) this is a blockade and there is less current through the resistor. When the equilibrium is reached, the incoming electron flow has been decreases to equal the electron flow through the resistor. So after a while, there will be equal current at any point in a series circuit. Otherwise you can again use the argument that charge will built up in one spot and repel more charge to arrive, thus slowing down the flow until inflow again equals outflow at every point in the circuit. $\endgroup$ – Steeven Mar 1 '16 at 12:19
  • $\begingroup$ thnx a lot for the help $\endgroup$ – codetalker Mar 1 '16 at 12:22
  • $\begingroup$ @Siddhantinf A follow-up comment: We are talking about two stages here, when a circuit is turned on; A stage where current is built up (points 1-3) and a stage where we have steady current (from point 4). Only in the latter stage do we have equilibrium and only then do the current and voltage laws from Kirchhoff apply. $\endgroup$ – Steeven Mar 1 '16 at 12:23
  • $\begingroup$ Okay... Thnx for the help $\endgroup$ – codetalker Mar 1 '16 at 12:26

These equations are held to be true since we postulated that in the DC limit, the electromotive force of the source must be exactly compensated by the load.

Usually, when drawing circuits with lumped components, people do not think too much about the shapes of the fields. But they, of course, exist in every circuit and have characteristic shapes.

The electric and magnetic field, as well as the energy flux, are well discussed here: https://electronics.stackexchange.com/questions/170765/nature-of-electromagnetic-waves/170771, and its didactic aspects are here http://www.abc.net.au/science/articles/2014/02/05/3937083.htm.


It's true because of ohms law. If you have 1 amp travelling through 3 resistors in series, and each resister is 2 ohms, you are going to have 2 volts across each resistor. They're in series, you add them up, to get a total of 6 volts that would be across the whole string of 3 resistors. The same thing works in reverse... a battery is connected across 3 resistors in series, then each resistor would have 1/3 the battery voltage. ...minus anything lost in the wires, since they have some resistance also. You might lose 0.1 volts across a long wire.

If they're in parallel, well they're all connected right across the battery terminals, so they will all have whatever voltage is coming out of the battery at that moment. ...minus losses in the wires, which is usually negligible unless high currents are involved. The current through each resistor would be 1/3 the current coming out of the wire connected to the battery.

  • $\begingroup$ No, Ohm's law has nothing to do with this: in a series circuit, the voltage drops across each element add up to 0 even if the elements do not satisfy Ohm's law. If Ohm's law holds for an element, then the current through it is proportional to the voltage across it and the proportionality constant is the resistance: that's all it tells you. $\endgroup$ – NickD Mar 8 '18 at 5:38

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