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Consider the following situation - A simple series circuit with a 10 V power source and two 5 Ω resistors

I know that a voltage drop of 5 V would occur at each resistor due to Ohm's law and the current remaining constant at 1 A. But I am much more interested in the logical/intuitive part of this scenario rather than the theoritical situation according to Ohm's law.

As fas as I know, resistance is only concerned with obstructing the flow of current, so how do two resistors in series (logically/intuitivelly) end up causing a voltage drop? Is there any analogy to explain this?

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  • \$\begingroup\$ How do you move an electron? Imagine being a conduction band electron, free to move about. If all of the electric fields around you "look the same" then what's the motivation to move? You'd sit there. To move to the right, you'd need to see somewhat more negative field to the left than to the right. And once you moved a little, you'd still need to see a similar situation (more negative left than right.) In short, you'd need electric field gradient. Stretched out over some distance this means "voltage difference" from one end to the other. Currents require a voltage gradient: i.e. "drop." \$\endgroup\$ Jan 17, 2023 at 3:16
  • \$\begingroup\$ For what it's worth, I have a solid understanding of the fundamentals of electronic circuits, I know the mathematics behind them pretty well, I've designed several circuit boards, I have a basic understanding of Maxwell's equations and how they lead to inductors and capacitors working, and I know a lot of the differential equations governing circuits and how to solve them, and I have no idea how resistors produce a voltage. For all I know, maybe there's a tiny person who lives in the resistor, counts how many electrons go by, and casts a magic spell that produces a voltage. \$\endgroup\$ Jan 17, 2023 at 3:26
  • \$\begingroup\$ My point in saying all this is to reassure you that if you find it hard to understand how a resistor produces a voltage, then that's completely okay—you'll be perfectly fine if you don't manage to figure it out. And if you do find the answer, all the better! \$\endgroup\$ Jan 17, 2023 at 3:30
  • \$\begingroup\$ Why do you consider it needs two resiators to drop voltage? Each resistor drops voltage all by itself. There just happens to be two resistors in a voltage divider circuit. It could be a single potentiometer but just set to half voltage where both resistances from center tap is equal. Or any other piece of resistive material. We just draw resistors as ideal lump with two terminals. \$\endgroup\$
    – Justme
    Jan 17, 2023 at 5:43
  • \$\begingroup\$ The think model of a fluid flowing still gives insight. Water will only flow if there is a level (pressure) difference, which is the analogy of voltage. \$\endgroup\$ Jan 17, 2023 at 6:35

3 Answers 3

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Start by considering the 10 V supply without any resistors:

schematic

simulate this circuit – Schematic created using CircuitLab

Now we will add those two 5 ohm resistors. There's 1 amp flowing.

schematic

simulate this circuit

That conjecture that

resistance is only concerned with obstructing the flow of current

seems to actually be the other way around. There's no current until we add the resistors. So the resistors are allowing some current to flow.

Another thing to note is that the voltage between V1 and V2 (and voltage is always the difference) is always 10 V whether we put in resistors or not. And if we were to put ten 1 ohm resistors in series where the two 5's are, and make nodes for each (V1, V2, V3, V4, ...) we would see the voltage change to 1 V as we checked each one in succession. Remember that we will always put one of our voltmeter's leads on node V1, so we get the difference between that and what we are measuring. The voltage "drop" is the change in voltage, which is less, the closer we measure to V1.

So you could say that the voltage is "spread out" across all of the series resistances.

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  • \$\begingroup\$ Thanks for the comprehensive answer. I understand that the voltage is spread out across the resistors. But why does this "spreading" of voltage even occur, why doesn't it occur when we have a single resistor in a series circuit? Ohms's law seems to allow for this phenomenon, and it does indeed occur, but to me, it makes 0 logical sense as how can resistance generate voltage \$\endgroup\$ Jan 17, 2023 at 8:53
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    \$\begingroup\$ This "spreading" does in fact occur inside the resistor, since it is a length of resistive material. If you could cut one open and probe it, you would observe the voltage changing along its length. \$\endgroup\$
    – gbarry
    Jan 17, 2023 at 9:05
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Ohm's law.

If the question is "resistors" the answer is "Ohm's law".

So the bad news is to understand how voltage dividers work you're going to need to apply Ohm's law. but the good news is that Ohm's law will also tell you about how they don't work, that is the ways in which voltage dividers are non-ideal.

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I am much more interested in the logical/intuitive part of this scenario

Consider the following circuit:-

schematic

simulate this circuit – Schematic created using CircuitLab

Ohm's law says that if a material draws a current proportional to the applied voltage then it has the property of resistance, whose value in ohms is equal to the voltage divided by the current.

In the circuit above R1 is a resistor, so it must follow Ohm's law. It has 5 V across it so it must be drawing 1 A, and it has 1 A going through it so it must have 5 V across it. Not only does it have 5 V applied to it, it is also dropping 5 V.

Now let's examine your circuit:-

schematic

simulate this circuit

Because R1 and R2 are in series you know that the same current must be flowing though both, and since they have the same resistance they must each drop the same voltage. So 'intuitively' you would expect each resistor to drop half the total voltage, ie. 10 V / 2 = 5 V.

This can be extended to any number of resistors. Imagine you had ten 1 Ω (or any equal value) resistors in series. Each resistor would drop a tenth of the total voltage (ie. 1 V) so the voltage to ground would step down 1 V at a time as you went down the chain.

Or consider a length of wire which has a resistance of 1 Ω per meter. Roll out 10 meters and connect 10 V across the ends, then measure the voltage across any part. You will see a voltage proportional to the length you are measuring. Now take that wire and cut it in half to make two 5 Ω resistors, then join it again in the middle. This is the same as your circuit. Half of the total length (and resistance) drops half the total voltage.

Is there any analogy to explain this?

Be warned - analogies sometimes help, but often cause more confusion when they break down. What people 'intuitively' know about something familiar like eg. water going through a pipe, may not even be accurate, let alone properly understood.

The analogy that springs to mind here is the length of a piece of string, which everyone knows is equal to the distance from the middle to either end. Useless information? Not if you want to cut it exactly in half...

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