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To my understanding, if a resistor reduces the potential energy of the current across it, then the current that leaves the resistor will have less potential energy and thus, less pressure or voltage.

If one resistor only drops one volt from the original voltage (only drops one volt because of its resistance value) and we add another resistor in series with the same value, then that second resistor would also drop one volt. Then, the current leaving the second resistor would be 2 volts less. In reality, this is not the case and somehow, in the middle of the two resistors the voltage would be dropped to half the original voltage. How is this possible if each resistor is made to only drop 1 volt?

My image below shows how each resistor would lower the potential energy of the current in order to end up with 4 volts, since each only drops one. enter image description here

I know this is incorrect in reality and I also know how to calculate the voltage drops, but physically it doesn't make sense to me.

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    $\begingroup$ Your problem is utterly ill posed. Use a full circuit instead, don't make assumptions, and everything will be clearer... $\endgroup$
    – TZDZ
    Jan 6, 2015 at 8:50

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The starting point of this question makes it hard to give a short, satisfying answer; maybe someone will come along and do a standout one with diagrams, but I don't have time for it. So I'll kick off with this:

With two resistors in series, there's more total resistance on the circuit, which means that less current flows. With less current flow, a given resistor drops less voltage because V=IR. Voltage drop isn't an intrinsic property of the resistor; resistance is.

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    $\begingroup$ thanks for your reply. The voltage dropped would then be proportional to the resistors values and it seems like it isn't since two resistors of 50omhs would drop the voltage the same as two resistors with 1 ohms each, which seems counter intuitive and weird. $\endgroup$
    – reddead
    Jan 5, 2015 at 21:01
  • $\begingroup$ @reddead it's proportional to the value of each resistor, but inversely proportional to the sum of all of the resistors (or other loads), because 100% of the voltage has to be dropped somewhere. $\endgroup$
    – hobbs
    Jan 5, 2015 at 21:02
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    $\begingroup$ what do you mean by "inversely proportional to the sum of all of the resistors" can you explain this? $\endgroup$
    – reddead
    Jan 5, 2015 at 21:04
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I know I'm a little late, but I'll take a shot at answering this for you. I'm actually very much a beginner at understanding electronics myself, so everyone: please keep me honest!

There has been some criticism of your question, as it does not show a complete circuit. I need to agree with this, as any reliable calculations within a circuit require knowledge of the entire circuit.

In order to know how much voltage will drop across a resistor, we need to know the value of the current flowing through said resistor. In order to know that, however, we need to go back even further and figure out the voltage being supplied to the circuit and the total resistance within the circuit. Once we have all our prerequisite information, we can say that the voltage drop across a resistor is equivalent to the charge's rate of flow (current) multiplied by the resistance it encounters. This holds true whether you are looking at a series circuit, a parallel circuit, or a mix of the two.

Take for example a circuit where a 12V battery serves as the power supply, and a 6 ohm resistor is placed in between the high and low potential terminals of the battery. The total current supplied by the battery will be equivalent to the voltage divided by the resistance (I = V / R = 12 / 6), resulting in a current of 2A. Now that we know the total resistance of the circuit and the value of the current, we can deduce that the voltage drop will be 12V (V = I * R = 2 * 6 = 12).

Now let's add a second resistor to the circuit (in series, right after the previous one), again with a resistance value of 6 ohms; the total resistance in the circuit is now 12 ohms. Since the resistance has changed, the amount of current will also change. As with our previous calculation, we will determine the total current by dividing the supplied voltage by the total resistance in the circuit. This time, our current is 1A (I = V / R = 12 / 12 = 1). Since the total current has changed, we need to recalculate our the amount of voltage consumed by each resistor. For each resistor (since they have identical resistance values), the total voltage drop is now 6V (V = I * R = 1 * 6 = 6). Since there are two resistors, the total voltage drop is 12V, which is equivalent to the supplied voltage.

This is an example of an ideal circuit, of course. In the real world, you'll see a lot more decimal places as the result of wires having their own internal resistance, resistors being accurate within certain tolerance levels, etc.

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a resistor reduces the potential energy of the current across it, then the current that leaves the resistor will have less potential energy and thus, less pressure or voltage.

The current does not have potential energy, the electric field does (the potential being exactly its potential as in the sense of differential forms).

If one resistor only drops one volt from the original voltage (only drops one volt because of its resistance value) and we add another resistor in series with the same value, then that second resistor would also drop one volt.

That is not true, as it depends on the current that enters the second resistor. Given any resistor with resistance $R$ placed between two points $A$ and $B$, the difference of potential measured between the two points once a current $i$ passes within the resistor is (in the simple cases of Ohm law): $V_A - V_B = Ri$; this means that all three parameters can in principle change however they want, with the only constraint that their product must satisfy the above. In particular (if we assume the resistance to be constant) both the difference of potential and the current may change, their rate being constant though. This said, according to the characteristics of the circuit, especially according to whether you keep potential or current as fixed, the other variable changes in response. You observe a certain drop through the first resistor because you have an initial current, say $i_1$, entering in. Once you plug another resistor the total current flowing in the circuit is in principle different from $i_1$ and the voltage drops have to be calculated accordingly, you cannot just sum two pieces as they were independent of each other. This is due to the fact that if you run the circuit with a fixed battery which ensures you a difference of potential $V_A - V_B$, the overall current must adapt so that along the entire circuit the voltage drop is never bigger than $V_A - V_B$, otherwise no more electrons will move, by definition of circuit.

Bottom line is, as already pointed out above, resistors only make sense and satisfy the laws when included in a circuit: Kirchoff's law is nothing but rephrasing the conservation of electrical charge within a loop (which is not fulfilled in your example, because the difference of potential in the latter case would be bigger than in the former, which may not be true).

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To my understanding, if a resistor reduces the potential energy of the current across it, then the current that leaves the resistor will have less potential energy and thus, less pressure or voltage.

Actually, the resistor produces a drop in potential (that is, potential energy per unit charge).

If one resistor only drops one volt from the original voltage (only drops one volt because of its resistance value) and we add another resistor in series with the same value, then that second resistor would also drop one volt.

Not necessarily, because with the second resistor added, you now have a greater (double) the total resistance. This means that if you have the resistor(s) supply is from a constant voltage source, the current will be halved when the second resistor is added, since $I=\frac{V}{R}$.

For example, of you have a $1V$ source across a $1\Omega$ resistor, you would get $1A$. When you add a second $1\Omega$ resistor in series, the current is reduced to $0.5A$.

On the other hand, if the resistor(s) supply is from a constant current source, then each of the two resistors will have the same voltage across them as the single resistor case.

Eg: A $1A$ supply across a $1\Omega$ resistor will result in a $1V$ drop across the resistor. When you add a second $1\Omega$ resistor in series, it too will have a $1V$ drop across it, because the constant current source will increase the voltage supply to $2V$ to maintain a constant current.

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    $\begingroup$ The resistor produces a drop in both potential and potential energy. Also, the OP says they already know how to calculate voltage drops, they just want to have an intuitive understanding of why adding another identical resistor doesn't double the voltage drop. $\endgroup$
    – Brionius
    Jan 6, 2015 at 2:51
  • $\begingroup$ If you have a constant voltage source of say $1V$ across a $1\Omega$ resistor, you get a current of $1A$. The resistor has a voltage drop of $1V$. If you add a second $1R$ resistor in series whilst keeping the source at a constant voltage of $1V$, the two resistors will each have a $0.5V$ drop across them, that gives a $1V$ total drop. You can't expect to get a $1V$ drop across EACH resistor if the supply is only $1V$! Of course, increasing the supply to $2V$ across 2 identical resistors in series results in $1V$ drop across each resistor. $\endgroup$
    – theo
    Jan 6, 2015 at 2:56
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    $\begingroup$ I...know that. Both I and the OP understand that his idea is not correct, he just wants to know why. You keep trying to explain how to calculate voltage drops, but that's not what the OP is looking for. $\endgroup$
    – Brionius
    Jan 6, 2015 at 3:04
  • $\begingroup$ The explanation I gave is that drop in potential across the load(s) can never be larger than the source voltage! This is a simple statement of conservation of energy, or more formally, Kirchhoff's Voltage Law. $\endgroup$
    – theo
    Jan 6, 2015 at 3:10
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    $\begingroup$ Well, I'll let this go with one final attempt to explain what I mean. The OP says "I know this is incorrect in reality and I also know how to calculate the voltage drops". He is, one would assume, aware of conservation of energy, Kirchhoff's Voltage Law, etc. If he can calculate voltage drops, he presumably knows they must add up to the source voltage. So it isn't helpful to explain that to him. That's all I'm saying. $\endgroup$
    – Brionius
    Jan 6, 2015 at 3:13
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One thing that may help you is an analogy. A popular one is between electrical circuits and water pipe systems (here's one explanation of it). Read that (scroll down - there's more below, too) and see if it helps you understand how resistors reduce voltage.

Here's my attempt at an explanation.

A battery holds the voltage at each terminal constant. The electrons leave one terminal with a particular high potential energy $V$ at one end, and the end up at the other terminal with a particular low potential energy (usually chosen to be 0).

What you put between the terminals of the battery determine the manner in which the electrons lose that energy.

For example, if you have a 6 V battery, the electrons are going to start with $6V$ and end with $0V$, thereby losing $6V$ total, no matter what, as they go from one end of the battery to the other.

  • You could lose all $6V$ in one $100 \Omega$ resistor (resulting in a current of $60 mA$
  • You could lose the $6V$ in two $100 \Omega$ resistors in series. In that case the electrons lose $3V$ in each resistor.
  • You could lose some in a $300 \Omega$ resistor, then more in a $200 \Omega resistor$ in series, then more in a $100 \Omega resistor$ in series, but you're going to lose $6V$ total. In that case, if you do the math, you'd lose $3 V$ in the first, $2 V$ in the second, and $1 V$ in the third.
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