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If ohms law pertains to voltage, current, resistance across a resistor, in order for voltage to drop, and current to remain the same, does resistance drop as well across the resistor?

To explain what I mean, if we have the circuit:

enter image description here


The voltage across the first resistor drops linearly from 5v to 1.667v. Let’s say we picked a point around the middle of the resistor and read that the voltage at that point was about half of the difference across the resistor, so 5v - [(5v - 1.667v) / 2] = 3.3335v.

From what I as told, current across resistors in series is constant/equal.

So if the voltage at that point is 3.3335v, the current is fixed at 1.667A at that point and every other point across the resistor, does this mean that resistance is dropping throughout the resistor as well to compensate for the voltage drop and so ohms law remains valid?

Because at that point if voltage = 3.3335v and current is 1.667A, resistance at that point would have to be 1.999700059988002 ohms, given by V=IR.

At a point somewhere between the middle and end if we measured the voltage and got 2v for example, R = 1.199760047990402 ohms.

Overall, this would show a downward trend in voltage, a constant current, but also a downward trend in resistance.

But is this correct? And if resistance is also fixed, how is it possible current can remain constant when resistance is also fixed and voltage is dropping while obeying ohms law?

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  • \$\begingroup\$ Pressure? Don't get caught up in analogies. \$\endgroup\$ – DKNguyen Jul 21 at 21:18
  • \$\begingroup\$ @DKNguyen no intention to insinuate the use of analogies, I updated the question to replace all uses of “pressure” with voltage. \$\endgroup\$ – Ietpt123 Jul 21 at 21:20
  • \$\begingroup\$ voltage across the first resistor drops linearly ... how do you know this? \$\endgroup\$ – jsotola Jul 22 at 0:05
  • \$\begingroup\$ it seems that you think that if you have a 10 ohm resistor, then the resistance between the end of the resistor and the middle of the resistor is 10 ohms \$\endgroup\$ – jsotola Jul 22 at 0:11
  • \$\begingroup\$ Have you consider to use a water analogy to help you with this? Just like a water in a pipe. If 1 litre per second enters the pipe, the same amount of water must leave the pipe at the other end. Water cannot magically disappear from the pipe. This is why in series circuit current must be the same everywhere. physicsforums.com/threads/… Also do not forget about the fact that that if one electron is entering the "wire" , at the same time another electron must leave the "wire". Wire cannot accumulate a charge. \$\endgroup\$ – G36 Jul 22 at 22:26
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So if the pressure at that point is 3.3335 V, the current is fixed at 1.667 A at that point and every other point across the resistor, does this mean that resistance is dropping throughout the resistor as well to compensate for the voltage drop and so ohms law remains valid?

The resistance will be proportional to the length. The resistance will be the same even if no current is flowing through it.

Because at that point if voltage = 3.3335 V and current is 1.667 A, resistance at that point would have to be 1.999700059988002 ohms, given by V=IR.

Let's call that 2 Ω. (15 decimal places is just silly.) You are measuring the voltage across the right half of the 2 Ω resistor and the 1 Ω resistor so you get \$ \frac {2}{1} + 1 = 2 \ \Omega \$.

Overall, this would show a downward trend in voltage, a constant current, but also a downward trend in resistance.

Yes. As you slide your measurement point across the resistors the voltage will reduce in proportion to the resistance.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. A potentiometer as an adjustable voltage divider.

This is the principle of operation of a potentiometer. By sliding the wiper up from the bottom to the top the voltage out will vary from 0 V to V+.


Note, we say current through a resistor and voltage across a resistor.


From the comments:

Also to confirm, so across a resistor and nothing but a resistor (taking the first resistor in the attached circuit as an example), voltage would drop from 5v at the entrance of the resistor -> 1.667v at the exit of the resistor ...

Yes 5 V at the left side and 1.667 V at the right. (SI units named after a person have their symbols capitalised but are lowercase when spelled out. 'V' for volt, 'A' for ampere, 'K' for kelvin, 'Ω' (capital omega) for ohm, etc. Meanwhile 'k' is for kilo. There's also a space between the number and the unit.)

... and resistance would drop from 2 ohms at the entrance of the resistor down -> 1 ohm at the exit of the resistor ...

No, resistance of the left resistor would decrease from 2 Ω to 0 Ω relative to the 1.667 V point. Resistance relative to the 0 V point would decrease from 3 Ω to 1 Ω.

... given by (V = IR, 1.667v/1.667A), therefore keeping current at a constant 1.667A from the entrance to exit of the resistor.

What goes in must come out. Provided your voltage measurement device has a very high impedance (typically 10 MΩ for a digital multimeter) it won't divert a significant current so if you have 1.667 A in then you get 1.667 A out.


Update 2:

But if resistance decreases from left to right of the first resistor, from 2 ohms -> 0 ohms, how could current exit the first resistor at 1.667A? If at that exit of that first resistor, like you said, resistance dropped to 0 or for arguments sake 0.01, by ohms law, the exiting voltage at that point = (1.667V/0 ohms) (1.667V/0.01 ohms) is not equal or close to 1.667A?

schematic

simulate this circuit

Figure 2. Measuring the voltage between the tap-off point and the right end of R1.

You have forgotten that the voltage will decrease between your measurement point and the right end of R1. (Re-read the explanation of Figure 1.) If you move the measurement point 3/4 way across R1 you will measure 3.33 / 4 = 0.84 V as shown on the voltmeter of Figure 2. If you move it all the way to the right of R1 you have \$ V = IR = 1.667 \times 0 = 0 \$. Everything works.

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  • \$\begingroup\$ Also to confirm, so across a resistor and nothing but a resistor (taking the first resistor in the attached circuit as an example), voltage would drop from 5v at the entrance of the resistor -> 1.667v at the exit of the resistor and resistance would drop from 2 ohms at the entrance of the resistor down -> 1 ohm at the exit of the resistor given by (V = IR, 1.667v/1.667A), therefore keeping current at a constant 1.667A from the entrance to exit of the resistor. @Transistor \$\endgroup\$ – Ietpt123 Jul 21 at 21:39
  • \$\begingroup\$ See the update. \$\endgroup\$ – Transistor Jul 21 at 21:46
  • \$\begingroup\$ but if resistance decreases from left to right of the first resistor, from 2 ohms -> 0 ohms, how could current exit the first resistor at 1.667A? If at that exit of that first resistor, like you said, resistance dropped to 0 or for arguments sake 0.01, by ohms law, the exiting voltage at that point = (1.667V/0 ohms) (1.667V/0.01 ohms) is not equal or close to 1.667A? \$\endgroup\$ – Ietpt123 Jul 21 at 21:55
  • \$\begingroup\$ See my update 2. \$\endgroup\$ – Transistor Jul 21 at 22:09
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Resistivity is a physical property of material and it will not adjust itself to satisfy Ohm's law.
So, the value of a resistor or resistance itself has nothing to do with Ohm's law.

Resistance can be expressed by $$ R = \rho \frac{\ell}{A}$$ where \$ \ell \$ is the length of the resistor, A is the cross-sectional area of the resistor (in m²) and ρ is the electrical resistivity (also called specific electrical resistance) of the material.

Depending of the point of view, the resistance decreases (or increases) with length.

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  • \$\begingroup\$ @ThePhoton You're right. Ill correct it. \$\endgroup\$ – Huisman Jul 21 at 21:59
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Yes. Large ceramic wirewound resistors have a slidable center tap to make them adjustable. Potentiometers also make use of this fact. It's the difference between a lump sum and distributed model.

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  • \$\begingroup\$ To confirm, in general, voltage and resistance decreases across resistors while current remains the same? \$\endgroup\$ – Ietpt123 Jul 21 at 21:23
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    \$\begingroup\$ Yes. It's like splitting up one resistor into an infinite number of smaller ones which is more accurate what you physically have in real life (specific resistivity of a material). \$\endgroup\$ – DKNguyen Jul 21 at 21:26
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Don't mix rounded and unrounded values! You are looking at rounded numbers, not exact values. And then you are trying to use those rounded numbers in exact calculations, and wondering why they don't match Ohm's law. Ohm's law is not broken, your calculations are.

The voltage is not 1.667V, it is exactly one third of 5V, my calculator says 1.666666667, but using that many, or infinite amount of numbers would be just plain silly. The current is also exactly 5/3 amps, or 1.666666667.

If you split the 2 ohm resistor into two 1 ohm resistor, the current does not change as there is still 5V over 3 ohms and voltage will be exactly two thirds of 5V at that point, 3.333333333, but the simulation software would just most likely round that to 3.333 for displaying. Of course, this assumes that wires you are using for connections are ideal and have no resistance, and the voltage source is ideal as well.

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