Electronic – Does resistance decrease over a resistor

voltagevoltage measurementvoltage-dropvoltage-reference

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?

Best Answer

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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