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I'm a software developer and I'm learning electronics for embedded systems.

As I understand circuit today when a voltage is applied across a wire then the wire have a potential difference equal to the voltage applied between the two ends of the wire.

We attached it to a battery, so we forced this side to be at zero volts and we forced this side to be at 12 volts based on our definition of the negative terminal being zero volts [...]

Physics 102 - Electric Potential and DC Circuits

And if I understand correctly there is a drop in potential going from highest to lowest potential (12,11,10,...)

because the charges can't build up and create an electric field. You'll still have a potential drop. Since we have an E applied, there's the 11 volt line and the 10, and the 9, the 8, 7, 6, 5, you get the idea, 4, 3, 2, 1 down to zero. This face is at 12, this is at zero, and a potential drops throughout the metal.

Physics 102 - Electric Potential and DC Circuits

Now why when I measure voltage in a circuit without a resistor the multimeter show a potential drop as expected but once a resistor is introduced in the wire the multimeter output zero ?

Without resistor:

enter image description here

With resistor:

enter image description here

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  • \$\begingroup\$ Not necessarily a duplicate, but see also this similar question that was asked recently, Voltage in a Short Circuit \$\endgroup\$ Jun 28, 2023 at 16:29
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    \$\begingroup\$ Does this answer your question? Voltage in a Short Circuit \$\endgroup\$ Jun 28, 2023 at 16:31
  • \$\begingroup\$ I think this story will be useful for you. \$\endgroup\$ Jun 28, 2023 at 17:05
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    \$\begingroup\$ @Circuitfantasist, The article draws a really good picture and analogy of voltage drop across a resistor. I think I have not considered the circuit as one big wire with different resistances across it. Thus, I was thinking more like multiple parts connected together and independent of each other. So my guess now is that the voltage drop due to the negligeable resistivity of the conductor, is distributate accross all the resistors. \$\endgroup\$
    – mba
    Jun 29, 2023 at 15:33

2 Answers 2

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With a wire directly connecting the two ends of the battery, you will have (in theory) an extremely large current. The simulator apparently assumes the wire has some very low resistance which will cause an 0.25 volt drop across that section of wire.

In Real Life the wire and battery in your first circuit will get very hot - the wire may melt and the battery will be destroyed.

When you add a 1 Ohm resistor, the current will be limited to 1.5 Amp, which will not produce a measureable voltage drop on that section of wire.

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Electrical wires have very low resistance. So low that the resistance is considered to be zero for most calculations and designs.

Considering that, the current through the wire is very very high. The only limiting factor is the small amount of resistance actually present in the wire.

Example: Copper wire of length 5 cm and diameter 1 mm has resistance of about 1.07 mΩ. This is very very low.

This very high current in the wire is causing some potential drop in the wire across those 2 points.

V = IR

If you can calculate the total resistance of the wire, you can use the above formula to calculate the current. Now calculate the resistance of the wire between you two points of measurement. Applying the above formula you should get a value close the the that shown in the meter.

In the second case, you are adding a significant amount of resistance with the resistor (significant in comparison to the resistance of the wire). This causes the current to drop significantly too. Less current causes less potential drop between those two points. The potential difference now is too low for the meter to measure. More sensitive voltmeters may be able to read the drop and give you a reading.

In both the cases there is a potential drop. In the first case, it is big enough for the meter to read and display. In the second case it is not big enough.

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  • \$\begingroup\$ So if im understanding correctly in the first case it is expected to have a potential drop that goes from 0 up to 1.5 volt as I measure points from one end of the wire to the other end of the wire ? A potential drop similar to a linear gradient. But when I introduce a resistor it is not the case anymore. \$\endgroup\$
    – mba
    Jun 29, 2023 at 10:51

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