# Voltage dropping in resistor

Discussion in 'General Electronics Discussion' started by alireza, May 19, 2017.

1. ### alireza

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May 19, 2017
drop of voltage in resistor has confused me. I know base on V=kq/r formula voltage depend on distance and the charge. and I know resistor don't change these two quantities. so what is reason of this voltage dropping ?
for show you what is in my mind let imagine a simple circuit with one battery and one resistor. we determine 3 points on this circuit like image below:

I know there is a voltage between A and C , and B and C , but there is no potential difference between A and B. my question is why there is no voltage between these two point ? A has some charge ( so we have q ) and distance to B (so we have r ). my question is why V(A)=V(B) but V(A) is not equal to V(C) ?????????

2. ### shrtrnd

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Jan 15, 2010
Points A & B are effectively an electrical short-circuit.
You won't read a voltage across those two points.
You would read voltage only between Points C and either points A or B.

davenn likes this.
3. ### alireza

2
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May 19, 2017
I understand what you say. my question is why ??
please explain based on electron movement in wire or V=kq/r formula. these are confusing me .
pleassssssssssssse

4. ### BobK

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1,688
Jan 5, 2010
A wire cannot have an excess of charge at one end, it will simply flow through the wire to neutralize.

Of course, wires do have some resistance, and a small diameter wire carrying a large current will show a voltage drop dependent on the resistance of the wire and the current.

Bob

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5. ### Cannonball

191
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May 6, 2017
You can only have a voltage reading across a resistor. A resistor is used to limit current. In your schematic without the resistor you would either burn the wire into or kill the battery. I hope this helps.

6. ### AnalogKid

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Jun 10, 2015
Yes, there is.

There is nothing wrong with your theory or logic, except that your conclusion is based on false data. Do not confuse what you can see or measure with what is real. Assuming none of your wires are superconductors, then each connecting wire has resistance of its own, and there *is* a voltage drop between A and B. With a 5- or 6-digit voltmeter you can measure it easily.

ak

7. ### davennModerator

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Sep 5, 2009
but for the sake of this basic learning the very tiny drop across that wire is irrelevant unless it is very long and or has significant resistance
... don't make it confusing for the guys when just learning the basics
...for this basic learning, the two wires between the battery and the ends of the resistors can be treated as ideal

8. ### Cannonball

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May 6, 2017
Yes what you say is true. If there wasn't

Yes what you say is true. Even the best conductor has some resistance. If you take out the resistor in this schematic and conect the wire staaight across the battey for a short time it will drop the entire voltage of the battery until the wire burns into or the battery explodes or drains very fast. But there will be a voltage drop.

Last edited by a moderator: May 20, 2017
9. ### davennModerator

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Sep 5, 2009
but you are not understanding the point I am making
that is .... when we are building the avg circuit in a radio, amplifier, microcontroller project or what ever else, there is NO NEED to consider the resistance of the interconnecting wires pcb tracks.
Its only when you have interconnecting cables that extend metres, 10's of metres, kilometres that the cable resistance needs to be taken into account

10. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,489
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Jan 21, 2010
I think I am just echoing what others have said, but for most purposes, the connections between components are considered to have zero length, zero resistance, etc.

Practically they're not, but the difference is normally so small as to be inconsequential.

For those of us with the appropriate equipment, we can measure the resistance of (say) a few cm of wire and knowing the current through it we could calculate the voltage drop across it..

For example, if your battery is 1.5V, the resistor is 1kΩ and the wire between a and b has a resistance of 0.0035Ω (that's based on a calculation of the resistance of 5cm of 24SWG wire). The current through the resistor will be 0.0015A (1.5mA), and at this current, the voltage across the wire will be 0.0000053V (or 5.3μV).

You may have noticed that I calculated the current through the resistor without taking into account the voltage drop across the wire. I did that knowing that the resistance would be so small as to be insignificant. If I had taken that into account I would probably also have needed to consider didn't of other things like the internal resistance of the battery, the resistance of the leads and connections inside the battery, the resistance of the leads of your resistor, the reusable of your joints, and even the rate of discharge of the battery. Most likely, all of these taken together would not have resulted in a significant change in the answer (let's say to 4 significant digits). Further to this point, if you connect a 6.5 digit meter to a battery you'll notice that the voltage is slowly changing, it's undergoing slow self discharge, temperature change, and even slight discharge from the meter.

In practical circuits, things should be designed so these things are insignificant.

If however that resistor was 0.01Ω, then all of these things would come into play. The resistance of the leads and connections may well exceed 10mΩ in total; the internal resistance of the battery may cause the voltage to drop from 1.5V to some number of millivolts (that's TO, not BY). In this case your circuit now has a stack more resistances in it, and you would probably show them as resistors n your circuit for simulation purposes.

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11. ### Cannonball

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May 6, 2017
Well said and you covered all of the basses. All is in toerance.

12. ### AnalogKid

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Jun 10, 2015
That is not the way I read posts #1 and #3. He's asking about charge balancing and electron movement, and in that context his statement about no voltage difference between A and B is incorrect. He knew it was incorrect, but didn't know why, which is the question he was asking. Given his statement of Ohm's Law, he looks to have significant education but little experience, which is why I framed my response to his level.

ak

13. ### davennModerator

13,810
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Sep 5, 2009
and I read it totally opposite as do the others

he states that there is no voltage drop between A and B

several of us have answered those Q's as they were stated

Last edited: May 20, 2017
14. ### Ratch

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Mar 10, 2013
I think that your problem is not knowing what voltage is. Voltage is the energy density of the charge, defined as joules/coulomb in the MKS system. It takes energy to move two charged particles of the same polarity close together. Even more energy if they are moved closer together. Still more energy if several charged particles are gathered. If it takes one joule of energy to corral one coulomb of charge together at a location, that location is defined to have an energy density per charge unit of one volt with reference to where the charges came from. The same polarity charges don't like to be together, and will scatter if a conduction path exists. The resistor and wire provide the conduction path. A higher energy density per unit charge (voltage) at one end of the resistor will cause the charges to move through the resistor to the lower voltage at the opposite end. While traveling through the resistor, the charges will encounter repulsive forces from the ionic cores of the resistor atoms and various fields from other charged particles of matter within the resistor. The charges will lose energy in the form of heat, and will arrive at the other end of the resistor with less energy than they started with. The same number of charges exit the resistor as entered, but with less energy. Therefore, the energy density per unit charge (voltage) will be less. This loss of voltage is called the voltage drop caused by the energy dissipation (heat) loss while transiting the resistor. The wire has much less resistance so the voltage drop is very low and is usually neglected.

Ratch