Why is the voltage across one resistor the same as supply?

[x86]

People say that the voltage across one resistor is equal to the voltage supplied.

I've drawn a picture of a simple circuit with two points near the positive terminal. Apparently, if we take the potential difference of these two points, it will equal 9V according to the above saying (which can be found all over google).

But then if the potential difference between the two red dots is 9 V, there will be no force driving the current to the negative terminal of the battery. So this one resistor essentially absorbs all of the energy supplied by our battery. I am confused about this.

 

[axmls]

We generally assume that the wires in the diagram have no resistance, and thus by ohm's law, have no voltage across them.

Anyway, certainly electrons are moving. Do you have a problem with that? Electrons are moving through the resistor because there's a voltage difference across it.

Now, I don't know if this is simplistic or what, but if there is a flow of electrons across the resistor, then certainly these electrons move through it and in turn push the ones in front of them. But maybe I'm wrong. It seems logical to me.

 

[davenn]

But then if the potential difference between the two red dots is 9 V, there will be no force driving the current to the negative terminal of the battery. So this one resistor essentially absorbs all of the energy supplied by our battery. I am confused about this.

HI again x86

I touched on this in your other thread with those drawing of the open circuit, resistor and capacitor

Yes the difference is 9V ... so you tell me what the voltages are on each side of the resistor
That is what are those 2 sides of the resistor connected to ??

when you work out the answer to the last part, you will see there is a force driving the current


Yes, now if you give the resistor a value, you can work out how much current is flowing through the resistor
and then how much power is absorbed by the resistor


cheers
Dave

 

[NascentOxygen]

But then if the potential difference between the two red dots is 9 V, there will be no force driving the current to the negative terminal of the battery.

In circuits it is assumed that components are connected by near ideal conductors. These will have vanishingly small resistance so require a voltage of almost zero to conduct reasonable currents, according to Ohm's Law. The practical outcome being that we can consider all of the battery voltage appears across the resistor.

 

[jim hardy]

But then if the potential difference between the two red dots is 9 V, there will be no force driving the current to the negative terminal of the battery. So this one resistor essentially absorbs all of the energy supplied by our battery. I am confused about this.

We exaggerate to perfection in order to grasp a concept
then we back up to reality
and judge case by case how far toward perfection it is reasonable to push for each task at hand.

For thinking purposes your simple circuit is actually four resistors and a perfect voltage source in series.
The perfect voltage source is 9.000000etc volts and could deliver infinite current.The four resistors are:
Internal resistance of your 9 volt battery, let's just say 5.00000etc ohms. That's why a 9V carbon-zinc battery only makes a couple amps,
Call that one Rs, for R of source.

R1, your resistor. Let's just say it's 4.00000etc ohms.

Resistance of your two wires, let's call them Rwl and Rwr for resistance of your wires from left and right side of resistor to battery,.
Let's just say they're 0.00100000etc ohm each, about a foot of #10.

What is total resistance of the loop?
The sum of the individual resistances.
Rs + R1 + Rwl + Rwr
5,0000 + 4.000000 + 0.001000000 + 0.00100000 = 9.00200000 ohms.

Now i'll leave it to you to figure the voltage drop across each individual resistance.

What that'll show you is that voltage drop across a wire isn't really zero, it just rounds off to zero.
I grew up before calculators so it's intuitive to me that all calculations are approximate. We always rounded to 3 digits because that is practical limit of slide rule accuracy.I really recommend you work that problem longhand writing every step with all the digits
then again rounding to three digits
that should get your thinking straight.

... drum roll...Using three digit accuracy did your voltage drop across the wires round to zero ?
And if your source really had zero internal resistance like you drew it, would voltage across load round off to same volts as source?

 

[x86]

Thanks for your help guys. I think that now I have the idea model of how everything happens. The electrons flow from the (-) to the resistor with no resistance, by the electric field created by the battery. Now we have a ton of electrons on one side of the resistortrying to get through it, of course, we have a region of low potential here. Once the electrons get through the resistor, they move really fast to the (+) terminal, so in this region there is high potential. This potential difference is 9V. Now we can find out the current through the resistor and how much energy is being lost to heat.

 

[nsaspook]

Thanks for your help guys. I think that now I have the idea model of how everything happens. The electrons flow from the (-) to the resistor with no resistance, by the electric field created by the battery. Now we have a ton of electrons on one side of the resistor trying to get through it, of course, we have a region of low potential here. Once the electrons get through the resistor, they move really fast to the (+) terminal, so in this region there is high potential. This potential difference is 9V. Now we can find out the current through the resistor and how much energy is being lost to heat.

You are still head-butting electrons and talking about speeds.
OK, where do you think the electron average drift velocity would be higher in the circuit and why?
https://www.physicsforums.com/threa...arriers-when-going-through-a-resistor.497058/

 

[maline]

x86, read Jim Hardy's post. The voltage across the resistor really is less than 9V, like you originally assumed. It's just that the difference is small enough to ignore.

 

[x86]

You are still head-butting electrons and talking about speeds.
OK, where do you think the electron average drift velocity would be higher in the circuit and why?
https://www.physicsforums.com/threa...arriers-when-going-through-a-resistor.497058/

At first, probably not where the resistor is. But then after electrons buildup, according to that thread, it will be equal all over the circuit.

x86, read Jim Hardy's post. The voltage across the resistor really is less than 9V, like you originally assumed. It's just that the difference is small enough to ignore.

Yes. I know that the voltage across the resistor will be a little less than 9V (As Jim Hardy was saying, because the wires have resistance too). But the thing that confuses me is how the voltage across a resistor is near 9 volts. Like, close to it. I know that the voltage between the two battery plates is 9V, but this doesn't explain why the voltage between the resistor (the two red dots in my first post) will be even close to 9V. Therefore, my prior post talking about electrons was me trying to find an explanation for this occurrence.

 

[russ_watters]

Yes. I know that the voltage across the resistor will be a little less than 9V (As Jim Hardy was saying, because the wires have resistance too). But the thing that confuses me is how the voltage across a resistor is near 9 volts. Like, close to it. I know that the voltage between the two battery plates is 9V, but this doesn't explain why the voltage between the resistor (the two red dots in my first post) will be even close to 9V. Therefore, my prior post talking about electrons was me trying to find an explanation for this occurrence.

I=V/R
The amperage (I) has to be the same everywhere in the circuit otherwise electrons will spill-out onto the floor. So every section of the circuit has to obey I=V/R and the I's are the same. The resister obeys I=V/R with its almost 9 volt drop and the wire obeys I=V/R with its almost no voltage drop.

 

[x86]

I=V/R
The amperage (I) has to be the same everywhere in the circuit otherwise electrons will spill-out onto the floor. So every section of the circuit has to obey I=V/R and the I's are the same. The resister obeys I=V/R with its almost 9 volt drop and the wire obeys I=V/R with its almost no voltage drop.

I realize that if we measure it with an ohmmeter, it will drop 9V. But is there an actual explanation for this (besides using V=IR with the experimental knowledge that V=9V and R = the resistance of the resistor)? Do we know why this happens, and say, why it doesn't drop by say 3 V and then short the battery instead?

 

[davenn]

it will only drop by 3V if ...
1) the supply is 3 V or ...
2) there is another resistor in circuit that is dropping the voltage by 6V

The voltage drop across any single or series resistors MUST add up to equal the total supply voltage

 

[davenn]

answer me this ...

what is the voltage at point A relative to the battery negative ?
what is the voltage at point B relative to the battery negative ?

 

[x86]

answer me this ...

what is the voltage at point A relative to the battery negative ?
what is the voltage at point B relative to the battery negative ?

View attachment 79344

Well, it is known that A-(-) is 10V and that B-(-) is 0V. Although I'm not entirely sure how this can be explained through physics. (Unless we just say that its because experiments told us so). I'm also not entirely sure why A is 10V (but I know it is true, of course).

 

[davenn]

I'm also not entirely sure why A is 10V (but I know it is true, of course).

where is point A connected to ( other than the resistor) ?

 

[x86]

where is point A connected to ( other than the resistor) ?

It is connected to the +10V terminal.

 

[davenn]

It is connected to the +10V terminal.

Yes ! , so, ignoring that tiny bit of resistance in the wire ( that we spoke about yesterday), the voltage at point A cannot be anything other than 10V

so on the top of the resistor we have 10V and on the bottom it's 0V
so the voltage drops from 10V at the top, to 0V at the bottom

 

[x86]

Yes ! , so, ignoring that tiny bit of resistance in the wire ( that we spoke about yesterday), the voltage at point A cannot be anything other than 10V

so on the top of the resistor we have 10V and on the bottom it's 0V
so the voltage drops from 10V at the top, to 0V at the bottom

Ah I see. Is it because of ohms law too? V=IR (for voltage drop). We know that R of the wire = 0, thus the voltage drop is 0. So by ohms law, if it is 10V at the positive terminal, minus 0 V drop, then it has to also be 10V at that point too? It confused me, thinking about potential difference as two points in space, and then implementing that into circuits. But I guess I have to remember circuits are much more complex than just charges in space :)

 

[davenn]

exactly

and for the same reason, the 0V at the battery terminal is also 0V at point B on the bottom of the resistor

 

[x86]

exactly

and for the same reason, the 0V at the battery terminal is also 0V at point B on the bottom of the resistor

Ah, thank you and everyone else so much for your help. It is so satisfying to not be plagued by this anymore :)

 

[davenn]

You're welcome

OK, would you now like to take this learning a step further and learn about current flowing in a circuit like this ?

 

[russ_watters]

[deleted -- should have read the next few posts...]

 

[x86]

You're welcome

OK, would you now like to take this learning a step further and learn about current flowing in a circuit like this ?

Yep. Soon enough though, as reading week finishes this Sunday. (Gotta start studying for some other stuff now too).

If you drop a rock from 5' off the ground, why doesn't it stop 1' off the ground?

The battery is a 9V battery: it has a voltage of 9V. The only way for the voltage not to be 9V is for the battery to be shorted by a wire with almost no resistance so the internal resistance of the battery takes over. But we already have a scenario with resistor with a known resistance that follows V=IR, so...

Ah yes. That is true too.