Relationship of voltage and resistance

[Googl]

Would you say voltage drop across a resistor is directly proportional to the resistance.

I am trying to understand the relationship between voltage and resistance.

Thanks.

 

[Dale]

V=IR by Ohms law. So if you keep I fixed and vary R then you will indeed find that V is proportional to R.

 

[Andrew Mason]

Would you say voltage drop across a resistor is directly proportional to the resistance.

Yes. But that is just Ohm's law: V=IR. If you have several resistors connected in series, the current is the same through each. So the voltage drop across each resistor is directly proportional to R.

AM

 

[epenguin]

But isn't all this just the definition of resistance?

There is no physics really in a definition.

There is some phenomenological physics in saying that in some common laboratory and technological situations (little else) resistance, so defined, is relatively independent of voltage and current. Up to a point. There is some simple physics I suppose - evidence of uniformity - with the (phenomenological) 'laws' of how it depends on the length and cross sectional area of uniform material. Though you can imagine situations and mechanisms where the simple laws break down

 

[Dale]

But isn't all this just the definition of resistance?

There is no physics really in a definition.

I agree. V is proportional to R if you keep I fixed precisely because R is defined as the ratio between V and I. No physics, just application of a definition.

 

[Googl]

Thanks a lot all of you, you've just strengthened my confidence in the idea. I know what voltage, current and resistance is but when put in certain scenarios it can be quite difficult to imagine, yes I understand the law is there but the imagination.

To understand further the relationship of resistance with voltage, we would have to keep current constant and adjust the resistance and observe the voltage across the resistor (not emf).

A variable resistor would be appropriate; Suppose I get a resistor that is adjustable from 0-20KΩ then I get a 20V battery and combine them together to form a circuit. My main goal is adjusting the resistance to observe what happens to the voltage across the resistor. How would you describe voltage when the resistance changes.

Am I correct in saying that as the resistance increases the voltage across the resistor will be higher. Would you say that at 20KΩ the voltage across the resistor would be 20V? It is also quite had to imagine the ratio, I am thinking that I cannot just say that at 19KΩ the voltage would be 19V OR am I wrong?

Also how would you describe the voltage across the resistor with the original voltage from the power supply (e.m.f)

 

[epenguin]

Noo. You are getting into a different question, the question of the supplies of voltage or current. A battery has an approximately constant voltage, so it's the current that mainly varies according to the resistance. Same thing with mains supply as usually engineered.

 

[Googl]

Noo. You are getting into a different question, the question of the supplies of voltage or current. A battery has an approximately constant voltage, so it's the current that mainly varies according to the resistance. Same thing with mains supply as usually engineered.

So when the emf is 20V and I connect it with a variable resistor, when I adjust the resistance to let's say 20KΩ and I put a voltmeter parallel to the variable resistor the voltmeter would read 20V? and when I change the resistance to 3KΩ the voltmeter would read 20V as well? I am referring to voltage drop across the resistor.

 

[sophiecentaur]

Would you say voltage drop across a resistor is directly proportional to the resistance.

I am trying to understand the relationship between voltage and resistance.

Thanks.

This is true but only true if a constant current is passed. In most cases (bulbs and batteries etc.) it is the voltage supplied that is constant. So beware; this relationship you seek will all depend upon the context.
If you just consider R as the ratio of V/I, you can't go far wrong.

 

[dlgoff]

Just thought I'd throw this in.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmlaw.html"

 

[Googl]

Thanks for the illustration and link. I understand that. However this does not describe the relationship between the voltage measured across a resistor and it's resistance in my scenario above. Let's not focus on emf here just the voltage drop across the resistor.

Suppose I use a variable resistor to contrast the resistance, connected to a source or battery that provides 20V. The variable resistor can be adjusted from 0KΩ to 20KΩ.

What relationship would you observe when you connect a voltmeter in the circuit parallel to the variable resistor if you keep adjusting the variable resistor resistance?

By my understanding is that as the resistance of the variable resistor is set to 20KΩ the voltmeter would read 20V (voltage drop) since resistance is directly proportional to voltage. But it would not make sense to say that when the resistance of the variable resistor is set to 0KΩ the voltage read from the resistor would be 0V. Am I missing something in the relationship here? If the above is true then this implies that the ratio between the two is a 1:1 ratio...

 

[dlgoff]

As the resistance approaches zero, the current would approach infinity. Do you know of any sources that can provide an infinite amount of power? In reality, power sources have "internal" resistance, so as the current increases, the voltage of the source lowers.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/dcex6.html"

 

[Googl]

Hello Don,

I am referring to voltage drop across the resistor, not current and not the voltage, V of the source. The resistor is a sliding variable resistor. I am trying to understand the relationship of sliding the contact to adjust the resistance and the voltage across.

 

[sophiecentaur]

I seriously suggest that you don't try to 'look at it your way' because it really is a bit of a dead end. All you can say about the relationship between resistance and voltage is that R=V/I. That works everywhere because R is defined that way; it is just a ratio of two measurable quantities. If the resistance is ohmic or non ohmic then the relationship still applies - you have an instantaneous resistance which varies as a signal varies.
If I asked you whether speed is proportional to distance then you would, rightly, say that it depends upon the time involved, too. It's the same for R and V.

 

[sophiecentaur]

Hello Don,

I am referring to voltage drop across the resistor, not current and not the voltage, V of the source. The resistor is a sliding variable resistor. I am trying to understand the relationship of sliding the contact to adjust the resistance and the voltage across.

OK. Once you have decided on the value of your variable resistor then treat the network as a potential divider and do the sums to find the volts across it. These sums, of course,involve (implied) current and volts - the current being the same through each and volts being shared between the two resistors in proportion to the R values. BUT, it is the combination of the two resistors that determines the voltage across them and not a simple rule involving just one resistor. As I said ten miles back, it's a matter of context; there is not a relationship between just any old resistor on its own and the voltage across it.

If you want to "understand" what happens, do some calculations for yourself and plot values for a range of variable resistance values. The graph you get will show you what happens; you don't get a straight line for V against R.

 

[epenguin]

Hello Don,

I am referring to voltage drop across the resistor, not current and not the voltage, V of the source. The resistor is a sliding variable resistor. I am trying to understand the relationship of sliding the contact to adjust the resistance and the voltage across.

To a first approximation you will not change the voltage at all between (let me say it carefully as you may have some setup in your head) between the terminals of the battery by changing the resistance. Not because it is absurd to think, but because (approximately) constant voltage is a property of batteries. What you will change essentially by changing resistance is what you seem not to look at - the current!

All this is in any beginning textbook of physics that deals with current electricity.

Edit: "I am referring to voltage drop across the resistor, not current and not the voltage, V of the source." But the the voltage drop across the resistor when current flows is the voltage of the source. OK, so you connect the battery to the resistor through copper wires. These are excellent conductors - that's why they are used for wires - then, by Ohm's law already quoted, because their resistance is small there is negligible voltage drop across them so the voltage drop across the resistor is that of the battery.

 

[Googl]

OK. Once you have decided on the value of your variable resistor then treat the network as a potential divider and do the sums to find the volts across it. These sums, of course,involve (implied) current and volts - the current being the same through each and volts being shared between the two resistors in proportion to the R values. BUT, it is the combination of the two resistors that determines the voltage across them and not a simple rule involving just one resistor. As I said ten miles back, it's a matter of context; there is not a relationship between just any old resistor on its own and the voltage across it.

If you want to "understand" what happens, do some calculations for yourself and plot values for a range of variable resistance values. The graph you get will show you what happens; you don't get a straight line for V against R.

Thanks for this. This is what I was referring to.

it is the combination of the two resistors that determines the voltage across them and not a simple rule involving just one resistor

I am not so sure what you meant here. I have a variable resistor that is adjustable. The total resistance for this variable resistors is let's say 20KΩ.

Let's consider proportionality or ratio here. Never mind current it is just a constant value. It's just the ratio or proportionality for this situation.

Let's say I get a battery of 20V connected with a VARIABLE resistor with a sliding contact. The total resistance of this resistor is 20KΩ. I would like to imagine this ratio, for let's say where will I set the resistance to output the maximum voltage or what resistance value will I have to set to output a 10V

 

[Andrew Mason]

...
Let's say I get a battery of 20V connected with a VARIABLE resistor with a sliding contact. The total resistance of this resistor is 20KΩ. I would like to imagine this ratio, for let's say where will I set the resistance to output the maximum voltage or what resistance value will I have to set to output a 10V

If you place a 20 V battery across a 20KΩ resistor, a 10KΩ resistor or a 1KΩ resistor, you will always measure 20V across the resistor (assuming the battery has a comparatively low internal resistance).

I am having difficulty picturing the particular configuration you are suggesting. It would be helpful if you could provide a schematic drawing.

AM

 

[Dale]

Googl, as I said in my first response you need to use a constant current source, not a constant voltage source. If you use a constant voltage source then the voltage you measure will be that of the constant voltage source.

 

[Googl]

Okay. Thanks a lot.

I think now I have found what I have been looking for.

So am I correct by saying this or is my diagram of maximum voltage correct?

 

[Dale]

V1 = 9V regardless of R. You need a constant current source, not a constant voltage source.

 

[sophiecentaur]

Okay. Thanks a lot.

I think now I have found what I have been looking for.

So am I correct by saying this or is my diagram of maximum voltage correct?
View attachment 32045

What you have drawn is a dead short circuit of the + terminal to the - terminal. Did you just want to make smoke?

 

[Dale]

What you have drawn is a dead short circuit of the + terminal to the - terminal. Did you just want to make smoke?

V1 didn't look like a short to me. There was an arrow and a small gap rather than a node. That said it didn't look like a standard circuit diagram either.

 

[sophiecentaur]

You could be right but the position of the slider produces a very low resistance which will be 'caning' any real battery. But, as the diagram doesn't actually put a voltmeter anywhere, then we don't know where the volts are being measured. I think Googl is hanging on to something untenable here, rather than doing some 'real' Physics and applying some very elementary rules.
Arguments involving immoveable objects and irresistible forces seldom lead anywhere.

 

[Dale]

Googl, here is an example of the simplest circuit diagram where V is proportional to R.

 

[Googl]

I should have added this as well.

The voltmeter is where the Vs are (V1 and V2). (The current is constant, I think I have confirm this before) The black box/bar is a potential meter and is adjustable to obtain a desired voltage at V₁.

My goal is to observe how the maximum and minimum voltage would be set in the images (that might help).

By my understanding, when the knob is moved in centre the voltage drop measured would 1/2. This means the output would be 10V if the emf is 20V and set resistance 10KΩ (half of 20KΩ). At 1/4 the voltage would be 5V (resistance 5KΩ) and at 3/4, 6.75V (resistance 5KΩ)

I think my question has been very clear throughout, try re-reading my original post.

 

[sophiecentaur]

The problem is that your question has not been made clear. You talk of constant current yet draw a battery and mention emf. This just makes for confusion. If you want to show V=IR then just set up a circuit and measure under varying voltage or resistance. What more do you want?

 

[Googl]

I understand that. I just want to examine it in more complex situations.

 

[sophiecentaur]

Then calculate it.
Simples

 

[sophiecentaur]

btw, on your circuit, is the continuous line, joining the ground to the wiper supposed to be a solid wire os is there supposed to be a voltmeter (Circle with a V in it) inserted?

Do you not have access to any basic information on circuit theory? Wikkers is usually great as a start.

 

[Googl]

There is supposed to be a voltmeter. This is a better image.

 

[sophiecentaur]

So, after pages and pages of discussion, we have arrived at the Potential Divider (/potentiometer). Halleluya. Isn't this in every elementary textbook with all the sums laid out for you?

 

[Googl]

sophiecentaur,

Isn't a variable resistor also known as a Potential Divider? I remember using the keywords voltage divider/variable resistor/potentiometer somewhere. These are all similar terms, am I not correct?

Anyway to not change subject and over complicate things, how would you describe voltage across the resistor as the knob is adjusted along the potentiometer? This is a good analysis of the relationship/proportionality of voltage against resistance as opposed to having a fixed resistor. I have given my idea above somewhere would you say this is true?

Logically it would seem that as the knob is adjusted to the full 20kΩ of the potentiometer, the voltage measured across would be 0V but this isn't true. So with a 20V battery when the knob is moved to the full end of the potentiometer at 20KΩ the voltage measured would be 20V. In fact here we could use the current formula to find the current at this total, keep it constant then work out what is the voltage as the knob is moved across (from 20KΩ, 18KΩ, 16KΩ, 14KΩ, 12KΩ, 10KΩ, 8KΩ, etc). I hope I have been clear there. I understand this is not current which is what you might think I am assuming here.

The idea that at a 20KΩ resistor will provide a 20V when connected to a 20V battery is what is hard to grasp. This would take me back to my original post;

"Would you say voltage drop across a resistor is directly proportional to the resistance.

I am trying to understand the relationship between voltage and resistance."

I think I have been constant throughout with my question...

 

[dlgoff]

I hate the water analogy but let's give it a try

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/watcir2.html#c1"

just for the heck of it.

 

[Andrew Mason]

...In fact here we could use the current formula to find the current at this total, keep it constant then work out what is the voltage as the knob is moved across (from 20KΩ, 18KΩ, 16KΩ, 14KΩ, 12KΩ, 10KΩ, 8KΩ, etc). I hope I have been clear there. I understand this is not current which is what you might think I am assuming here.

Stop right there. A potentiometer has three terminals: one on one end of the resistor, one on the other end, and a sliding contact that slides along the resistor. If you do not connect a load between the sliding contact and one of the other contacts, the voltage measured from the sliding contact to one end will be proportional to the distance the sliding contact is from that end to the total length of the resistor.

The idea that at a 20KΩ resistor will provide a 20V

Stop right there. A resistor does not provide voltage. A battery provides voltage, which is energy per unit charge. A resistor uses energy. The energy it uses is the potential drop (change in potential energy/unit charge) x charge passing through the resistor. The rate of energy use (power consumption) is the potential drop x current (charge/unit time).

AM