relationship between V, I and R in graphical terms

[brownt]

I can't make sense of this graph


I understand the in written terms as in proportionality and the inverse. But I can't see how the graph relates. Can someone explain it please.

[agehall]

U=R*I

If U is constant, you get something like above. If R is high, I needs to be lower and vice versa.

[IanB]

The graph is incorrect in showing a linear relationship. Therefore it cannot be explained.

[Nitrousoxide]

Usually, you will have Voltage on the x-axis (dependent variable) and current on the Y axis (independent variable) and R is the "controlled variable". But this does depend on device/experimental setup

So you will have:
V = I*R
R = V/I (can be seen as dV/dI or gradient).

Alternatively you can switch the axies, that case the gradient will be 1/R.

[brownt]

I see. well that's good.

What about these graphs then. I think the first one makes sense, but the other one does not?

[agehall]

I realise my previous post was a bit unclear - the graph is not showing U, which is what is somewhat confusing. It is showing R as a function of I. So for every point on the x-axis, multiplying it with the value of the y-axis should produce a constant product.

Edit: To me, both of the last two graphs makes sense. Those are much easier to understand as they plot U as a function of R and I...

[firewalker]

Edit: The diagrams are correct it shows what you should do in order to maintain a constant of something.


The diagram is wrong. It should be a hyperbolic line. The only correct aspect of it is that it shows that when the resistor of a circuit goes down, the current will increase.

Why don't you try it your self? Make a diagram by experiment.

It will ake about 10 1 kOhms, a voltage source (a 9 volts battery will do) and a multimeter.

Alexander.

[agehall]

The diagram is wrong. It should be a hyperbolic line.

One slight problem - there are no units on the diagram, so it is very hard to say that it is wrong.

The way I look at it is that the creator of the diagram is illustrating the point that as current rises, resistance has to go down to maintain a constant voltage. The exact shape is irrelevant.

This could have been shown in much easier ways though...

[vk6zgo]

I can't make sense of this graph

I understand the in written terms as in proportionality and the inverse. But I can't see how the graph relates. Can someone explain it please.

It isn't  a very well thought out graph---- although correct, it is a bit anti-intuitive.

The more usual way to show this is with R as the independent variable.
In that case, R would increase from left to right, & I would fall from high values for the lowest values of R, to low values for the highest values of R.

[firewalker]

The diagram is wrong. It should be a hyperbolic line.

One slight problem - there are no units on the diagram, so it is very hard to say that it is wrong.

The way I look at it is that the creator of the diagram is illustrating the point that as current rises, resistance has to go down to maintain a constant voltage. The exact shape is irrelevant.

This could have been shown in much easier ways though...

You are correct. I misread the text of the diagram. It shows what the relation between two quantities should be in order to have a constant something.

[tautech]

[brownt]

I think i have it then. are the comments correct on the graphs attached.

Basically the centre one is upside down, and all in all its a weird way to represent ohms law.

or are they all correct, but its just a weird way to explain it

[agehall]

Your comment on the second graph is wrong.

Just do the math - U=I*R. Let's set U to 10 and see what happens:
With R=1 => I=10
With R=5 => I=2
With R=10 => I=1

Not exactly linear but since there are no units on the graph, you can assume it just shows you the general idea - as resistance increases, current has to go down to maintain a constant voltage.

[hamster_nz]

If I recall the name correctly the graph of R vs current should be a 'hyperbola', not a line.

Any small section of it will look a linear relationship, but  R approaches infinity when current is 0, and current is 0 when R approaches infinity.

[Shock]

Homework completed good job team.

[vk6zgo]

If I recall the name correctly the graph of R vs current should be a 'hyperbola', not a line.

Any small section of it will look a linear relationship, but R approaches infinity when current is 0, and current is 0 when R approaches infinity.

In most practical situations, it is near as dammit linear.
The graph is correct, as it did not specify "for all possible values of I or R".

[Rick Law]

I can't make sense of this graph

I understand the in written terms as in proportionality and the inverse. But I can't see how the graph relates. Can someone explain it please.

Homework completed good job team.

Mr./Ms. BrownT,

If I may... allow me to make a suggestion.  Please don't take offense...

If this is indeed homework for a school course, you may want to make time to review some math and algebra.  There are two ways to get credit for a course, one is to (#1) "just get by and put the credits in the bag", or (#2) "besides the credits in the bag, fill the brain with corresponding knowledge."

The algebra knowledge to comprehend a linear graph is foundational for mathematics.  Many other things you will need will be build on top of the foundation in use here.  In almost any field of science and in many "humanity fields" including even MBA, you will run across the need for such foundational knowledge again and again.  If this foundation is weak, it will make whatever built on top of this foundation weak (and thus more difficult).

So, do try to review the math and the algebra getting to and beyond this point.  Otherwise, it will hinder your future progress again and again.

Please understand, my directness here is done in the hope that you will be more successful.  So please don't take offense.

Rick

[hamster_nz]

If I recall the name correctly the graph of R vs current should be a 'hyperbola', not a line.

Any small section of it will look a linear relationship, but R approaches infinity when current is 0, and current is 0 when R approaches infinity.

In most practical situations, it is near as dammit linear.
The graph is correct, as it did not specify "for all possible values of I or R".

I'll just have to disagree, with an example.

I would say that 10V and resistances in the 1000 and 3000 ohms range is a practical situation - for example, maybe you are trying to light an LED from a 12V battery.

At 10V, the current through a 1000 ohm resistor is 10mA.

Change that for a 2000 ohm resistor and the current is 5mA.

The linear model gives the following equation:
   current = 15mA - R * 0.005

So if the relationship is linear (and we all know that it isn't), current through a 3000 ohm resistor should result in 0mA of current.

And it isn't only extrapolation that fails - a linear model predicts current of 7.5mA with a 1500 ohm resistor - a value that is off by 12.5%.

[vk6zgo]

If I recall the name correctly the graph of R vs current should be a 'hyperbola', not a line.

Any small section of it will look a linear relationship, but R approaches infinity when current is 0, and current is 0 when R approaches infinity.

In most practical situations, it is near as dammit linear.
The graph is correct, as it did not specify "for all possible values of I or R".

I'll just have to disagree, with an example.

I would say that 10V and resistances in the 1000 and 3000 ohms range is a practical situation - for example, maybe you are trying to light an LED from a 12V battery.

At 10V, the current through a 1000 ohm resistor is 10mA.

Change that for a 2000 ohm resistor and the current is 5mA.

The linear model gives the following equation:
   current = 15mA - R * 0.005

So if the relationship is linear (and we all know that it isn't), current through a 3000 ohm resistor should result in 0mA of current.

And it isn't only extrapolation that fails - a linear model predicts current of 7.5mA with a 1500 ohm resistor - a value that is off by 12.5%.

Can you run that past me again?

Where does the 15mA come from?

Ohm's Law is a simple  proportion .

In its most familiar form ,  we have. I=V/R.

To obviate any problems with  battery internal resistance, let's postulate an ideal one volt battery, with an
ideal one Ohm resistor connected across it between its output terminal.

With that resistor kept constant:-
For V=1, I =V/R  gives I=1/1 =1amp
Now let us increase V in one volt increments:-
For V=2, I=2/1= 2amp

For V=3, I=3/1=3amp

For V=4, I=4/1=4amp

For V=5, I=5/1=5amp

For V=6, I=6/1=6amp

For V=7, I=7/1=7amp

For V=8, I=8/1=8amp

For V=9, I=9/1=9amp

For V=10,I=10/1=10amp.

Graphing current (I)versus voltage (V) for a fixed R gives a straight line------no logarithms, exponentials, or anything else.

[AG6QR]

If I recall the name correctly the graph of R vs current should be a 'hyperbola', not a line.

Any small section of it will look a linear relationship, but R approaches infinity when current is 0, and current is 0 when R approaches infinity.

In most practical situations, it is near as dammit linear.
The graph is correct, as it did not specify "for all possible values of I or R".

I'll just have to disagree, with an example.

I would say that 10V and resistances in the 1000 and 3000 ohms range is a practical situation - for example, maybe you are trying to light an LED from a 12V battery.

At 10V, the current through a 1000 ohm resistor is 10mA.

Change that for a 2000 ohm resistor and the current is 5mA.

The linear model gives the following equation:
   current = 15mA - R * 0.005

So if the relationship is linear (and we all know that it isn't), current through a 3000 ohm resistor should result in 0mA of current.

And it isn't only extrapolation that fails - a linear model predicts current of 7.5mA with a 1500 ohm resistor - a value that is off by 12.5%.

Can you run that past me again?

Where does the 15mA come from?

Ohm's Law is a simple  proportion .

In its most familiar form ,  we have. I=V/R.

To obviate any problems with  battery internal resistance, let's postulate an ideal one volt battery, with an
ideal one Ohm resistor connected across it between its output terminal.

With that resistor kept constant:-
For V=1, I =V/R  gives I=1/1 =1amp
Now let us increase V in one volt increments:-
For V=2, I=2/1= 2amp

For V=3, I=3/1=3amp

For V=4, I=4/1=4amp

For V=5, I=5/1=5amp

For V=6, I=6/1=6amp

For V=7, I=7/1=7amp

For V=8, I=8/1=8amp

For V=9, I=9/1=9amp

For V=10,I=10/1=10amp.

Graphing current (I)versus voltage (V) for a fixed R gives a straight line------no logarithms, exponentials, or anything else.

But that's not the graph you were talking about.

Look at the very first post in this thread.  It is a graph with current on the horizontal, and resistance on vertical axis, and claiming that the line represents a line of constant voltage.  Remember, Voltage is what's being held constant.

V = I * R

The graph text says that V is a constant.  Just arbitrarily, pick V to be 12V.

When R is 12 ohms, I is 1 amp.
When R is 6 ohms, I is 2 amps
When R is 3 ohms, I is 4 amps
When R is 4 ohms, I is 3 amps
When R is 2 ohms, I is 6 amps
When R is 1 ohm, I is 12 amps.

Draw those points on a graph.  There is no way you get anything like a straight line. 

The last two graphs, showing V vs R for constant I, and V vs I for constant R, are correct, as those are linear relationships.

V = I * R

There is a direct, linear relationship between V and I, (with constant of proportionality R) and a direct, linear relationship between V and R (with constant of proportionality I).  But the relationship between I and R is an inverse relationship, which will not produce a straight line graph.

[ebastler]

Geez, guys. I realize that this is the Beginners board, but how long can one debate over a trivial 1/x relationship?!

[bson]

I think it's better as an introductory teaching aid to plot current as a function of resistance, keeping voltage constant.  Then current as a function of voltage, keeping resistance constant.  Skip the variable currents; it's better to learn Kirchoff's laws first, circuit analysis, and basic semiconductor models (since these are current devices) before tackling current as anything other than a byproduct that just needs to be kept reasonable.  Semiconductors in particular are often greatly simplified through a current-centric view.