relationship between V, I and R in graphical terms - Page 1
Forum section: electronics / beginners

author: brownt
author_rank: Regular Contributor
timestamp: July 12, 2018, 06:19:16 am
content: I can't make sense of this graphI 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.

author: agehall
author_rank: Frequent Contributor
timestamp: July 12, 2018, 06:27:09 am
content: U=R*IIf U is constant, you get something like above. If R is high, I needs to be lower and vice versa.

author: IanB
author_rank: Super Contributor
timestamp: July 12, 2018, 06:28:41 am
content: The graph is incorrect in showing a linear relationship. Therefore it cannot be explained.

author: Nitrousoxide
author_rank: Regular Contributor
timestamp: July 12, 2018, 06:31:37 am
content: 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 setupSo you will have:V = I*RR = V/I (can be seen as dV/dI or gradient).Alternatively you can switch the axies, that case the gradient will be 1/R.

author: brownt
author_rank: Regular Contributor
timestamp: July 12, 2018, 06:36:42 am
content: I see. well that's good.What about these graphs then. I think the first one makes sense, but the other one does not?

author: agehall
author_rank: Frequent Contributor
timestamp: July 12, 2018, 06:37:01 am
content: 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...

author: firewalker
author_rank: Super Contributor
timestamp: July 12, 2018, 06:37:08 am
content: 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.

author: agehall
author_rank: Frequent Contributor
timestamp: July 12, 2018, 06:41:00 am
content: Quote from: firewalker on July 12, 2018, 06:37:08 amOne 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...

author: vk6zgo
author_rank: Super Contributor
timestamp: July 12, 2018, 06:41:47 am
content: Quote from: brownt on July 12, 2018, 06:19:16 amIt isn't a very well thought out graph---- although correct, it is a bit anti-intuitive.The more usual way to show this is withR 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.

author: firewalker
author_rank: Super Contributor
timestamp: July 12, 2018, 06:45:06 am
content: Quote from: agehall on July 12, 2018, 06:41:00 amYou 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.

author: brownt
author_rank: Regular Contributor
timestamp: July 12, 2018, 07:17:48 am
content: 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

author: agehall
author_rank: Frequent Contributor
timestamp: July 12, 2018, 07:22:02 am
content: 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=10With R=5 => I=2With R=10 => I=1Not 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.

author: hamster_nz
author_rank: Super Contributor
timestamp: July 12, 2018, 08:10:19 am
content: 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.

author: Shock
author_rank: Super Contributor
timestamp: July 12, 2018, 09:14:18 am
content: Homework completed good job team.

author: vk6zgo
author_rank: Super Contributor
timestamp: July 12, 2018, 09:24:10 am
content: Quote from: hamster_nz on July 12, 2018, 08:10:19 amIn 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".

author: Rick Law
author_rank: Super Contributor
timestamp: July 12, 2018, 07:47:24 pm
content: Quote from: brownt on July 12, 2018, 06:19:16 amQuote from: Shock on July 12, 2018, 09:14:18 amMr./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

author: hamster_nz
author_rank: Super Contributor
timestamp: July 12, 2018, 08:59:37 pm
content: Quote from: vk6zgo on July 12, 2018, 09:24:10 amI'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 * 0.005So 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%.

author: vk6zgo
author_rank: Super Contributor
timestamp: July 14, 2018, 04:37:25 am
content: Quote from: hamster_nz on July 12, 2018, 08:59:37 pmCan 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 anideal 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 =1ampNow let us increase V in one volt increments:-For V=2, I=2/1= 2ampFor V=3, I=3/1=3ampFor V=4, I=4/1=4ampFor V=5, I=5/1=5ampFor V=6, I=6/1=6ampFor V=7, I=7/1=7ampFor V=8, I=8/1=8ampFor V=9, I=9/1=9ampFor 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.

author: AG6QR
author_rank: Frequent Contributor
timestamp: July 14, 2018, 05:11:53 am
content: Quote from: vk6zgo on July 14, 2018, 04:37:25 amBut 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 constantvoltage. Remember, Voltage is what's being held constant.V = I * RThe 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 ampsWhen R is 3 ohms, I is 4 ampsWhen R is 4 ohms, I is 3 ampsWhen R is 2 ohms, I is 6 ampsWhen 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 * RThere 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.

author: ebastler
author_rank: Super Contributor
timestamp: July 14, 2018, 05:59:35 am
content: Geez, guys. I realize that this is the Beginners board, but how long can one debate over a trivial 1/x relationship?!

author: bson
author_rank: Unknown
timestamp: July 14, 2018, 09:19:01 pm
content: 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.