Electronics > Beginners

relationship between V, I and R in graphical terms

<< < (4/5) > >>

vk6zgo:

--- Quote from: hamster_nz on July 12, 2018, 08:10:19 am ---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.

--- End quote ---
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:

--- Quote from: brownt on July 12, 2018, 06:19:16 am ---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.

--- End quote ---


--- Quote from: Shock on July 12, 2018, 09:14:18 am ---Homework completed good job team.

--- End quote ---

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:

--- Quote from: vk6zgo on July 12, 2018, 09:24:10 am ---
--- Quote from: hamster_nz on July 12, 2018, 08:10:19 am ---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.

--- End quote ---
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".

--- End quote ---
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:

--- Quote from: hamster_nz on July 12, 2018, 08:59:37 pm ---
--- Quote from: vk6zgo on July 12, 2018, 09:24:10 am ---
--- Quote from: hamster_nz on July 12, 2018, 08:10:19 am ---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.

--- End quote ---
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".

--- End quote ---
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%.

--- End quote ---

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:

--- Quote from: vk6zgo on July 14, 2018, 04:37:25 am ---
--- Quote from: hamster_nz on July 12, 2018, 08:59:37 pm ---
--- Quote from: vk6zgo on July 12, 2018, 09:24:10 am ---
--- Quote from: hamster_nz on July 12, 2018, 08:10:19 am ---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.

--- End quote ---
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".

--- End quote ---
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%.

--- End quote ---

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.


--- End quote ---

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.

Navigation

[0] Message Index

[#] Next page

[*] Previous page

There was an error while thanking
Thanking...
Go to full version
Powered by SMFPacks Advanced Attachments Uploader Mod