Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

cant make sense of current, voltage, resistance relationship, graph

Status
Not open for further replies.

brownt

Member level 3
Joined
Sep 25, 2017
Messages
63
Helped
0
Reputation
0
Reaction score
0
Trophy points
6
Activity points
597
I can't make sense of this graph

Untitled.png


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

Hi,

You're not the only one. I feel like I'm walking into a trick question...

It looks like the relationship has been drawn upside down. I've yet to see a circuit where less resistance equals less current...
 

well thanks. It's not a trick. I can't make sense of it, and I guess it is wrong. But, what do i know? That is why I am asking. Is the graph upside down, or is it not possible to represent what its trying to represent with a single line?

- - - Updated - - -

what about these graphs. I think the first one makes sense, but not the second. correct?graphs.png
 

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

Hi. This looks pretty normal to me. If you take the voltage constant, say a constant voltage source with a resistor attached to ground then the current going through the resistor decreases as the resistance increases. Maybe we're used to seeing the current in the y axis but if you look closely, it does follow the V=IR relationship. if current increases (going left) resistance decreases (going down). But to control the current we need to adjust resistance that is why it is better to put current in the y-axis. the same applies to the other graphs.
 

Hi,

Post#1
I agree the graph doesn't make much sense.

There is Ohm's Law. R = V / I
And it says "constant voltage"

Let's calculate some values:
Let's assume V = 12V
Then: if
I = 1A --> R = 12 Ohms
I = 2A --> R = 6 Ohms
I = 4A --> R = 3 Ohms
I = 6A --> R = 2 Ohms
I = 12A --> R = 1 Ohms
You see: the higher the current, the lower the resistance.
This is what the graph shows.
Mind: the graph has no X values and no Y values. In most cases the axis are linear --> but there is no need for this.
Thus one may expect that the graph shows a hyperbola --> but there is no need for this. With non linear axis the graph may look like a straight line.

Post#3:
Both graphs show the correct relationship:
Constant R: the higher the current, the higher the voltage
Constant current: The higher R, the giher V.

Both are directly proportional. In both casess: with both linear axis you will get a straight line graph.

I recommend to do some example calculations and graphs with Excel to get a feel.

Klaus
 
  • Like
Reactions: d123

    d123

    Points: 2
    Helpful Answer Positive Rating
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.

graph2.png
 

Hi,

You are correct with figure 1.21 and figure 1.23.

But you are not correct with figure 1.22.
See my example of post #5.

Mathematical solution:
V = constant, thus V1 = V2
V1 = R1 x I1
V2 = R2 x I2
because V1 = V2 we can say:
R1 x I1 = R2 x I2

Now we increase R2. Lets say R2 = 2 x R1
Then we get
R1 x I1 = 2 x R1 x I2 ; we divide both sides by R1
--> I1 = 2 x I2
I2 = 1/2 x I1
we see that I2 is decreased.
This is named "indirect proportionality"
The higher R the lower I

Klaus

Added:
Chart with non linear "R" axis.
RVI.png
 
  • Like
Reactions: d123

    d123

    Points: 2
    Helpful Answer Positive Rating
Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top