# Help resistance graphWatch

Announcements
#1
..
0
4 years ago
#2
The gradient is change in y over change in x, or in your case change in I over change in V, so...

Posted from TSR Mobile
0
4 years ago
#3
(Original post by iamspiderman)
I have plotted a I/V graph .. so current and voltage; and it's not a straight line..
I need to use the graph to determine the resistance of the resistor.. How do I do this?
I've attached the graph ( it's a rough drawing, but I just want you to see the shape of the line)
Draw the best straight line you can through the points.
That's the best I can suggest without the background to what you have done and where the data has come from.
0
#4
..
0
4 years ago
#5
(Original post by iamspiderman)
I have uploaded the question for you to see and I have redone the graph
The graph needs a (0,0) origin.
You have not started the vertical axis at 0.

When you've changed this, draw the best straight line you can.
Also I'm not sure your 1st point is plotted correctly but it's very difficult to read the graph.
0
#6
..
0
4 years ago
#7
(Original post by iamspiderman)
Thanks for replying I have redone the graph and put 0 on both axis.. I've attached this
Yes but you are still not plotting the graph correctly.
The 0,0 point should be in the corner. The two zeros should coincide and be the 1st points on each axis.
Your vertical axis is then not correct. The point you have called 0 is actually 0.030 and where the origin should be is 0.020
0
#8
..
0
4 years ago
#9
(Original post by iamspiderman)
I have done te graph correctly now
I just need to know how to work out the resistance of the resistor
So you have a graph of I vs V
You know that R=V/I. If you make this V=IR then this is the equation of a straight line with y axis V, x-axis I and gradient R. This is what you've plotted, thus the resistance is given by the gradient of the line.

Edit: Hang on you've drawn the wrong graph it should by V vs I
0
#10
..
0
4 years ago
#11
(Original post by iamspiderman)
Thanks for replying. I've done a new graph with y axis as voltage and x axis as current
I forgot to mention that they give this value :
47omhs +- 10% (the plus minus are on top of each other, I don't know how to do that sign in here)
this is supposed to be the reistance of the carbon resistor given by the manufactor
Ok so find the gradient of the line to give the resistance and then compare the value you get with the manufacturer's value. If it lies within the +-10% range then the rating given by the manufacturer is accurate
0
X

new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

Thu, 24 Oct '19
• Cardiff University
Sat, 26 Oct '19
• Brunel University London
Sat, 26 Oct '19

### Poll

Join the discussion

Yes (65)
23.13%
No (216)
76.87%