Diagram about potential difference over current

[Krashy]

Homework Statement

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Homework Equations

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I = G * U

The Attempt at a Solution

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Hello, i know the solution is 5.0 A/V but i don't know if my way of solving this was correct. I rearranged it to G= I/U and solved this equation for all the values in the table and then added them together, so:

1.0/1.0 = 1
1.7/2.0 = 0.85
2.7/3.0 = 0.9
3.7/4.0 = 0.925
4.4/5.0 = 0.88

1 + 0.85 + 0.9 + 0.925 + 0.88 = 4.555 = 5.0 (2 s.f)

I thought this is directly proportional so y/x will always be the same number but its not, so I am kinda confused and i don't think i did this the right way. Also i don't even know if my graph is accurate, because i drew it again on a seperate, big piece of paper and its not really a straight graph like in the picture.
I hope i described my problem well enough, thanks for every answer.

 

[Decimal]

If you take a look at the equation they give you in the question, can you see how you might use that to find $G$ from the graph? What does $G$ represent in the relation between $I$ and $U$?

 

[Krashy]

I thought, because G = I / U, I need to solve this equation for every value in the table and take the average but this answer wouldn't come near 5.0.
Also the text says "If the conductance G of the resistor is constant, the following should apply..." but if i take I / U the solutions are not constant, so I am kinda confused.

 

[Decimal]

Are you sure the solution is supposed to be 5.0? This doesn't seem to be consistent with the measured data.

Also the question asks you to use the graph in determining $G$. What does $G$ represent in the graph you drew?

 

[gneill]

I thought this is directly proportional so y/x will always be the same number but its not, so I am kinda confused and i don't think i did this the right way. Also i don't even know if my graph is accurate, because i drew it again on a seperate, big piece of paper and its not really a straight graph like in the picture.

Hi Krashy. Why do you suppose they asked you to plot the graph?

Suppose that the "measured" values had some small errors associated with the measuring process. You expect, from the theory, that the points will form a perfectly straight line on the graph, but due to those measurement errors (or possibly other factors), they aren't perfectly aligned. What do you you do in such a case? You draw a "best fit" straight line by eye. That's a surprisingly powerful "mathematical tool", where you don't use equations but your best judgement to "fit" the line to the points. It's a sort of visual best curve fit to the data.

Now, once you have drawn your best fit straight line, what property of that line corresponds to the quantity that you seek?

 

[Krashy]

Yeah the solution says 5.0 A/V but sometimes the solutions are incorrect so i´m not 100% sure its right.
I think G represents the gradient in the graph, so to solve it i need to calculate delta y/ delta x, is that correct?

Ah all right i thought it seems kind of odd that its not a straight line, thanks

 

[Decimal]

so to solve it i need to calculate delta y/ delta x, is that correct?

Yes, the graph you drew is almost linear, but as Gneill stated, not perfectly linear. Our expectation is that the relation will behave according to the equation, so we compensate for any measurement errors by making a best "fit" to the data. Since the equation represents a straightforward linear relationship, the gradient of the graph will then be equal to $G$.

 

[Krashy]

So the solution is 4.4 - 0/5.0 - 0 = 0.88, is that right?

 

[Decimal]

So the solution is 4.4 - 0/5.0 - 0 = 0.88, is that right?

Yes

 

[Krashy]

all right thanks