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Graph of 1/R against E/V 
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#1
May214, 08:21 AM

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Can someone explain to me the concept of this graph and how it works with relation to y = mx + c ?
Usually the gradient is something simple but this time I have no idea... Also, the c intercept, what is the c intercept? As far as I am concerned it is 1 but what does the fact it is 1 mean,... surely it is emf / emf so can I just say it is emf or what do i say. I know I am asking about the answers, but hints are also welcome and some analysis/ evaluation in regard to this graph? I am doing ALevel Physics. Thank you for all the help you can give me 


#2
May214, 10:04 AM

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"This graph" is not a helpful phrase. It would be useful it you had posted the graph you are talking about.
I'm a bit puzzled but perhaps it's because you are, I think, in Great Britain and I don't know the school system there. I would take "A level Physics" to mean at least the equivalent of high school physics here in the US and I don't think you can get that far without understanding the graph of a line. It would have been helpful had you said a bit more about where you are, but again, I guess "A Level Physics" is well understood where you are as as indication of your level in school. ANYWAY ... for the equation y = mx + c When you set x to zero, what's the value of Y? How does that relate to where the graph intercepts the Y axis? if you go over one unit in the X direction, how many units have you gone UP in the Y direction? How do those two numbers relate to the slope (what you are calling the gradient) of the line? 


#3
May214, 12:41 PM

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I don't see any graphs



#4
May214, 02:24 PM

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Graph of 1/R against E/V



#5
May214, 05:25 PM

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We definitely need to see "this graph" before we can go any further.



#6
May514, 09:52 AM

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Is what I plotted in class but I don't have the original one so this is from memory. 


#8
May514, 11:48 AM

P: 10

Gradient is 8.6. (0.30/0.035) So y = 8.6x + 1 But what does that tell me in general, what is this equation for when it comes to the physics side of it. Like e.g. V = E  Ir and so V = rI + E if we look at y = mx + c of a E against I graph. What I am trying to find out is the equation that my values could possibly relate to such that I know what the gradient tell me (in the example above it tell me r), what y tells me (V in example above) and that c tells me (Emf in the example above). So that I can predict what section 3 of my test (this is section 2) could possibly be about. Gradient is E/V over 1/R but what does that tell me... Like in a velocity time graph the gradient would be acceleration. 


#9
May514, 12:11 PM

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#10
May514, 12:37 PM

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You can write Ohm's law in a form that fits the variables on the graph (E/V and 1/R).
You just need to eliminate the current (I) and rearrange a little. Can you write I in terms of V and R? Then plug into E=V+Ir. 


#11
May514, 04:51 PM

P: 10

I am not very sure, but if I look at this and E = IR + Ir, does this mean, because there is negligible internal resistance, that 1 = Ir? But then what about 8.6 x 1/R... How does that become IR and how does E/V become just E? I tried this: E/V = 8.6 x 1/R + 1 E/V = 8.6/R +1 E/V = 8.6/(I/V) +1 'Because R = I/V E/V = 8.6I/V + 1 The Vs then cancel out if I times everything by V but the +1 just ruins everything because it means that the equation becomes E = 8.6I + 1/V So I'm still stuck... If I had to take a quess I'd say 8.6 is the current but that's just guessing, I don't understand anything When I plug I = RV, I just get E = V + RVr... did you mean something else? 


#12
May614, 10:16 AM

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Could please someone help me? I really dont know what the gradient tells me nor nothing...



#13
May614, 10:28 AM

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If so, which quantities did you measure/control  and how did you accomplish that? Which quantities are you trying to determine? From the replies of DrGreg & phinds, and from the expression you're somewhat trying to match (## E = IR + Ir \ ## or one of the others) we can sort of guess what you're trying to do. Please give more details if you want more help. 


#14
May614, 11:23 AM

P: 10

This graph was plotted from results Ive obtained in an experiment. Emf was controlled, ive been changing the resistance and calculating the pd on the volt meter and so R is the independent and V was my dependent variable. 


#15
May614, 02:03 PM

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That is somewhat better a description, but still some guessing remains. Correct me if I'm wrong: You had an ideal voltage source with a very reliable output of E volts.  This is the EMF. You had some small resistance, r, in series with that. You measured (not calculated) the voltage, V, across that series combination. Right ? You then had some variable resistance, or a set of known resistances which were in series with the previously described combination. This resistance you called R. Right? You may have calculated E/V . Graphing E/V versus 1/R gave you a linear plot of the form y = mx + b, Where y is E/V and x is 1/R . So, I suppose you are trying to determine how m and b are related to the quantities in the circuit. Is that the issue here ? 


#16
May614, 03:19 PM

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#17
May614, 03:36 PM

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How did you determine each ? 


#18
May614, 04:33 PM

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