How Does the Graph of 1/R Against E/V Relate to the Equation y = mx + c?

[Tangent100]

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 A-Level Physics.

Thank you for all the help you can give me :thumbs:

 

[phinds]

"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?

 

[dauto]

I don't see any graphs

 

[DrGreg]

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.

For the benefit of non-UK readers, A-Level courses are typically 2-year courses taken by 16-18 year-olds, just before university.

 

[sophiecentaur]

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

 

[Tangent100]

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

 

[phinds]

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

OK. Now go back and read post #2. What do you know?

 

[Tangent100]

OK. Now go back and read post #2. What do you know?

c is 1 so intercept is 1.
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.

 

[DrGreg]

So y = 8.6x + 1

OK, but your graph isn't y against x, it's E/V against 1/R. So you should be able to write down an equation which includes E, R, V, 8.6 and 1.

 

[nasu]

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.

 

[Tangent100]

OK, but your graph isn't y against x, it's E/V against 1/R. So you should be able to write down an equation which includes E, R, V, 8.6 and 1.

Okay so... E/V = 8.6 x 1/R + 1
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

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.

Where do I need to eliminate the current there is no current in my equation.
When I plug I = RV, I just get E = V + RVr... did you mean something else?

 

[Tangent100]

Could please someone help me? I really don't know what the gradient tells me nor nothing...

 

[SammyS]

Could please someone help me? I really don't know what the gradient tells me nor nothing...

Did you get this graph by fitting data that you obtained from measurements?

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.

 

[Tangent100]

Did you get this graph by fitting data that you obtained from measurements?

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.

What I am trying to do is analyse the graph, because the next section of test will be about analysing our measurements. Things like how does this graph relate to y = mx + c are common questions. So I am trying to quess what kind of questions could come up on the test, such as what does the gradient of this graph tells me. Besically pre predicting the questions they could ask me...

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

 

[SammyS]

What I am trying to do is analyse the graph, because the next section of test will be about analysing our measurements. Things like how does this graph relate to y = mx + c are common questions. So I'm trying to guess what kind of questions could come up on the test, such as what does the gradient of this graph tells me. Basically predicting the questions they could ask me...

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

What is r (lower case R) ?


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 ?

 

[Tangent100]

What is r (lower case R) ?


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 ?

Yes that's pretty much it. I had 3 resistors in series and I am guessing the internal resistance was negligible since I had nothing for that... I knew the resistance of all 3 resistors, but I was asked to get results for all possible values of R and so I tried singualr resistances, 2 resistance and all 3. I also got the emf before that to be 1.616.

 

[SammyS]

Yes that's pretty much it. I had 3 resistors in series and I am guessing the internal resistance was negligible since I had nothing for that... I knew the resistance of all 3 resistors, but I was asked to get results for all possible values of R and so I tried singualr resistances, 2 resistance and all 3. I also got the emf before that to be 1.616.

How is your EMF any different from your V ?

How did you determine each ?

 

[Tangent100]

How is your EMF any different from your V ?

How did you determine each ?

Emf was calculated without the resistors in circuit. Then the circuit switch was placed on and the current had to pass through resistors. The voltage was then measured with a volt meter attached in parallel, and so it decreased as more energy was lost through the resistors, depending on what resistors were in the circuit at that instance, so R1,R2,R3,R1+R2,R1+R3,... and so on.

 

[SammyS]

Emf was calculated without the resistors in circuit. Then the circuit switch was placed on and the current had to pass through resistors. The voltage was then measured with a volt meter attached in parallel, and so it decreased as more energy was lost through the resistors, depending on what resistors were in the circuit at that instance, so R1,R2,R3,R1+R2,R1+R3,... and so on.

OK.

Now it's making more sense.

I'll try to get back to this soon. In the meantime, someone else may jump in.


In your circuit, the battery itself is usually modeled as an ideal voltage source providing emf, E, in series with a resistor, having (usually small) resistance, r.

Thus the equation:

E = IR + Ir​

The R is whatever you used, R₁, R₂ , ... , R₂ + R₃, ...

 

[chihanghanh]

Yh i have my test tomorrow on the 7th May. AQA Physics Unit 03X task 3.

 

[SammyS]

Take $\ E=IR+Ir \$ and divide by $\ IR\ .$

 

[Tangent100]

Take $\ E=IR+Ir \$ and divide by $\ IR\ .$

Oh my... that would mean the gradient is internal resistance, could someone check if what I done here is correct?

 

[nasu]

Yes, the slope of the curve is the internal resistance.

 

[SammyS]

Oh my... that would mean the gradient is internal resistance, could someone check if what I done here is correct?

[ Img]http://s24.postimg.org/ro4owqox1/2014_05_07_11_15_16.png[/PLAIN]

Yes, that's it.It strikes me as kind of a strange way to combine two (or is it three?) very common quantities.

Substituting in other orders can make it very difficult to see this result.

 

[peterspencers]

No. I don't believe you can deduce the internal resistance from that gradient.

gradient = EMF/V by 1/R

(IR + Ir) / IR = 1 + r/R

(1 + r/R) x R = R+r

given the gradient is a constant, however the sum of the two resisitances (R+r) the total resistance of the circuit is not constant. I don't see how this can be accurate?

 

[sophiecentaur]

No. I don't believe you can deduce the internal resistance from that gradient.

gradient = EMF/V by 1/R

(IR + Ir) / IR = 1 + r/R

(1 + r/R) x R = R+r

given the gradient is a constant, however the sum of the two resisitances (R+r) the total resistance of the circuit is not constant. I don't see how this can be accurate?

What would be changing?
Given that 'internal resistance' is not Ohmic over the whole range of currents, it is considered to be, in most situations. When it isn't, you don't get a straight line graph.

 

[peterspencers]

your right I am completely wrong, I took the same exam today, luckily I realized before I answered the question.

 

[peterspencers]

I couldn't see that all you needed to do was factor out the 1/R as x