Internal resistance - Edexcel a-level physics practical

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Bigflakes
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So basically I recently did the required practical and something went wrong. I tried plotting voltage against current and then using the gradient as it would be -r but I got -0.0513 which seems too small. I had the results as voltage in V and current in mA. Any help?

Thanks
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learningizk00l
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(Original post by Bigflakes)
So basically I recently did the required practical and something went wrong. I tried plotting voltage against current and then using the gradient as it would be -r but I got -0.0513 which seems too small. I had the results as voltage in V and current in mA. Any help?

Thanks
This is internal resistance, you would expect it to be low
We don't want to be losing lots of energy to wires
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Bigflakes
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(Original post by learningizk00l)
This is internal resistance, you would expect it to be low
We don't want to be losing lots of energy to wires
But I looked it up on google and it said that it should be somewhere between 0.1 and 0.9. Also should it be volts against amps or is it fine that it is volts against milliamps?
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Callicious
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(Original post by Bigflakes)
But I looked it up on google and it said that it should be somewhere between 0.1 and 0.9. Also should it be volts against amps or is it fine that it is volts against milliamps?
As long as your units are consistent then you're fine (though if you are calculating the gradient, you will have to be diligent about this, since the gradient of a V/I graph with I in mA is given by volts per milliamp, for example.)
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Bigflakes
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(Original post by Callicious)
As long as your units are consistent then you're fine (though if you are calculating the gradient, you will have to be diligent about this, since the gradient of a V/I graph with I in mA is given by volts per milliamp, for example.)
So what should I do? If I use volts per amps I get a gradient of -51 which I know is too big.
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Callicious
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(Original post by Bigflakes)
So what should I do? If I use volts per amps I get a gradient of -51 which I know is too big.
I have no clue about if your readings/etc are correct, nor do I know what the resistances involved are. All I can say is that if you know that the y-axis is say, 2 volts per centimetre, and the x-axis is say, 10 milliamps per centimetre, and you have a gradient of dy/dx = 2cm/2cm = 1, then your gradient (in units) is given by 2 volts per 10 milliamps. Just take your dy in cm and multiply it by the unit sensitivity of the y-axis, dx by the unit sensitivity of the x-axis, then divide them.

If your answer is -51, then that is your answer. If your answer is -1000000, that is your answer. As long as the method is correct, there's nothing else that can be said (unless your data is faulty and/or your graphing is incorrect to begin with.)

I would really recommend using a spreadsheeting tool like Excel or LibreOffice Calc to do this sort of thing- we did that most of A-Level and it worked fine.
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Bigflakes
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(Original post by Callicious)
I have no clue about if your readings/etc are correct, nor do I know what the resistances involved are. All I can say is that if you know that the y-axis is say, 2 volts per centimetre, and the x-axis is say, 10 milliamps per centimetre, and you have a gradient of dy/dx = 2cm/2cm = 1, then your gradient (in units) is given by 2 volts per 10 milliamps. Just take your dy in cm and multiply it by the unit sensitivity of the y-axis, dx by the unit sensitivity of the x-axis, then divide them.

If your answer is -51, then that is your answer. If your answer is -1000000, that is your answer. As long as the method is correct, there's nothing else that can be said (unless your data is faulty and/or your graphing is incorrect to begin with.)

I would really recommend using a spreadsheeting tool like Excel or LibreOffice Calc to do this sort of thing- we did that most of A-Level and it worked fine.
Yes I’m using excel to graph, I guess that the results are faulty, should I just plot the graph in volts against amps and state that during the experiment we might have had some faulty equipment? (Those were my results) Name:  E8930054-D883-4438-BDDC-6C00FF011A37.jpg.jpeg
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Callicious
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(Original post by Bigflakes)
Yes I’m using excel to graph, I guess that the results are faulty, should I just plot the graph in volts against amps and state that during the experiment we might have had some faulty equipment? (Those were my results) Name:  E8930054-D883-4438-BDDC-6C00FF011A37.jpg.jpeg
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I don't recommend artificially fudging your results if you've already got them (changing mA to A when you know it's mA for example.)

Just do the working as you normally would and discuss it as you normally would. If your results aren't as expected, who cares? You can use that in discussion tbf.
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Bigflakes
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(Original post by Callicious)
I don't recommend artificially fudging your results if you've already got them (changing mA to A when you know it's mA for example.)

Just do the working as you normally would and discuss it as you normally would. If your results aren't as expected, who cares? You can use that in discussion tbf.
When I said plotting volts against amps I meant taking my results in mA and divide them by 1000 to get A and then plotting that, because I was sceptical whether plotting V against mA would give me the gradient in ohms or milli-ohms.
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Callicious
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(Original post by Bigflakes)
When I said plotting volts against amps I meant taking my results in mA and divide them by 1000 to get A and then plotting that, because I was sceptical whether plotting V against mA would give me the gradient in ohms or milli-ohms.


\begin{align*}

V[\textrm{V}] = I[\textrm{A}]R[\Omega] \\ 

\displaystyle\frac{V}{I}\frac{[\textrm{V}]}{[\textrm{A}]} = \displaystyle{R}[\Omega] \\ 

\displaystyle\frac{V}{I}\frac{[\textrm{V}]}{[\textrm{mA}]} = \displaystyle{R}[\textrm{k}\Omega] 

\end{align*}

^ Dimensionally speaking what you're dealing with. Four volts divided by two milliamps gives 2 kilo-ohms. If you get a dimensionless value of the gradient of a V/I V/mA graph, that value is in units of kilo ohms.
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Bigflakes
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(Original post by Callicious)


\begin{align*}

V[\textrm{V}] = I[\textrm{A}]R[\Omega] \\ 

\displaystyle\frac{V}{I}\frac{[\textrm{V}]}{[\textrm{A}]} = \displaystyle{R}[\Omega] \\ 

\displaystyle\frac{V}{I}\frac{[\textrm{V}]}{[\textrm{mA}]} = \displaystyle{R}[\textrm{k}\Omega] 

\end{align*}

^ Dimensionally speaking what you're dealing with. Four volts divided by two milliamps gives 2 kilo-ohms. If you get a dimensionless value of the gradient of a V/I V/mA graph, that value is in units of kilo ohms.
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