learningizk00l
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I'm plotting a graph of V against I ( to calculate EMF and Internal resistance), I have taken the uncertainty of both the voltmeter and ammeter to be +-0.01V/A. I've been when regressing y on x the error bars should be plotted vertically (so the +-0.01V) but this gives error bars too small to see. I am unsure whether I should just take the final absolute error to be 0.01V or if I should plot horizontal error bars which I can take a gradient from.

Also I'm doing this on excel and to put in the error bars I am going to custom error bars and just putting in 0.01 for +ve and -ve error- is this correct?
Thank you!
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Nagromicous
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(Original post by learningizk00l)
I'm plotting a graph of V against I ( to calculate EMF and Internal resistance), I have taken the uncertainty of both the voltmeter and ammeter to be +-0.01V/A. I've been when regressing y on x the error bars should be plotted vertically (so the +-0.01V) but this gives error bars too small to see. I am unsure whether I should just take the final absolute error to be 0.01V or if I should plot horizontal error bars which I can take a gradient from.

Also I'm doing this on excel and to put in the error bars I am going to custom error bars and just putting in 0.01 for +ve and -ve error- is this correct?
Thank you!
What does the plot look like?
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learningizk00l
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(Original post by Nagromicous)
What does the plot look like?
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Size:  37.6 KB (This contains vertical error bars)
Last edited by learningizk00l; 1 month ago
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Nagromicous
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(Original post by learningizk00l)
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Clearly, your error bars are wrong. Roughly two-thirds of your data should be within one error bar of the trendline. It would help if you thought about why your errors would be more than 0.01V, e.g. because the current was changing as the circuit was heating up, etc.
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learningizk00l
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(Original post by Nagromicous)
Clearly, your error bars are wrong. Roughly two-thirds of your data should be within one error bar of the trendline. It would help if you thought about why your errors would be more than 0.01V, e.g. because the current was changing as the circuit was heating up, etc.
But surely it doesn't matter if the heat changed the current seeing as current is something I am measuring and not just assuming constant?
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Nagromicous
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(Original post by learningizk00l)
But surely it doesn't matter if the heat changed the current seeing as current is something I am measuring and not just assuming constant?
By drawing a straight line with equation V = e - Ir, you assume that the resistance is constant in time (for each measurement). With larger currents, the battery and wires will heat up, increasing the resistance in a way you haven't accounted for. The discrepancy between your results and the expected results could be caused by the heat changing the resistance. If I were you, I'd redo the experiment, making sure that I let the equipment cool down for each measurement, only leaving the circuit closed for long enough to take the measurement.
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