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# Physics coursework - improving uncertainties watch

1. The aim of the investigation was to calculate, as accurately and precisely as possible, the internal resistance (r) and the Electromotive Force (EMF) of two cells; an AA Cell and a D Cell, both of which were rated to supply a voltage of 1.5V. I proceeded to do this by measuring the curent of the circuit and the voltage supplied by the cells.

The multimeter has a resolution of 0.01A and 0.01V so an uncertainty of ±0.01. I have also found that I am uncertain due to the multimeters flickering on the last digit of the readout; this flickered by 0.01 so does that make my uncertainty ±0.02?

Also, how can I adjust the experiment to improve the uncertainty in the results and therefore the uncertainty in the calculated EMF and internal resistance?
2. (Original post by TekWarfare)
The multimeter has a resolution of 0.01A and 0.01V so an uncertainty of ±0.01. I have also found that I am uncertain due to the multimeters flickering on the last digit of the readout; this flickered by 0.01 so does that make my uncertainty ±0.02?
I think you can take the average of the two results and take the uncertainty to be .

Also, how can I adjust the experiment to improve the uncertainty in the results and therefore the uncertainty in the calculated EMF and internal resistance?
Make sure to work with the smallest possible range. That is, if your result is milliampers, do not set the range to tens of ampers. Do all the measurements several times. Make sure the measurements are made at the same temperature of the wire/resistor - wait long enough between each measurement, and either take them quickly after letting the current go (i.e. at low temperature) or long after that (i.e. when temperature is stable at a high level).

3. (Original post by TekWarfare)
The aim of the investigation was to calculate, as accurately and precisely as possible, the internal resistance (r) and the Electromotive Force (EMF) of two cells; an AA Cell and a D Cell, both of which were rated to supply a voltage of 1.5V. I proceeded to do this by measuring the curent of the circuit and the voltage supplied by the cells.

The multimeter has a resolution of 0.01A and 0.01V so an uncertainty of ±0.01. I have also found that I am uncertain due to the multimeters flickering on the last digit of the readout; this flickered by 0.01 so does that make my uncertainty ±0.02?

Also, how can I adjust the experiment to improve the uncertainty in the results and therefore the uncertainty in the calculated EMF and internal resistance?
How do you intend to calculate the internal resistance?
What other components are there in the circuit?

Generally you reduce uncertainty by repeating results.
Draw a graph if you can. Is there any quantity you can vary in the circuit and measure change in current?
Can you find internal resistance from a slope or intercept?

They are not the results obtained, I think I've wrote it incorrectly. That's the resolution of the multimeter, I had to take the readings of voltage and current of the circuit when a certain resistor is connected to the circuit, I then change the fixed resistor for a lower rated one and repeat.

I did work in the smallest range initially, as I've stated, I worked in 0.01V and 0.01A for voltage and current respectively.
I could incorporate a rheostat into the circuit,
5. Hey Stonebridge, thanks for the reply. Are they questions you want me to answer or ideas for me to consider?
6. (Original post by TekWarfare)

They are not the results obtained, I think I've wrote it incorrectly.
You had written it correctly and I did understand you. I just meant that as what you could see on the voltmeter was flickering between two values, you could take the average of the two values, the uncertainty remaining the same.
7. (Original post by TekWarfare)
Hey Stonebridge, thanks for the reply. Are they questions you want me to answer or ideas for me to consider?
Both.
If the object of the experiment is to find the internal resistance of the batteries, then you must have a way of getting the answer from your measurements.
How do you intend to calculate the internal resistance from the readings you take?
The uncertainty in that final answer will depend on that.
If you are using a given resistor, do you use its stated value in the calculation?
If so, do you know the uncertainty in its value? (There is one.)
This will contribute to the uncertainty in the final result.
You may know the precision of the meters, but do you know their accuracy?
If you have access to a number of different resistors you can plot a suitable graph with current and resistance and find the emf and internal resistance from it.
8. I'm using the equation E = V + Ir and rearranging it to suit, i.e. I'll use V = -rI +E to draw the graph as it is the form y = mx + c - the straight line equation. The gradient gives me the internal resistance, but unfortunately the graph doesn't cross the y-axis so I need to use the equation: E = V + Ir
I've used the stated value of the fixed resistors, but I have also measured their resistance with a multimeter and it corresponds.
I don't know the uncertainty in the fixed resistor's resistance, no. How would I find that out?
For the second test, I have to improve upon the uncertainties so I will ensure that they are calibrated correctly and do not have a systematice or zero error. Is that their accuracy?
9. Also, my physics teacher told me to use the followqing equations for the uncertainties:

For the resistances:
dR’ = dR + dr

and for the emf:
dE= dI + dR’
.E......I......R’

Where d is Delta - meaning "change in", but which values do I sbstitute into the equations?
10. (Original post by TekWarfare)
I'm using the equation E = V + Ir and rearranging it to suit, i.e. I'll use V = -rI +E to draw the graph as it is the form y = mx + c - the straight line equation. The gradient gives me the internal resistance, but unfortunately the graph doesn't cross the y-axis so I need to use the equation: E = V + Ir
I've used the stated value of the fixed resistors, but I have also measured their resistance with a multimeter and it corresponds.
I don't know the uncertainty in the fixed resistor's resistance, no. How would I find that out?
For the second test, I have to improve upon the uncertainties so I will ensure that they are calibrated correctly and do not have a systematice or zero error. Is that their accuracy?
If you are drawing a graph of V against I, how are you varying V and I? (I assume this is for one of the batteries.)
What are you connecting the meter to, in order to find V and I and how are you connecting in the fixed resistors? I don't see any value for the fixed resistors in your equation, just the internal resistance. If they are in the circuit then their resistance will affect the current.
I'm just a little puzzled as to exactly what you are doing here.

Depending on their construction, fixed resistors usually have a "tolerance" that is specified as a %. This is the accuracy to which their rated value is quoted.
It was a good idea to measure them with the meter.
However, as I asked before, where does the value of this resistor figure in your calculation?
The current in the circuit will be from E = I(R + r) where R is the "external" resistance. I assume this is what the standard resistor is for?
So is the V in your equation = IR where R is this standard resistor?
11. V and I are varying as I change the fixed resistor, I use an 8.2, 5.6, 3.9, 2.7, 1.5 and a 1 ohm fixed resistor.
I have a voltmeter connected to the cell in parallel and the ammeter in series with the circuit.
The many fixed resistors are connected to the circuit board and I connect the leads to the desired resistor.
I have to measure the emf and internal resistance of the cell
Yep, I think the fixed resistors are incorporated, and changed, in the circuit so I can get values for V , I and R so that I can calculate E and r using the equation: E = V + Ir -> E = IR + Ir
12. Ok now I'm clear what you are doing. There are a couple of ways of doing this so I needed to check.
You should get a graph with negative gradient equal to r as you say.
You just need to extend it back to the y axis (V - axis) and the intercept will be E. It's numerically the pd you measure across the cell when the current is zero.

If this is the method, it doesn't actually matter what those resistances are. They are just a way of varying the current in the circuit. You could have used a rheostat.

The accuracy in the result depends on the accuracy of the V and I measurements.
The uncertainty in the final result is obtained from the graph by drawing a best and worst gradient.
Normally, the points drawn on the graph would have error bars showing the uncertainty in the individual measurements. These error bars then help you to draw the two gradients.

The meters do have a high precision. The manufacturer will often quote the instrument's "accuracy" (as a %), though this is usually pretty good. Very often it's the same as the precision with a good quality instrument. You could ask your teacher if the instruments have an accuracy quoted in their user guide or look to see if there is anything printed on the back.
The formula your teacher gave you is what is used for combining errors in measurements.
If you measured two resistances and added them together, the error in the answer would be the sum of the error in the two. dR means error in R here.
If you have an error in two values, say I and R that you multiply together for V = IR, then the fractional error in V is dV/V and is equal to the sum of the fractional errors in I and R. dI/I + dR/R
You will often see this stated in the alternative form of % error in V is the sum of the % errors in I and R.
% error is just the fractional error x 100.

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