Lesson Element uncertainties.
This brief guide is produced to give teachers and candidates a better idea of how
uncertainties could be treated in experimental Physics, specifically as it relates to the OCR
Physics A and B specification for first teaching from 2015. This is not intended to be a
guide on how the PAGs should be carried out for the practical endorsement but rather,
just provides the context for the example.
The following sets out an example from PAG 3.1
Investigation to determine the resistivity of a Metal
+
S
R
A
V
metal wire
L
Fig. 1
The following shows some sample data and the subsequent treating of the uncertainties.
This is not prescriptive in that you have to follow this exact model for every experiment, it
merely points out a possible approach bearing in mind that any of this manipulation can
come up on the question papers
The student used a voltmeter with a resolution of 0.01V and an ammeter with a resolution of
0.01A. The length of wire was varied at 10cm intervals and the rheostat adjusted so that a
current of 0.35A was maintained throughout the investigation. The length of nichrome wire
was mounted on to a metre rule.
The following data was collected and the experiment was not repeated.
The variables in the data are linked by the equation
V I

l
A
Where V is the voltage across the wire
I is the current through the wire
l is the length of wire
A is the cross sectional area of the wire
 is the resistivity of the wire
In order to find a value for resistivity ρ, all of the uncertainties in the equation have to be
considered
Current 0.35A
l/m
0.100
0.200
0.300
0.400
0.500
0.600
0.700
0.800
0.900
pd/V
0.68
1.30
1.75
2.39
2.80
3.65
4.00
4.67
5.52
Absolute uncertainties were as follows
Uncertainty in voltmeter 0.01v
Uncertainty in ammeter 0.01A
Uncertainty in the cross sectional area of wire- The wire was measured three times along its
length with a digital micrometer and a mean taken. The readings were as follows
d1/mm d2/mm d3/mm dmean/mm
0.248 0.252 0.254 0.251
Uncertainty in a range of readings is half the range of the spread of results so
max  min
2
0.254  0.248
 0.003mm
2
This is the absolute uncertainty in the diameter of the wire ±0.003mm. We need the
diameter in the area and the area is
𝐴 = 𝜋𝑟 2
so radius is 𝐴 = 𝜋(0.000251/2)2= 4.95x10-8 m2
-because the radius is squared we have to take the uncertainty into account twice. However,
we can’t just add absolute uncertainties; rather, we have to add the percentage
uncertainties.
0.003
x100  1.2% x 2  2.4%
0.251
Therefore the uncertainty in the cross sectional area of the wire is ±2.4%
The uncertainty in length presents a couple of challenges. The mm ruler has an analogue
scale, therefore, the uncertainty in the instrument is ± half the smallest division 0.5mm. If we
were just measuring the length of the wire, then the uncertainty would be ± 1mm as we have
the uncertainty in the measurement at each end. However, because there is an element of
judgement due to the difficulty in measuring where the crocodile clip is in contact with the
wire, we have to use our judgement and increase the uncertainty. In this case it could be
logical to say it is anything between 2-4mm. Therefore we will call the uncertainty in the
length of wire ±3mm.
Now that we have considered the uncertainties, we now process the data to determine the
resistivity of the wire and quote its uncertainty.
Below is the graphed data.
7.00
6.00
5.00
pd/V
4.00
3.00
2.00
1.00
0.00
0.00
0.20
0.40
0.60
0.80
1.00
Length/m
Applying the equation to the data then, we have plotted V/l therefore the equation from
above becomes
V I

l
A
So is the gradient of the line =
V
l
1.20
7.00
6.00
5.00
pd/V
4.00
3.00
2.00
1.00
0.00
0.00
0.20
0.40
0.60
0.80
1.00
1.20
Length/m
Gradient
(6.00  1.20)
 6.00
(1.00  0.20)
V
I
= gradient =
A
l
Therefore:
gradientxA

I
6.00 x 4.95 x10 8
 8.49 x10 7 Ω m
0.35
The experimental value for resistivity then, is 8.49 x 10-7 Ω m
We now have to compare this value to the book value given in some textbooks as 1.00x10-6
Ω m
Comparing this to our value of 8.49 x 10-7 Ω m and calculating the percentage difference
between the two
1.00 x10 6  8.49 x10 7
x100  15%
1.00 x10 6
That means there is a 15% difference between the experimental value and the accepted
book value. We now have to process our uncertainties to see if ‘within the uncertainties’ our
value is in agreement
Uncertainties in our data
We used this equation to calculate our experimental value; therefore we need to consider the
uncertainties in each quantity and add the percentage uncertainties to calculate the total
uncertainty in the resistivity
gradientxarea

length
We calculated the uncertainty in area before and determined it was 2.4%
The uncertainty in the current was 0.01A so percentage uncertainty is 0.01/0.35 x 100 =
2.9%
The uncertainty in the gradient is calculated as follows. A line of worst fit is drawn on the
extremes of the data, that is, the worst possible line of fit you can draw as shown by the
dotted line below. Calculate the gradient of that line and calculate the percentage difference
between the best and worst fit. This is the percentage uncertainty in the gradient.
7.00
6.00
5.00
pd/V
4.00
3.00
2.00
1.00
0.00
0.00
0.20
0.40
0.60
0.80
1.00
1.20
Length/m
Worst fit gradient
6.00
=6.82
(0.96  0.08)
Percentage difference in the gradient
6.82  6.00
X 100 = 13.7%
6.00
Therefore:
% uncertainty in current = 2.9%
% uncertainty in area =2.4%
% uncertainty in gradient =13.7%
Total uncertainty in our experimental value = ±19%
The percentage difference between the book value and our experimental value was 15%
We can say then, that our experimental value lies within the experimental uncertainties and
the two values are in agreement.
The example used above doesn’t have error bars on the graph. The reason for this is that
they are just too small to display, therefore the worst fit line is just drawn using judgement on
the extremes of data. That is, the worst possible line you could draw considering the spread
of data.
Below is an example of some data with the error bars plotted with best and worst fit lines.
It is reasonable to expect the worst fit line to extend from the top extremity of the largest data
point to the lowest point on the smallest data point as shown.
The uncertainty in the gradient can be determined by the method explained earlier. We can
take this a step further if you are asked for the uncertainty in the y-intercept.
In the example above, the y-intercept of the best fit line is 2.2. The uncertainty in this is the
difference in y-intercepts of best and worst fit divided by the best fit intercept x 100.
So in this example:
±32%
2.2  1.5
 100 = 32% Therefore the uncertainty in the y-intercept is
2.2