Physics experiment #1: Analysis of data by graphical techniques
Nicolai Solheim, 2IBG
Physics 2HL
Purpose: To determine the relationship between the resistance (
R
), the length ( l
) and the diameter ( d
) of an electrical conductor.
Procedure: 1) plot the resistance (
R
) in ohms versus the diameter ( d
) in centimetres for a piece of nichrome wire 100 cm long (i.e.
R
on the vertical axis and d on the horizontal axis)
2) Using the graph, find, by interpolation, the values of of 0.03 and 0.06 cm.
R
for diameters
3) Using extrapolation, find the resistances when the diameters are
0.02 cm and 0.15 cm.
4) Analyse the shape of the graph and write a proportionality statement relating
R
and d
.
5) Choose a function of the diameter such that, when plotted against
R
, a straight line results. Plot Graph 2 of
R
versus the rearranged data for d ( l =100 cm). Scale the resistance axis so that the maximum resistances for the other lengths can eventually be plotted on the same graph.
6) When a straight line is achieved, write a proportionality statement relating
R
and d
for a constant length. Find the slope of the graph and write an equation for the relationship between
R
and d
.
7) Using the equation, determine the resistance for a diameter of 0.06 cm. Compare your results with those determined in step 2.
8) On the same graph, plot
R
versus rearranged data for d
for the other three lengths, 140 cm, 170 cm, and 220 cm. Label each line according to its length.
9) Using Graph 2, predict the relationship between
R
and l
.
10) R versus l for a constant diameter 2.5 x 10
If the line is extrapolated, does it pass through the origin?
-2 cm.
11) Combine the proportionality statements for R and d and R and l to form one statement relating
R
, d
and l
. Express this as a general equation, using the values from the data chart (i.e. use a set of values of
R
, d
and l
to find the value of the constant for equation.). Using the equation, determine the value of R for d =5.0 x 10 -2 cm and l =200 cm.
I

Physics experiment #1: Analysis of data by graphical techniques
Experimental Data: The data given below was recorded in an investigation to measure the electrical resistances of nichrome wires of various lengths and diameters.
Length (cm): 100 140 170 220
Resistance ( Ω ) Diameter (cm):
2.5 x 10 -2
4.0 x 10 -2
7.5 x 10 -2
12.5 x 10 -2
15.7 22.0 26.8 34.7
1.9 2.5 3.0 3.9
0.6 0.9 1.1 1.4
Conclusion: 1) below:
II

Physics experiment #1: Analysis of data by graphical techniques
2) The values of R when diameters are:
0.03 cm ⇒ 10.0 Ω
0.06 cm ⇒ 3.0 Ω
3) The values of
R
when diameters are:
0.02 cm ⇒ ≈ 23.0 Ω
0.15 cm ⇒ ≈ 0.5 Ω
4) The shape of the graph is hyperbola, and the proportionality statement is the following:
R ∝ l
R
∝ 1/ d
2
⇒
R ∝ l
/ d
2
5) Rearranged data ( l =100 cm) ( I ):
1/ d
2
1/(2.5 x 10
1/(4.0 x 10
1/(7.5 x 10
-2
-2
-2
1/(12.5 x 10
)
)
)
-2
2
2
2
) 2
R
( Ω ) 1/ d
2
15.7 ⇒ 1 / 0,000625 = 1600
6.1
1.9
0.6
R
(
15.7
⇒ 1 / 0,0016 = 625 6.1
⇒ 1 / 0,005625 = 177,777778 1.9
⇒ 1 / 0,0156 = 64,1025641 0.6
Ω )
≈ 1/ d
2
R
( Ω )
1600 15.7
625 6.1
178 1.9
64 0.6
Graph 2: see page 5.
6) The proportionality statement relating
R
and d
for a constant length
⇒
R ∝ 1/ d
2
7) Ω ) for a di ameter of 0.06 cm.
R
=
( 1
-2 ) 2
)
= 2.777…
Compa rison: Result in step
Result in step
2: 0.06 cm ⇒ 3.0 Ω
7: 0.06 cm ⇒ 2.777… Ω
The slope ( s
) for the straight line is:
∆
∆ y y x
2
– y x
2
– x
1
1 =
15.7 – 6.1
1600 – 625
=
9.6
975
=
The equation for the relationship between
R
and d
is:
R
= s
x (1/ d
2
⇒ d
) = 0.01 x (1/
= √ d
2 )
((1 x 0.01)/R)
0.009846153846
≈ 0.010
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Physics experiment #1: Analysis of data by graphical techniques
8) Rearranged data ( l =140 cm):
1/ d
2
1/(2.5 x 1 0
1/(4.0 x 10
1/(7.5 x 10
-2
-2
-2
1/(12.5 x 10
)
)
)
-2
2
2
2
) 2
R
( Ω ) 1/ d
2
22.0 ⇒ 1 / 0,000625 = 1600
8.5
2.5
0.9
R
(
22.0
⇒ 1 / 0,0016 = 625 8.5
⇒ 1 / 0,005625 = 177,777778 2.5
⇒ 1 / 0,0156 = 64,1025641 0.9
Ω )
1/ d
2
1/(2.5 x 10
1/(4.0 x 10
1/(7.5 x 10
-2
-2
-2 data
)
)
)
2
2
2
1/(12.5 x 10 )
R
( Ω )
26.8
20.4
3.0
1.1
⇒ 1 / 0,000625 = 16 00
⇒ 1 / 0,0016 = 625
⇒
1/ d
2
1 / 0,005625 = 177,77777 8
R
(
3.0
⇒ 1 / 0,0156 = 64,1025641 1.1
Ω
26.8
10.4
)
1/ d
1/(2.5
1/(4
2
x 10
.0 x 10
1/(7.5 x 10
-2
-2
-2
1/(12.5 x 10
)
)
)
-2
2
2
2
) 2 l
=220
R (
34.7
13.3
3.9
1.4
Ω ) 1/ d
2
⇒ 1 / 0,000625 = 1600
R ( Ω
34.7
⇒ 1 / 0,0016 = 625 13.3
⇒ 1 / 0,005625 = 177,777778 3.9
⇒ 1 / 0,0156 = 64,1025641 1.4
)
II l
=140 cm: III
≈ 1/ d
2 l
=170 cm
R
( Ω ) ≈ 1/ d
2
IV l
=220 cm
R
( Ω ) ≈ 1/ d
2
R
( Ω )
1600 22.0 1600 26.8 1600 34.7
625 8.5
178 2 .5
1 3.9
64 0 .9
6 4 64 1.4
See d
for the other three le ngths, 140 cm,
170 cm, and 220 cm on page 5 .
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Physics experiment #1: Analysis of data by graphical techniques
V

Physics experiment #1: Analysis of data by graphical techniques
9) The relationship between R and l is: s
*
R
= d
2
I s
1
∆
∆ y y x x
2
2
– y
– x
1
1
=
15.7 – 6.1
1600 – 625
=
II s
2
∆
∆ x
= 2
– y
1 x
2
– x
1
=
III s
3
∆
∆ y y x x
2
2
– y
– x
1
1
=
22.0 – 8.5
1600 – 625
26.8 – 10.4
1600 – 625
=
=
9.6
975
13.5
975
=
=
0.009846153846
0.013846153850
16.4
975
= 0.016820512820
IV s
4
∆
∆ x
= 2
– y
1 x
2
– x
1
=
34.7 – 13.3
1600 – 625
=
21.4
975
= 0.021948717950
∆ s
∆ l
=
⇒ s
=
⇒ s
=
* Slope s l
–
– l s
1
1 s
1
– s
2 l
1
– l
2
100
=
–
· ( l
220 s l
–
1
1 l
–
–
1
0.0098 – 0.022 l s
2
2
) + s
1
· ( l
–
100
) + 0.0098
VI

Physics experiment #1: Analysis of data by graphical techniques
⇒ s
= 0.00010166 · – 100 ) + 0.0098
⇒ s
= 0.00010166
l
– 0.010166 + 0.0098
⇒ s
= 0 .00010
166
l
– 0.00
0366
R =
0.00010166
l
– 0.000366 d
2
?
0.00010166 l + 0.000366 d
2 eq ti d li a ha p ie to solve the task, the right equation to use.
10) Rearranged data: d
2.5 x 10
2.5 x 10
2.5 x 10
-2
-2
-2
2.5 x 0 -2
R
( Ω ) l
15.7 100
22.0 140
3
.8
4.7
170
220
The line does not go through the origin, even though it goes very close
(, the line crosses the y-axis at approximately (0,0.97) and the x-axis at approximately (-0.15,0)), see generated graph beneath.
See Graph 3 on pag e 8 .
VII

Physics experiment #1: Analysis of data by graphical techniques
VIII

Physics experiment #1: Analysis of data by graphical techniques
11)
R ∝ l
R ∝ 1/ d
2
⇒
R ∝ l
/ d
2
Equation: R = c l / d
2 d
= 5.0 x 10 -2 cm l
= 200 cm
R = unknown c
(constant) = unknown
Finding c
:
15.7 Ω = ( c x 100 cm)/0.000625 c c
= (15.7 Ω x 0.000625 cm 2 )/100 cm c
= 0.000098125 Ω cm check: R = (0.000098125 x 100)/0.000625
R = 15.7
Finding R :
R
= ( c
x l
)/( d
) 2
R
= (0.000098125 x 200)/(5.0 x 10 -2 ) 2
R = 0.019625/0.0025
R
= 7.85
Conclusion: From what we have done, we can see that there is a connection between the length, resistance and diameter, which allows us to determine the relationship between the resistance ( R ), the length
( l
) and the diameter ( d
) of an electrical conductor.
IX