RESULTS
The resistance of each graphite pencil is obtained from the slope of the I-V characterisitics
curves. This is given by
𝑆𝑙𝑜𝑝𝑒 ⟹ 𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 =
△𝑉
△𝐼
(2)
With the resistance obtained from equation (2) and utilizing equation (1), the resistivity is
determined as follows
𝐶𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟 𝑥 𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑅𝑒𝑠𝑖𝑠𝑡𝑖𝑣𝑖𝑡𝑦 =
𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟
(3)
The cross-sectional area of the graphite pencil is obtained from the diameter using the
relation
𝐶𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎, 𝐴 = 𝜋
𝑑2
4
, 𝑤ℎ𝑒𝑟𝑒 𝑑 𝑖𝑠 𝑡ℎ𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑟𝑎𝑝ℎ𝑖𝑡𝑒
It is obtained as underlisted for the various pencil ranges.
(4)
(i)
Figure 3 shows the voltage versus current characteristic curve for the pencil B as obtained
from the given data.
Pencil B
1,8
1,6
Voltage [V]
1,4
y = 1,6527x + 0,0125
1,2
1
0,8
0,6
0,4
0,2
0
0
0,2
0,4
0,6
0,8
1
Current [A]
Figure 3: Voltage-current characteristics of Pencil B
It is noted from the plot that the slope and hence the resistance of this kind of pencil is 1.6527
ohms. Given that the diameter of the pencil B is 2 [mm], its cross-sectional area is then
obtained from (4) as
(2 𝑥 10−3 )2
𝑑2
𝐴=𝜋
=𝜋
= 3.142 𝑥 10−6 𝑚2
4
4
With the provided data of the length and the calculated cross-sectional area of the pencil, its
resistivity is determined from equation (3) as
𝜌=
3.142 𝑥 10−6 𝑚2 𝑥 1.6527
= 4.71 𝑥 10−5 Ω ∙ 𝑚
110.25 𝑥 10−3 𝑚
(ii)
Figure 4 shows the voltage againt current graph for this kind of pencil as obtained from the
given dataset.
Pencil 4B
0,7
0,6
y = 0,6102x + 0,0035
Voltage [V]
0,5
0,4
0,3
0,2
0,1
0
0
0,2
0,4
0,6
0,8
1
1,2
Current [A]
Figure 4: Voltage-current characteristics of Pencil 4B
It is noted from the plot that the slope and hence the resistance of this kind of pencil is 0.6102
ohms. Given that the diameter of this pencil is 2 [mm], its cross-sectional area is obtained
from (4) as
(2 𝑥 10−3 )2
𝑑2
𝐴=𝜋
=𝜋
= 3.142 𝑥 10−6 𝑚2
4
4
With the provided data of its length and the calculated cross-sectional area, the resistivity of
this pencil is determined from equation (3) as
𝜌=
3.142 𝑥 10−6 𝑚2 𝑥 0.6102
= 2.044 𝑥 10−5 Ω ∙ 𝑚
93.8 𝑥 10−3 𝑚
(iii)
Figure 5 shows the voltage versus current characteristic curve for the pencil 2B as obtained
from the given dataset.
Pencil 2B
1,4
1,2
y = 1,168x + 0,0053
Voltage [V]
1
0,8
0,6
0,4
0,2
0
0
0,2
0,4
0,6
0,8
1
1,2
Current [A]
Figure 5: Voltage-current characteristics of Pencil 2B
It is noted from the plot that the slope and hence the resistance of this kind of pencil is 1.168
ohms. Given that the diameter of the 2B pencil is 2 [mm], its cross-sectional area is then
obtained from (4) as
(2 𝑥 10−3 )2
𝑑2
𝐴=𝜋
=𝜋
= 3.142 𝑥 10−6 𝑚2
4
4
With the provided data of the length and the calculated cross-sectional area of the pencil, its
resistivity is determined from equation (3) as
𝜌=
3.142 𝑥 10−6 𝑚2 𝑥 1.168
= 3.551 𝑥 10−5 Ω ∙ 𝑚
103.35 𝑥 10−3 𝑚
(iv)
Figure 6 shows the voltage versus current characteristic curve for the F pencil as obtained
from the given dataset.
Voltage [V]
Pencil F
2
1,8
1,6
1,4
1,2
1
0,8
0,6
0,4
0,2
0
0,00E+00
y = 1,9513x + 0,0111
2,00E-01
4,00E-01
6,00E-01
8,00E-01
1,00E+00
Current [A]
Figure 6: Voltage-current characteristics of Pencil F
It is noted from the plot that the slope and hence the resistance of this kind of pencil is 1.9513
ohms. Given that the diameter of the F pencil is 2 [mm], its cross-sectional area is then
obtained from (4) as
(2 𝑥 10−3 )2
𝑑2
𝐴=𝜋
=𝜋
= 3.142 𝑥 10−6 𝑚2
4
4
With the provided data of the length and the calculated cross-sectional area of the pencil, its
resistivity is determined from equation (3) as
𝜌=
3.142 𝑥 10−6 𝑚2 𝑥 1.9513
= 4.478 𝑥 10−5 Ω ∙ 𝑚
136.9 𝑥 10−3 𝑚
(v)
Figure 7 shows the voltage versus current characteristic curve for the 2F pencil as obtained
from the given dataset.
Pencil 2F
1,6
y = 1,5162x + 0,0071
1,4
Voltage [V]
1,2
1
0,8
0,6
0,4
0,2
0
0
0,2
0,4
0,6
0,8
1
Current [A]
Figure 7: Voltage-current characteristics of Pencil 2F
It is noted from the plot that the slope and hence the resistance of this kind of pencil is 1.5162
ohms. Given that the diameter of the 2F pencil is 2 [mm], its cross-sectional area is then
obtained from (4) as
(2 𝑥 10−3 )2
𝑑2
𝐴=𝜋
=𝜋
= 3.142 𝑥 10−6 𝑚2
4
4
With the provided data of the length and the calculated cross-sectional area of the pencil, its
resistivity is determined from equation (3) as
𝜌=
3.142 𝑥 10−6 𝑚2 𝑥 1.5162
= 3.5605 𝑥 10−5 Ω ∙ 𝑚
133.8 𝑥 10−3 𝑚
(vi)
Figure 8 shows the voltage versus current characteristic curve for the HB pencil as obtained
from the given dataset.
Voltage [V]
Pencil HB
2
1,8
1,6
1,4
1,2
1
0,8
0,6
0,4
0,2
0
y = 2,1379x + 0,0115
0
0,2
0,4
0,6
0,8
1
Current [A]
Figure 8: Voltage-current characteristics of Pencil HB
It is noted from the plot that the slope and hence the resistance of this kind of pencil is 2.1379
ohms. Given that the diameter of the HB pencil is 2 [mm], its cross-sectional area is then
obtained from (4) as
𝐴=𝜋
(2 𝑥 10−3 )2
𝑑2
=𝜋
= 3.142 𝑥 10−6 𝑚2
4
4
With the provided data of the length and the calculated cross-sectional area of the pencil, its
resistivity is determined from equation (3) as
3.142 𝑥 10−6 𝑚2 𝑥 2.1379
𝜌=
= 6.7409 𝑥 10−5 Ω ∙ 𝑚
99.65 𝑥 10−3 𝑚