Graphs reducing non-linear laws to linear form 153
Gradient of straight line,
lg b =
AB
BC
=
2.13 − 1.17
3.0 − 1.0
=
0.96
2.0
= 0.48
Hence, b = antilog 0.48 = 10 0.48 = 3.0, correct to 2
significant figures.
Vertical axis intercept, lg a = 0.70, from which
a = antilog 0.70
= 10
0.70
= 5.0, correct to
2 significant figures.
Hence, the law of the graph is y = 5.0(3.0)
x
(a) When x = 2.1, y = 5.0(3.0) 2.1 = 50.2
(b) When y = 100, 100 = 5.0(3.0) x , from which
100/5.0 = (3.0) x
i.e. 20 = (3.0)
x
Taking logarithms of both sides gives
lg 20 = lg(3.0)
x
= x lg 3.0
Hence, x =
lg 20
lg 3.0
=
1.3010
0.4771
= 2.73
Problem 7. The current i mA flowing in a
capacitor which is being discharged varies with
time t ms, as shown below.
i (mA) 203 61.14 22.49 6.13 2.49 0.615
t (ms) 100 160 210 275 320 390
Show that these results are related by a law of the
form i = I e t / T , where I and T are constants.
Determine the approximate values of I and T
Taking Napierian logarithms of both sides of
i = I e
t / T
gives
ln i = ln(I e
t / T
) = ln I + ln e
t / T
= ln I +
t
T
ln e
i.e.
ln i = ln I +
t
T
since ln e = 1
or
ln i =
1
T
t + ln I
which compares with y = mx + c
showing that ln i is plotted vertically against t horizontally, with gradient
1
T and vertical-axis intercept
ln I .
Another table of values is drawn up as shown below.
t
100 160
210
275 320 390
i
203 61.14 22.49 6.13 2.49 0.615
ln i 5.31 4.11
3.11
1.81 0.91 −0.49
A graph of ln i against t is shown in Figure 18.7
and, since a straight line results, the law i = I e t/T is
verified.
A
B
C
ln i
5.0
4.0
3.0
3.31
2.0
1.0
100
200
300
400
t (ms)
D (200, 3.31)
21.0
1.30
0
Figure 18.7
Gradient of straight line,
1
T
=
AB
BC
=
5.30 − 1.30
100 − 300
=
4.0
−200
= −0.02
Hence,
T =
1
−0.02
= −50
Selecting any point on the graph, say point D, where
t = 200 and ln i = 3.31, and substituting
into
ln i =
1
T
t + ln I
gives
3.31 = −
1
50
(200) + ln I
from which, ln I = 3.31 + 4.0 = 7.31
and
I = antilog 7.31 = e 7.31 = 1495 or 1500
correct to 3 significant figures.
Hence, the law of the graph is i = 1500e −t/50
Gradient of straight line,
lg b =
AB
BC
=
2.13 − 1.17
3.0 − 1.0
=
0.96
2.0
= 0.48
Hence, b = antilog 0.48 = 10 0.48 = 3.0, correct to 2
significant figures.
Vertical axis intercept, lg a = 0.70, from which
a = antilog 0.70
= 10
0.70
= 5.0, correct to
2 significant figures.
Hence, the law of the graph is y = 5.0(3.0)
x
(a) When x = 2.1, y = 5.0(3.0) 2.1 = 50.2
(b) When y = 100, 100 = 5.0(3.0) x , from which
100/5.0 = (3.0) x
i.e. 20 = (3.0)
x
Taking logarithms of both sides gives
lg 20 = lg(3.0)
x
= x lg 3.0
Hence, x =
lg 20
lg 3.0
=
1.3010
0.4771
= 2.73
Problem 7. The current i mA flowing in a
capacitor which is being discharged varies with
time t ms, as shown below.
i (mA) 203 61.14 22.49 6.13 2.49 0.615
t (ms) 100 160 210 275 320 390
Show that these results are related by a law of the
form i = I e t / T , where I and T are constants.
Determine the approximate values of I and T
Taking Napierian logarithms of both sides of
i = I e
t / T
gives
ln i = ln(I e
t / T
) = ln I + ln e
t / T
= ln I +
t
T
ln e
i.e.
ln i = ln I +
t
T
since ln e = 1
or
ln i =
1
T
t + ln I
which compares with y = mx + c
showing that ln i is plotted vertically against t horizontally, with gradient
1
T and vertical-axis intercept
ln I .
Another table of values is drawn up as shown below.
t
100 160
210
275 320 390
i
203 61.14 22.49 6.13 2.49 0.615
ln i 5.31 4.11
3.11
1.81 0.91 −0.49
A graph of ln i against t is shown in Figure 18.7
and, since a straight line results, the law i = I e t/T is
verified.
A
B
C
ln i
5.0
4.0
3.0
3.31
2.0
1.0
100
200
300
400
t (ms)
D (200, 3.31)
21.0
1.30
0
Figure 18.7
Gradient of straight line,
1
T
=
AB
BC
=
5.30 − 1.30
100 − 300
=
4.0
−200
= −0.02
Hence,
T =
1
−0.02
= −50
Selecting any point on the graph, say point D, where
t = 200 and ln i = 3.31, and substituting
into
ln i =
1
T
t + ln I
gives
3.31 = −
1
50
(200) + ln I
from which, ln I = 3.31 + 4.0 = 7.31
and
I = antilog 7.31 = e 7.31 = 1495 or 1500
correct to 3 significant figures.
Hence, the law of the graph is i = 1500e −t/50