basic_engineering_mathematics0

122 Basic Engineering Mathematics
(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
Hence x =
lg 20 = lg(3.0) x = x lg 3.0
lg 20
lg 3.0 = 1.3010
0.4771 = 2.73
ln i
5.0 A
4.0
3.31
3.0
2.0
D(200, 3.31)
Problem 7. The current i mA flowing in a capacitor which
is being discharged varies with time t ms as shown below.
1.30
1.0
B
C
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 = Ie t/T , where I and T are constants. Determine the
approximate values of I and T.
0
1.0
Fig. 16.7
100 200 300 400 t (ms)
Taking Napierian logarithms of both sides of i = Ie t/T gives
ln i = ln (Ie t/T ) = ln I + ln e t/T
i.e. ln i = ln I + t (since ln e = 1)
T
( ) 1
or ln i = t + ln I
T
which compares with y = mx + c, showing that ln i is plotted
vertically against t horizontally. (For methods of evaluating
Napierian logarithms see Chapter 15.) 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 Fig. 16.7 and since a straight
line results the law i = Ie t/T is verified.
Gradient of straight line,
1
T = AB 5.30 − 1.30
=
BC 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
1
ln i = 3.31, and substituting into ln i = t + ln I gives
T
3.31 =− 1 (200) + ln I
50
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
Now try the following exercise
Exercise 61
Further problems on reducing non-linear
laws to linear form (Answers on page 277)
In Problems 1 to 3, x and y are two related variables and
all other letters denote constants. For the stated laws to be
verified it is necessary to plot graphs of the variables in
a modified form. State for each (a) what should be plotted
on the vertical axis, (b) what should be plotted on the
horizontal axis, (c) the gradient and (d) the vertical axis
intercept.
1. y = ba x 2. y = kx l y
3.
m = enx
4. The luminosity I of a lamp varies with the applied voltage
V and the relationship between I and V is thought to be
I = kV n . Experimental results obtained are:
I candelas 1.92 4.32 9.72 15.87 23.52 30.72
V volts 40 60 90 115 140 160
Verify that the law is true and determine the law of
the graph. Determine also the luminosity when 75V is
applied across the lamp.