basic_engineering_mathematics0

Graphs with logarithmic scales 129
1000
1000
Hence T = −27.7
2.3026 = −12.0, correct to 3 significant figures. from which, time t = (12.0)(4.2438) = 50.9 ms.
y
y ae kx
t
v Ve T
100
A
100
(36.5, 100)
A
10
B
C
10
B
C
0
2 1 0 1 2 3 4 5 6 x
1 0 10 20 30 40 50 60 70 80 90
Fig. 17.6
Time, t ms
Fig. 17.7
Problem 6. The voltage, v volts, across an inductor is
believed to be related to time, t ms, by the law v = Ve t/T ,
where V and T are constants. Experimental results obtained
are:
Since the straight line does not cross the vertical axis at t = 0
in Fig. 17.7, the value of V is determined by selecting any point,
say A, having co-ordinates (36.5, 100) and substituting these
values into v = Ve t/T . Thus
v volts 883 347 90 55.5 18.6 5.2
t ms 10.4 21.6 37.8 43.6 56.7 72.0
100 = Ve 36.5/−12.0
i.e V = 100 = 2090 volts.
e−36.5/12.0 Show that the law relating voltage and time is as stated and
correct to 3 significant figures.
determine the approximate values of V and T. Find also the
value of voltage after 25 ms and the time when the voltage Hence the law of graph is: υ = 2090e −t/12.0
is 30.0V
When time t = 25 ms, voltage υ = 2090e −25/12.0
= 260 V.
Since v = Ve t/T then ln v = 1 T t + ln V
which is of the form Y = mX + c.
When the voltage is 30.0 volts, 30.0 = 2090e −t/12.0 , hence
Using ‘log 3 cycle × linear’ graph paper, the points are plotted as
shown in Fig. 17.7.
e −t/12.0 = 30.0
Since the points are joined by a straight line the law v = Ve t/T 2090
is
and e t/12.0 = 2090
30.0 = 69.67
verified.
Gradient of straight line,
Taking Napierian logarithms gives:
t
1
T = AB ln 100 − ln 10
=
BC 36.5 − 64.2 = 2.3026
−27.7
= ln 69.67 = 4.2438
12.0
Voltage, v volts