Higher Engineering Mathematics

36 NUMBER AND ALGEBRA

Problem 25. In an experiment involving

Newton’s law of cooling, the temperatureθ(◦C)

is given by θ=θ 0 e−kt. Find the value of

constant k whenθ 0 = 56. 6 ◦C,θ= 16. 5 ◦C and

t= 83 .0 seconds.

Transposing θ=θ 0 e−ktgives

θ

θ 0

=e−kt

from which

θ 0

θ

=

1

e−kt

=ekt

Taking Napierian logarithms of both sides gives:

ln

θ 0

θ

=kt

from which,

k=

1

t

ln

θ 0

θ

=

1

83. 0

ln

(

56. 6

16. 5

)

=

1

83. 0

(1. 2326486 ...)

Hencek=1.485× 10 −^2

Problem 26. The current i amperes flow-

ing in a capacitor at timetseconds is given

by i= 8 .0(1−e

−t

CR), where the circuit resist-

anceRis 25× 103 ohms and capacitanceCis

16 × 10 −^6 farads. Determine (a) the currenti

after 0.5 seconds and (b) the time, to the near-

est millisecond, for the current to reach 6.0 A.

Sketch the graph of current against time.

(a) Currenti= 8 .0(1−e

−t

CR)

= 8 .0[1−e

− 0. 5

(16× 10 −^6 )(25× 103 )]= 8 .0(1−e−^1.^25 )

= 8 .0(1− 0. 2865047 ...)= 8 .0(0. 7134952 ...)

=5.71 amperes

(b) Transposingi= 8 .0(1−e

−t

CR)

gives

i

8. 0

= 1 −e

−t

CR

from which, e

−t

CR= 1 −

i

8. 0

=

8. 0 −i

8. 0

Taking the reciprocal of both sides gives:

e

t

CR=

8. 0

8. 0 −i

Taking Napierian logarithms of both sides gives:

t

CR

=ln

(

8. 0

8. 0 −i

)

Hence

t=CRln

(

8. 0

8. 0 −i

)

=(16× 10 −^6 )(25× 103 )ln

(

8. 0

8. 0 − 6. 0

)

wheni= 6 .0 amperes,

i.e. t=

400

103

ln

(

8. 0

2. 0

)

= 0 .4ln4. 0

= 0 .4(1. 3862943 ...)= 0 .5545 s

=555 ms, to the nearest millisecond

A graph of current against time is shown in

Fig. 4.8.

0.5 1.0 t(s)

0.555

i = 8.0 (1−e−t/CR)

8

6

4

2

0

5.71

i (A)

1.5

Figure 4.8

Problem 27. The temperatureθ 2 of a winding

which is being heated electrically at timetis

given by:θ 2 =θ 1 (1−e

−t

τ) whereθ 1 is the tem-

perature (in degrees Celsius) at timet=0 andτ

is a constant. Calculate,

(a)θ 1 , correct to the nearest degree, whenθ 2 is

50 ◦C,tis 30 s andτis 60 s

(b) the timet, correct to 1 decimal place, forθ 2

to be half the value ofθ 1.