Graphs reducing non-linear laws to linear form 153

Gradient of straight line,

lgb=

AB

BC

=

2. 13 − 1. 17

3. 0 − 1. 0

=

0. 96

2. 0

= 0. 48

Hence, b=antilog 0. 48 = 100.^48 =3.0, correct to 2
significant figures.

Vertical axis intercept, lga= 0. 70 ,from which

a=antilog 0. 70

= 100.^70 = 5. 0 ,correct to

2 significant figures.

Hence,the law of the graph isy= 5. 0 ( 3. 0 )x

(a) Whenx= 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

lg20=lg( 3. 0 )x=xlg3. 0

Hence,x=

lg20

lg3. 0

=

1. 3010

0. 4771

=2.73

Problem 7. The currentimA flowing in a

capacitor which is being discharged varies with

timetms, 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

formi=Iet/T,whereIandTare constants.

Determine the approximate values ofIandT

Taking Napierian logarithms of both sides of

i=Iet/T

gives lni=ln(Iet/T)=lnI+lnet/T

=lnI+

t

T

lne

i.e. lni=lnI+

t

T

since lne= 1

or lni=

(

1

T

)

t+lnI

which compares with y=mx+c

showing that lni is plotted vertically againstt hor-
izontally, with gradient^1 T and vertical-axis intercept
lnI.

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

lni 5.31 4.11 3.11 1.81 0.91 −0.49

A graph of lni against tis shown in Figure 18.7

and, since a straight line results,the lawi=Iet/Tis

verified.

A

B

C

ln

i

5.0

4.0

3.0

3.31

2.0

1.0

(^100200300400) t (ms)
D (200, 3.31)
2 1.0
1.30
0
Figure 18.7
Gradient of straight line,
1
T

AB
BC

5. 
   

   30 − 1. 30
   100 − 300

   
   


6. 0
   − 200
   =− 0. 02
   Hence, T=
   1
   − 0. 02
   =− 50
   Selecting any point on the graph, say pointD,where
   t=200 and lni= 3 .31, and substituting
   into lni=
   (
   1
   T
   )
   t+lnI
   gives 3. 31 =−
   1
   50
   ( 200 )+lnI
   from which, lnI= 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 isi= 1500 e−t/^50