0195136047.pdf

1.1 ELECTRICAL QUANTITIES 9

EXAMPLE 1.1.2

Figure E1.1.2 (a) gives a plot ofq(t)as a function of timet.

(a) Obtain the plot ofi(t).

(b) Find the average value of the current over the time interval of 1 to 7 seconds.

3

t, seconds

− 1

(^012345678910)
q(t), coulombs
(a)
t, seconds
i(t), amperes
−2.0
(b)
1.0
1.5
123 45 678910
Figure E1.1.2(a)Plot ofq(t).
(b)Plot ofi(t).
Solution
(a) Applying Equation (1.1.5) and interpreting the first derivative as the slope, one obtains
the plot shown in Figure E1.1.2(b).
(b)Iav=( 1 /T )
∫T
0 idt. Interpreting the integral as the area enclosed under the curve, one
gets:
Iav=
1
( 7 − 1 )
[( 1. 5 × 2 )−( 2. 0 × 2 )+( 0 × 1 )+( 1 × 1 )]= 0
Note that the net charge transferred during the interval of 1 to 7 seconds is zero in this case.