Page 240 - Basic Electrical Engineering
P. 240
−1
then i = I (1 − e )
0
= I (1 − 0.368)
0
= 0.632 I 0
Thus, i = 63.2 per cent of I . This has been shown in Fig. 2.161.
o
Now let us consider what happens when the switch S in Fig. 2.161 is
changed to position 2 as shown. We are of course assuming that the switch
was in position 1 and current had attained its steady-state value of I A. From
0
position 1, the switch is moved to position 2. Applying KVL we can write the
voltage equation as
Integrating with respect to time t
The value of K is determined by applying the initial conditions. Here the
initial condition is that at
Therefore,
Substituting the value of K in (iii)
