0195136047.pdf

3.2 TRANSIENTS IN CIRCUITS 127

Since we know that the inductor current cannot change its value instantaneously, as otherwise
the inductor voltagevL(t)=LdiL/dtwould become infinite, it follows that the inductor current
immediately after closing the switch,iL( 0 +), must be the same as the inductor current just before
closing the switch,iL( 0 −),

iL( 0 +)=iL( 0 −) (3.2.17)

In Figure 3.2.1 note thatiL( 0 −)is zero. While the inductor current satisfies Equation (3.2.17), the
inductor voltage may change its value instantaneously, as we shall see in the following example.

EXAMPLE 3.2.1

Consider theRLcircuit of Figure 3.2.1 withR= 2 , L=5H,andv(t)=V =20 V (a dc
voltage source). Find the expressions for the inductor currentiL(t)and the inductor voltagevL(t)
fort>0, and plot them.

Solution

It follows from Equation (3.2.16) that

i(t)=iL(t)=

[

iL

(

0 +

)

−iL,ss

(

0 +

)]

e−t/τ+iL,ss(t)

where the subscript ss denotes steady state, andτ=L/R.
In our caseiL,ss=V/R= 20 / 2 =10 A, andiL( 0 −)= 0 =iL( 0 +). Thus we have
iL(t)=( 0 − 10 )e−^2 t/^5 + 10 = 10 ( 1 −e−^2 t/^5 )A, fort> 0

Then the inductor voltage is obtained as

vL(t)=L

diL

dt

=+ 20 e−^2 t/^5 , fort> 0

iL(t)andvL(t) are plotted in Figure E3.2.1. The following points are worth noting.

1. iL( 0 −)=iL( 0 +)=0; the inductor current cannot change instantaneously.


2. vL( 0 −)=0 andvL( 0 +)=20; the inductor voltage has changed instantaneously.

05

t = 2.5

10(1 − e−^1 ) = 6.32

10(1 − e−^2 t/5)

iL(t), A

t, s

8.64

T = 2.5

2 T

10

(a)

2

4

6

8

10

15

Figure E3.2.1(a)Inductor current

iL(t)= 10 ( 1 −e−^2 t/^5 )A, fort>

0. (b) Inductor voltagevL(t) =
   20 e−^2 t/^5 V, fort>0.