Tutorials - Digital Signal Processing: An Introduction with MATLAB and Applications - page 251

Solution: Let the capacitor initial voltage be V o (volts). After the switch is closed,
a d.c. voltage of V volts will suddenly be applied, hence, v i (t) = Vu(t) volts.

i ð t Þ¼ i C ð t Þ¼ C dv C ð t Þ
dt

L

Tables I ð s Þ¼ C ½ sV C ð s Þ v C ð 0 Þ¼ C ½ sV C ð s Þ V o ;

|{z}
current through C

or simply:

I ¼ C ½ sV C V o ð 1 Þ

v c ¼ v i iR L V C ¼ V i IR ¼ V = s IR

ð 2 Þ

From ( 1 ), ( 2 ) we get:

I ¼ 1
R

V V o
s þ 1 = RC ¼ I o

1
s þ 1 = RC

ð 3 Þ

[where I o = (V - V o )/R.]
From Tables: i(t) = I o e -t/RC u(t) amperes.
Substituting ( 3 )in( 2 ) we get: V C ¼ V ½ s

s þ 1 = RC þ V o

1

s þ 1 = RC
From Tables: v c (t) = [V(1 - e -t/RC )+V o e -t/RC ]u(t) volts.

Tutorial 18

Q: The switch S in Fig. 1 below has been closed for a long time. Find the current
across the capacitor after the switch is opened at t = 0.

+ R 1 − + R 2 −

ν c

+

+

i ( t )

ν i

V (d.c.)

S

C

L

_

_

_

ν L

+

(1)

Solution: As the switch was closed for a long time, both v C (0 - ) and i L (0 - ) are
zero. Even if there was an initial voltage V o on C and an initial current I o in L when
S was closed, the final current and voltages in the right loop will all be zero. This is
due to the consumption of energy by R 2 . To prove this, consider the right loop
where the switch has been closed as shown in Fig. 2 .