0195136047.pdf

128 TIME-DEPENDENT CIRCUIT ANALYSIS

05

20 e−^1 = 7.36

20 e−^2 t/5

vL(t), V

t, s

T = 2.5^10

(b)

20

15

Figure E3.2.1Continued

3. After one time constant has elapsed, the inductor current has risen to 63% of its final or
   steady-state value. After five time constants, it would reach 99% of its final value.


4. After one time constant, the inductor voltage has decayed to 37% of its initial value. The
   inductor voltage, in this example, eventually decays to zero as it should, since in the steady
   state the inductor behaves as a short circuit for dc sources.

The resistor voltage and resistor current, if needed, can be found as follows:

iR(t)=iL(t)=i(t)= 10

(

1 −e−^2 t/^5

)

A, fort> 0

vR(t)=RiR(t)= 20

(

1 −e−^2 t/^5

)

V, fort> 0

Note that the KVL equationv(t)=V=vR(t)+vL(t)=20 is satisfied.

EXAMPLE 3.2.2

Consider theRCcircuit of Figure E3.2.2(a) withR= 2 , C=5F,andi(t)=I=10 A (a

dc current source). Find the expressions for the capacitor voltagevC(t) and the capacitor current

iC(t) fort>0, and plot them.

Solution

The KCL equation at the upper node fort>0 is given by

iC+iR=i(t), or C

dvC(t)

dt

+

vC(t)

R

=i(t) or

dvC(t)

dt

+

1

RC

vC(t)=

i(t)

C

This equation is the same in form as Equation (3.2.3). The total solution, which is of the form of

Equation (3.2.16), can be written as

vC(t)=

[

vC

(

0 +

)

−vC,ss

(

0 +

)]

e−t/(RC)+vC,ss(t)

whereRCis the time constantT, which is 10 s in this case.vC,ss(t)=RI=20 V in our case,

since the source is dc and the capacitor currentiC=C(dvC/dt)=0 in the steady state fort>0,

denoting an open circuit. Then the solution is given by

vC(t)=

[

vC

(

0 +

)

− 20

]

e−t/^10 +20 V, fort> 0