0195136047.pdf

130 TIME-DEPENDENT CIRCUIT ANALYSIS

vC(t)andiC(t) are plotted in Figures E3.2.2(b) and (c). The following points are noteworthy.

1.vC( 0 +)=vC( 0 −); the capacitor voltage cannot change instantaneously.

2.iC( 0 −)=0 andiC( 0 +)=10 A; the capacitor current has changed instantaneously.

3. After one time constant has elapsed, the capacitor voltage has risen to 63% of its final
   (steady-state) value. After five time constants, it would reach 99% of its final value.


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

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

vR(t)=vC(t)= 20

(

1 −e−t/^10

)

V, fort> 0

iR(t)=vR(t)/R= 10

(

1 −e−t/^10

)

A, fort> 0

Note that the KCL equationi(t)=I=iC(t)+iR(t)=10 is satisfied. The charge transferred to

the capacitor at steady-state is:

Qss=CVC,ss= 5 × 20 = 100 C

Examples 3.2.1 and 3.2.2 are chosen with zero initial conditions. Let us now consider nonzero

initial conditions for circuits still containing only one energy-storage element,LorC.

EXAMPLE 3.2.3

For the circuit of Figure E3.2.3(a), obtain the complete solution for the currentiL(t)through the

5-H inductor and the voltagevx(t) across the 6-resistor.

t = 0

(a)

iL(t)

vL(t)

4 Ω

S

5 Ω

6 Ω

10 V

5 A

5 H

b

a

−

+

+

−

vx(t)

+

−

Figure E3.2.3

(b)

4 Ω

S

5 Ω

6 Ω

RTh =

4 + 6 = 10 Ω

Open

Short

b

a