0195136047.pdf

194 TIME-DEPENDENT CIRCUIT ANALYSIS

where VSIN is the name of the source, NODE-

PLUS is the number identifying the positive node,

NODEMINUS is the number identifying the nega-

tive node, VDC is the dc offset, VPEAK is the peak

value of the ac component, FREQ is the frequency

in hertz, TD is the time delay in seconds, DF is a

damping factor, and PHASE is the phase angle in

degrees. The voltage generated is then given by

v(t)=VDC+VPEAK sin[PHASE],

for 0<t<TD

=VDC+VPEAK sin[2π

FREQ(t−TD)+PHASE]

×exp[−DF(t−TD)],

for TD<t

Also note that DF=0 for a constant-amplitude

sinusoid.

*3.5.3With the initial voltage across the capacitor being
vC( 0 )=10 V, obtain a plot ofvC(t)for the circuit
of Figure P3.5.3 by writing a PSpice program.
Hint:Use the following analysis request:

• TRAN 0.02 10 0 0.02 UIC
  3.5.4For the circuit shown in Figure P3.5.4, solve for
  i(t)by writing and executing a PSpice program,
  and obtain a plot ofi(t)by using PROBE.
  Take the hint from Problem 3.5.2 for specifying the
  sinusoidal voltage source in the PSpice program.
  Also use the following analysis request:


• TRAN 0.1 MS 100 MS 0 0.1 MS UIC
  3.5.5A dc source is connected to a seriesRLCcircuit by
  a switch that closes att=0, as shown in Figure
  P3.5.5, with initial conditions. For the values of
  R=20, 40, and 80, solve forvC(t)by writing
  and executing a PSpice program, and obtain plots
  by using PROBE.
  Hint:Use the following analysis request:


• TRAN IUS 2MS 0 1US UIC
  3.5.6In the circuit of Figure P3.5.6, solve for phasors
  V ̄ 1 andV ̄ 2 with peak magnitudes by writing and
  executing a PSpice program.

1 mA t^ =^0 v(t) R = 10 kΩ C = 1 μF

+

−

Figure P3.5.1

vs(t) = 2 sin(200 t)

1 R^ = 5 kΩ

i(t) C = 1 μF vC(t); vC(0) = 1 V

+

+

−

−

O

2 Figure P3.5.2

1

C = 1 μF R = 2 MΩ

vC(t) = v 1

vC(0) = 10 V

+

−

O

Figure P3.5.3