Page 57 - Circuit Analysis II with MATLAB Applications

Solutions to Exercises

From (3) and (4)
3ksin M = 40 (5)
and from (2) and (5)
3ksin M 40
----------------- = ------------
kcos M – 100
3tan M = – 0.4

M tan – 1 – = 0.4 3 = – 0.1326 rad = – 7.6q
e

The value of can be found from either (2) or (5). From (2)
k
kcos – 0.1236 = – 100

–
100
k = --------------------------------- = – 100.8
cos – 0.1236
and by substitution into (1)

v t = 100 100.8e – 0.4t cos 3t – 7.6q (6)
–
C
Since i t = i t = Cdv e dt , we use MATLAB to differentiate (6).

C
L
C
syms t; vC=100 100.8*exp( 0.4*t)*cos(3*t 0.1326); C=0.02183; iL=C*diff(vC)
iL =
137529/156250*exp(-2/5*t)*cos(3*t-663/5000)+412587/62500*exp(-

2/5*t)*sin(3*t-663/5000)
137529/156250, 412587/62500

ans =
0.8802
ans =
6.6014
i t = 0.88e – 0.4t cos 3t – 7.6q + 6.6e – 0.4t sin 3t – 7.6q

L

The plots for v t and i t are shown on the next page.
L
C

1-45 Circuit Analysis II with MATLAB Applications
Orchard Publications