Page 54 - Circuit Analysis II with MATLAB Applications

Chapter 1 Second Order Circuits

syms t k2
v0=100+exp( 25*t)*(k2*t 100); v1=diff(v0)

v1 =
-25*exp(-25*t)*(k2*t-100)+exp(-25*t)*k2

Thus,
dv – 25t – 25t
C
-------- = k e 25e – k t – 100
dt 2 2
and
dv
-------- C = k + 2500 (3)
2
dt
t = 0
dv i i
C
C
Also, -------- = ---- = ---- L and at t = 0
dt C C

dv i 0
L
C
-------- = --------------- = 0 (4)
dt C
t = 0
From (3) and (4) k + 2500 = 0 or k = – 2500 and by substitution into (2)
2
2
–
v t 100 e – 25t = 2500t + 100 (5)
C
We find i t = i t by differentiating (5) and multiplication by . Using MATLAB we get:
C
L
C
syms t
C=8*10^( 3);
i0=C*(100-exp( 25*t)*(100+2500*t)); iL=diff(i0)
iL =
1/5*exp(-25*t)*(100+2500*t)-20*exp(-25*t)

Thus,
i t = i t = 0.2e – 25t 100 + 2500t – 20e – 25t

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

1-42 Circuit Analysis II with MATLAB Applications
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