Page 336 - Solutions Manual to accompany Electric Machinery Fundamentals

v  v d
C 1  C v  0
R dt C
d v  1 v  1 v
dt C RC C RC 1

d v  1 v  V M sin t 
dt C RC C RC
The solution can be divided into two parts, a natural response and a forced response. The natural
response is the solution to the equation
d v  1 v  0
dt C RC C
The solution to the natural response equation is

t

v  t  A e RC
,
Cn
where the constant A must be determined from the initial conditions in the system. The forced response is
the steady-state solution to the equation
d v  1 v  V M sin t 
dt C RC C RC
It must have a form similar to the forcing function, so the solution will be of the form


v Cf  t  B 1 sin t B 2 cos t
,
where the constants B and B must be determined by substitution into the original equation. Solving for
1
2
B and B yields:
2
1
d  sin  B cos t B 1  t  sin t   B cos t B  V M sin t
dt 1 2 RC 1 2 RC
  cos   B s in t  B  1  t  sin t B  cos  t  B V M sin t
2
1
RC 1 2 RC
cosine equation:

1
 B  B  0  B   RC B
1 2 2 1
RC
sine equation:
1 V
  B  B  M
2 1
RC RC
1 V
 2 RC B  B  M
1
RC 1 RC
 2 1  V M
   RC  RC   B  1 RC

 1  2 R C   2 2 V M
  RC   B  RC
1

Finally,
330