Page 323 - Solutions Manual to accompany Electric Machinery Fundamentals

Therefore, the forced solution to the equation is
V  RC V
v Cf  t  1   2 M RC 2 sin t   1   2 R C 2 cos t 
M
,
2
2
and the total solution is
 
vt v C ,n  t  v C , f  t
C
t
 V  RC V
 
M
vt Ae RC  1   2 M RC 2 sin t   1   2 RC 2 cos t 
C
2
2
The initial condition for this problem is that the voltage on the capacitor is zero at the beginning of the
half-cycle:
0 V  RC V


v  0  Ae RC  M sin  0 M cos 0  0
C
1   2 RC 2 1   2 R C 2
2
2
 RC V
A M  0
2

1  2 RC 2
 RC V
A  M
2
1  2 R C 2

Therefore, the voltage across the capacitor as a function of time before the DIAC fires is
 RC V  t V  RC V
 
vt M e RC  M sin t   M cos 
t
1   RC 1   RC 1   R C
C 2 2 2 2 2 2 2 2 2
If we substitute the known values for R, C, , and V , this equation becomes
M
 
vt 42e  100t  11.14 sin t  42 cos t
C
This equation is plotted below:

It reaches a voltage of 30 V at a time of 3.50 ms. Since the frequency of the waveform is 60 Hz, the
waveform there are 360 in 1/60 s, and the firing angle  is

   3.50 ms    1/60 s 360    75.6 or 1.319 radians

317