Page 317 - Solutions Manual to accompany Electric Machinery Fundamentals

T/12
 V
vt M sin t
A
 
 V
vt M sin  t 2 /  3
B
 
 V
vt M sin  t 2 /  3
C
SOLUTION By symmetry, the rms voltage over the interval from 0 to T/12 will be the same as the rms
voltage over the whole interval. Over that interval, the output voltage is:
 2   2 
v  t  v C  t  v B  t  V M sin    t 3    V M sin     t  3  
 2 2    2  2  
vt   V M   sin t  cos 3  cos t  sin 3    V M   sin t  cos 3  cos t  sin 3  

 2 
vt   V M   2cos t sin 3    3cos t


Note that the period of the waveform is T  2/ , so T/12 is  /6 . The average voltage over the
interval from 0 to T/12 is

1 6  /6  6 3  /6
V DC   v () t dt   3V M cos  t dt  V M sin t 
T   0
0
33
V DC  V  M 1.6540 V
M

The rms voltage is

1 6  /6 
2
t
V   v () t 2 dt   3V 2 cos  dt
rms M
T 
0
311