Page 320 - Solutions Manual to accompany Electric Machinery Fundamentals

If switch S is shut, the charging time constant is increased, and the DIAC fires later in each half cycle.
1
Therefore, less power is supplied to the load.
S1-4. What would the rms voltage on the load in the circuit in Figure S1-1 be if the firing angle of the SCR
were (a) 0°, (b) 30°, (c) 90°?
SOLUTION The input voltage to the circuit of Figure S1-1 is
v ac  t  339 sin t  , where   377 rad/s

Therefore, the voltage on the secondary of the transformer will be

v ac  t  169 5 . sin t 

(a) The average voltage applied to the load will be the integral over the conducting portion of the half
cycle divided by /, the period of a half cycle. For a firing angle of 0°, the average voltage will be

/
1 T   1  /
V   v () t dt   V sin t dt    V cos t 
ave
T  M  M
0 0 0
1 2


V ave   V M     1 1   M V  0.637 169.5 V  108 V
(b) For a firing angle of 30°, the average voltage will be
/
1 T   1  /
V ave   v () t dt   V M sin t dt    V M cos t 
T  
 /6  / 6  /6
1  3  2  3


V ave  V M   1   M V  0.594 169.5 V  101 V
  2  2
(c) For a firing angle of 90°, the average voltage will be

/
1 T   1  /
V   v () t dt   V sin t dt    V cos t 
ave
T  /2   / 2 M  M  /2
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