Page 330 - Solutions Manual to accompany Electric Machinery Fundamentals

Solution

(a) When the SCR is turned on, it will remain on until the current flowing through it drops below I .
H
This happens when the capacitor charges up to a high enough voltage to decrease the current below I . If
H
we ignore resistor R (because it is so much larger than R LOAD ), the capacitor charges through resistor
R LOAD with a time constant  LOAD = R LOAD C = (250 )(150 F) = 0.0375 s. The equation for the voltage
on the capacitor as a function of time during the charging portion of the cycle is

t

 
vt A B e R LOAD C
C
where A and B are constants depending upon the initial conditions in the circuit. Since v (0) = 0 V and
C
v () = V DC , it is possible to solve for A and B.
C
A = v () = V DC
C
A + B = v (0) = V DC  B = -V DC
C
Therefore,

 t
 V
vt DC  V DC e R LOAD C V
C
The current through the capacitor is
d
 
it C v  t
C
dt C
d   t 
 
it C V   V e R LOAD C 
C
dt  DC DC   
V  t
 
it DC e R LOAD C A
C
R LOAD
Solving for time yields
 R
it it
  R
t  R C ln C 2   0.0375 ln C 2
LOAD
V V
DC DC
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