Introduction to Electric Circuits

8.4 The Laplace transform 199

1~ Cs /// (= 1/0.1s). The transform circuit thus takes the form shown in Fig. 8.36.
From this circuit we see that

i(s) = [(100/s) - (20/s)]/[R + (1/Cs)] = 80/s[R + (1/Cs)]

Now

Vc(S) = i(s)(1/Cs) = 80/{s[R + (1/Cs)lCs} = 80/s(CRs + 1)

Multiplying numerator and denominator by 1/CR we get

Vc(S) = 80(1/CR)/s[s + (1/CR)]

Let

(1/CR)/s[(s + 1/CR)] : [A/s] + [B/{s + (1/CR)}]

Then

1/CR - A[s + (1/CR)] + Bs =- (A + B)s + A/CR

Equating the coefficients of s we have that A + B = 0 ~ A = -B. Equating the
coefficients of s o we have that 1/CR = A/CR ~ A = 1 and B = -A = - 1.

Thus

Vc(S) = 8011/s - 1//(s + 1/CR)]

From Table 8.1 we see that 1Is is the transform of 1 and that 1/(s + 1/CR)is

the transform of exp (-t/CR)(transform pair number 1 with a = 1/CR). In the

time domain, therefore

Vc(t) - 8011 - exp (-t/CR)] = 8011- exp (-t/0.5)]

When Vc has reached 40 V (i.e. it is double its initial value) we have

40 : 8011 - exp(-t/0.5)l

exp (-2t) = 0.5

Therefore

-2t=ln0.5 and t=0.345s

Example 8.14

An inductor of 250 IxH inductance is energized from a 1 kV d.c. supply via two

thyristors connected in series. The thyristors are identical except that one of

them has a delay at turn-on of a few microseconds longer than the other.

Voltage sharing is assisted by placing an RC circuit in parallel with each

thyristor. Obtain an expression for the current in the RC circuit connected

across the 'slow' thyristor after the other one has turned on.