Introduction to Electric Circuits

198 Transient analysis

Let

(R/Ll/ [s + (R/Ell : [A/sl + + (R/L)}I

Then

R/L- A[s + (R/L)] + Bs

so that

(R//L)s ~ (A + B)s + (AR//L)s ~

Equating the coefficients of s we have that 0 - A + B ~ B - -A. Equating the

coefficients of s o we have that R/L-AR/L~A-1 and B--A--1.

Thus

(V/R){(R/L)/[s{s + (R/L)}]}- (V/R){[1/s] - [1/(s + (R/L))]}

From Table 8.1 we see that 1Is is the transform of i (transform pair number 1)

and that 1/[s + (R/L)] is the transform of exp (-Rt/L) (transform pair

number 1 with a = R/L). In the time domain, therefore, the current is given by

i(t) = (V/R){1 - exp (-Rt/L)} and since V/R - I,

i(t) - 111 = exp (-Rt/L)]

This is the same result as that obtained in Equation (8.2).

Example 8.13

In the circuit shown in Fig. 8.35, the capacitor is initially charged to 20 V when

the switch is closed at an instant t = 0. Obtain an expression for the voltage (Vc)

across the capacitor as a function of time. Hence determine the time taken for

vc to double its initial value.

i(t) ~ R = 5~ ,(s), R 1/Cs

o S"-o! I --~--I! I I

,00vC)

C=0.1F

(^11) II
20V
100
l
Figure 8.35 Figure 8.36
( )T^20
Solution
First we obtain the transform circuit. The step voltage of amplitude 100 V
resulting from the closing of the switch S transforms to 100/s (Table 8.1,
transform pair number 3 with A - 100). The capacitor is initially charged to
20 V so that there will be a source Vc(0)/s (=20/s)in series with the element