Introduction to Electric Circuits

8.4 The Laplace transform 197

Equating the coefficients of s 1 we have that

I=A+B

Equating the coefficients of s o we have that

3=2A+B

By subtraction, A = 2; by substitution, 1 = 2 + B ~ B = -1. Therefore

i(s) = [2/(s + 1)]- [1/(s + 2)]

From the Table 8.1 we see that the inverse transform of 2/(s + 1) is 2 exp (-t).
This is from pair number 1 with a = 1. Using the same inverse pair with a - 2
we see that the inverse transform of 1/(s + 2) is exp (-2t). Thus the time
response of the circuit is given by i(t) - (V/R)[2 exp (-t) - exp (-20].

Example 8.12

Obtain an expression for the current in the circuit of Fig. 8.33 following the

closing of switch S.

o'~ i(t)s....o__~ R L Ls

' ~! I 1

i(s) R

I

V

Figure 8.33 Figure 8.34

Solution

The circuit equation is V = iR + Ldi/dt, and earlier in the chapter we solved

this equation for i and obtained the result shown in Equation (8.2). We will now

obtain the same result using the Laplace transform method. First we obtain the

transform circuit.

At the instant of closing the switch S, the current is zero so that there is no

LI(O) source and the transform circuit takes the form shown in Fig. 8.34.

The closing of the switch is equivalent to there being a step function of

amplitude V applied, the transform of which is V/s (pair 3 from Table 8.1 with

A = V). From this circuit we see that

i(s) = (V/s)/(R + Ls) = V/sIR(1 + (L/R)s)]

Multiplying numerator and denominator by R/L we have

i(s) = V(R//L)//s[R(1 + (L//R)s)][R//L] = V(R/L)//Rs[s + (R/L)]

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