Solution:

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$$i(t) = \frac{V_s}{R} + \left(I_0 - \frac{V_s}{R}\right)e^{-t/\tau}$$

\n(7.60)

This is the complete response of the RL circuit. It is illustrated in Fig. 7.49. The response in Eq. (7.60) may be written as

$$i(t) = i(\infty) + [i(0) - i(\infty)]e^{-t/\tau}$$

(7.61)

Figure 7.49 Total response of the RL circuit with initial inductor current I0.

where i(0) and i(∞) are the initial and final values of i, respectively. Thus, to find the step response of an RL circuit requires three things:

• 1. The initial inductor current i(0) at t = 0. 2. The final inductor current i(∞).
• 3. The time constant τ.

We obtain item 1 from the given circuit for t < 0 and items 2 and 3 from the circuit for t > 0. Once these items are determined, we obtain the response using Eq. (7.61). Keep in mind that this technique applies only for step responses.

Again, if the switching tak es place at time t = t0 instead of t = 0, Eq. (7.61) becomes

$$i(t) = i(\infty) + [i(t_0) - i(\infty)]e^{-(t-t_0)/\tau}$$

\n(7.62)

If I0 = 0, then

$$i(t) = \begin{cases} 0, & t < 0 \\ \frac{V_s}{R}(1 - e^{-t/\tau}), & t > 0 \end{cases}$$

(7.63a)

or

$$i(t) = \frac{V_s}{R} (1 - e^{-t/\tau}) u(t)$$

\n(7.63b)

This is the step response of the RL circuit with no initial inductorcurrent. The v oltage across the inductor is obtained from Eq. (7.63) using v = L di∕dt. We get

$$v(t) = L\frac{di}{dt} = V_s \frac{L}{\tau R} e^{-t/\tau}, \qquad \tau = \frac{L}{R}, \qquad t > 0$$

or

$$v(t) = V_s e^{-t/\tau} u(t) \tag{7.64}$$

Figure 7.50 shows the step responses in Eqs. (7.63) and (7.64).

Step responses of an RL circuit with no initial inductor current: (a) current response, (b) voltage response.

For Example 7.12.

Find i(t) in the circuit of Fig. 7.51 for t > 0. Assume that the switch has been closed for a long time.

Solution:

When t < 0, the 3-Ω resistor is short-circuited, and the inductor acts like a short circuit. The current through the inductor at t = 0⁻ (i.e., just before t = 0) is

$$i(0^{-}) = \frac{10}{2} = 5 \text{ A}$$

Because the inductor current cannot change instantaneously,

$$i(0) = i(0^+) = i(0^-) = 5 \text{ A}$$

When t > 0, the switch is open. The 2- and 3- Ω resistors are in series, so that

$$i(\infty) = \frac{10}{2+3} = 2 \text{ A}$$

The Thevenin resistance across the inductor terminals is

$$R_{\rm Th} = 2 + 3 = 5 \ \Omega$$

For the time constant,

^τ = ___^L RTh = _1 3 __ 5 ⁼___1 15 s

Thus,

$$i(t) = i(\infty) + [i(0) - i(\infty)]e^{-t/\tau}$$

= 2 + (5 - 2)e^-15t = 2 + 3e^-15t A, t > 0

Check: In Fig. 7.51, for t > 0, KVL must be satisfied; that is,

$$10 = 5i + L\frac{di}{dt}$$ $$5i + L\frac{di}{dt} = [10 + 15e^{-15t}] + \left[\frac{1}{3}(3)(-15)e^{-15t}\right] = 10$$

This confirms the result.

At t = 0, switch 1 in Fig. 7.53 is closed, and switch 2 is closed 4 s later. Example 7.13 Find i(t) for t > 0. Calculate i for t = 2 s and t = 5 s.

Figure 7.53 For Example 7.13.

Solution:

We need to consider the three time intervals t ≤ 0, 0 ≤ t ≤ 4, and t ≥ 4 separately. For t < 0, switches S1 and S2 are open so that i = 0. Since the inductor current cannot change instantly,

$$i(0^-) = i(0) = i(0^+) = 0$$

For 0 ≤ t ≤ 4, S1 is closed so that the 4- and 6- Ω resistors are in series. (Remember, at this time, S2 is still open.) Hence, assuming for now that S1 is closed forever,

$$i(\infty) = \frac{40}{4+6} = 4 \text{ A}, \qquad R_{\text{Th}} = 4+6 = 10 \text{ }\Omega$$ $$\tau = \frac{L}{R_{\text{Th}}} = \frac{5}{10} = \frac{1}{2} \text{ s}$$

Thus,

$$i(t) = i(\infty) + [i(0) - i(\infty)]e^{-t/\tau}$$

= 4 + (0 - 4)e^-2t = 4(1 - e^-2t) A, 0 \le t \le 4

For t ≥ 4, S2 is closed; the 10-V voltage source is connected, and the circuit changes. This sudden change does not affect the inductor current because the current cannot change abruptly. Thus, the initial current is

$$i(4) = i(4^-) = 4(1 - e^{-8}) \simeq 4 \text{ A}$$

To find i(∞), let v be the voltage at node P in Fig. 7.53. Using KCL,

$$\frac{40 - v}{4} + \frac{10 - v}{2} = \frac{v}{6} \implies v = \frac{180}{11} \text{ V}$$ $$i(\infty) = \frac{v}{6} = \frac{30}{11} = 2.727 \text{ A}$$

The Thevenin resistance at the inductor terminals is

$$R_{\text{Th}} = 4||2 + 6 = \frac{4 \times 2}{6} + 6 = \frac{22}{3} \,\Omega$$

and

$$\tau = \frac{L}{R_{\text{Th}}} = \frac{5}{\frac{22}{3}} = \frac{15}{22} \text{ s}$$

Hence,

$$i(t) = i(\infty) + [i(4) - i(\infty)]e^{-(t-4)/\tau}, \qquad t \ge 4$$

We need (t − 4) in the exponential because of the time delay. Thus,

$$i(t) = 2.727 + (4 - 2.727)e^{-(t-4)/\tau}, \qquad \tau = \frac{15}{22}$$ $$= 2.727 + 1.273e^{-1.4667(t-4)}, \qquad t \ge 4$$

Putting all this together,

$$i(t) = \begin{cases} 0, & t \le 0 \\ 4(1 - e^{-2t}), & 0 \le t \le 4 \\ 2.727 + 1.273e^{-1.4667(t-4)}, & t \ge 4 \end{cases}$$

At t = 2,

At t = 5,

$$i(2) = 4(1 - e^{-4}) = 3.93 \text{ A}$$ $$i(5) = 2.727 + 1.273e^{-1.4667} = 3.02$$

A

Switch S1 in Fig. 7.54 is closed at t = 0, and switch S2 is closed at t = 2s.

Answer: ⁱ(t) = { 0, 4(1 − e⁻9*^t* ), 7.2 − 3.2 e⁻5(t−2) , t < 0 0 < t < 2 t > 2 i(1) = 4 A, i(3) = 7.178 A.

Calculate i(t) for all t. Find i(1) and i(3).

Figure 7.54 For Practice Prob. 7.13.