7.3 The Source-Free RL Circuit

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7.3 The Source-Free RL Circuit

Consider the series connection of a resistor and an inductor, as shown in Fig. 7.11. Our goal is to determine the circuit response, which we will assume to be the current i(t) through the inductor. We select the induc tor current as the response in order to take advantage of the idea that the inductor current cannot change instantaneously. At t = 0, we assume that the inductor has an initial current I0, or

$$i(0) = I_0 \tag{7.13}$$

with the corresponding energy stored in the inductor as

$$w(0) = \frac{1}{2} L I_0^2 \tag{7.14}$$

Applying KVL around the loop in Fig. 7.11,

$$v_L + v_R = 0 \tag{7.15}$$

But vL = L di∕dt and vR = iR. Thus,

$$L\frac{di}{dt} + Ri = 0$$

Figure 7.11 A source-free RL circuit.

$$\frac{di}{dt} + \frac{R}{L}i = 0\tag{7.16}$$

Rearranging terms and integrating gives

$$\int_{I_0}^{i(t)} \frac{di}{i} = -\int_0^t \frac{R}{L} dt$$

\n

$$\ln i \Big|_{I_0}^{i(t)} = -\frac{Rt}{L} \Big|_0^t \implies \ln i(t) - \ln I_0 = -\frac{Rt}{L} + 0$$

or

$$\ln \frac{i(t)}{I_0} = -\frac{Rt}{L} \tag{7.17}$$

Taking the powers of e, we have

$$i(t) = I_0 e^{-Rt/L}$$

\n(7.18)

This shows that the natural response of the RL circuit is an e xponential decay of the initial current. The current response is shown in Fig. 7.12. It is evident from Eq. (7.18) that the time constant for the RL circuit is

$$\tau = \frac{L}{R} \tag{7.19}$$

with τ ag ain ha ving the unit of seconds. Thus, Eq. (7.18) may be written as

$$i(t) = I_0 e^{-t/\tau}$$

\n(7.20)

With the current in Eq. (7.20), we can find the voltage across the resistor as

$$v_R(t) = iR = I_0 Re^{-t/\tau}$$

(7.21)

The power dissipated in the resistor is

$$p = v_R i = I_0^2 Re^{-2t/\tau}$$

(7.22)

The energy absorbed by the resistor is

$$w_R(t) = \int_0^t p(\lambda) d\lambda = \int_0^t I_0^2 Re^{-2\lambda/\tau} d\lambda = -\frac{\tau}{2} I_0^2 Re^{-2\lambda/\tau} \Big|_0^t, \qquad \tau = \frac{L}{R}$$

or

$$w_R(t) = \frac{1}{2} L \, I_0^2 (1 - e^{-2t/\tau}) \tag{7.23}$$

Figure 7.12 shows an initial slope interpretation may be given to τ.

Note that as t → ∞,wR(∞) → _1 2 L I² 0, which is the same as wL(0), the initial energy stored in the inductor as in Eq. (7.14). Again, the energy initially stored in the inductor is eventually dissipated in the resistor.

The smaller the time constant τ of a circuit, the faster the rate of decay of the response. The larger the time constant, the slower the rate of decay of the response. At any rate, the response decays to less than 1 percent of its initial value (i.e., reaches steady

state) after 5τ.

In summary:

The Key to Working with a Source-Free RL Circuit Is to Find:

• 1. The initial current i(0) = I0 through the inductor.
• 2. The time constant τ of the circuit.

With the tw o items, we obtain the response as the inductor current iL(t) = i(t) = i(0)e⁻t∕^τ . Once we determine the inductor current iL, other variables (inductor voltage vL, resistor voltage vR, and resistor current iR) can be obtained. Note that in general, R in Eq. (7.19) is the Thevenin resistance at the terminals of the inductor.

Assuming that i(0) = 10 A, calculate i(t) and ix(t) in the circuit of Example 7.3 Fig. 7.13.

Solution:

There are two ways we can solve this problem. One way is to obtain the equivalent resistance at the inductor terminals and then use Eq. (7.20). The other way is to start from scratch by using Kirchhoff’s voltage law. Whichever approach is taken, it is always better to first obtain the inductor current.

■ METHOD 1 The equivalent resistance is the same as the Thevenin resistance at the inductor terminals. Because of the dependent source, we insert a voltage source with vo = 1 V at the inductor terminals a-b, as in Fig. 7.14(a). (We could also insert a 1-A current source at the terminals.) Applying KVL to the two loops results in

$$2(i_1 - i_2) + 1 = 0 \qquad \Rightarrow \qquad i_1 - i_2 = -\frac{1}{2} \tag{7.3.1}$$ $$6i_2 - 2i_1 - 3i_1 = 0 \qquad \Rightarrow \qquad i_2 = \frac{5}{6}i_1 \tag{7.3.2}$$

Substituting Eq. (7.3.2) into Eq. (7.3.1) gives

$$i_1 = -3$$

A, $i_o = -i_1 = 3$ A

Figure 7.14 Solving the circuit in Fig. 7.13. When a circuit has a single inductor and several resistors and dependent sources, the Thevenin equivalent can be found at the terminals of the inductor to form a simple RL circuit. Also, one can use Thevenin’s theorem when several inductors can be combined to form a single equivalent inductor.

Hence,

The time constant is

$$R_{\text{eq}} = R_{\text{Th}} = \frac{v_o}{i_o} = \frac{1}{3} \Omega$$

^τ = ___^L Req = _1 2 _ _1 3 ⁼__3 2 s

Thus, the current through the inductor is

$$i(t) = i(0)e^{-t/\tau} = 10e^{-(2/3)t} \text{ A}, \qquad t > 0$$

■ METHOD 2 We may directly apply KVL to the circuit as in Fig. 7.14(b). For loop 1,

$$\frac{1}{2}\frac{di_1}{dt} + 2(i_1 - i_2) = 0$$

or

$$\frac{di_1}{dt} + 4i_1 - 4i_2 = 0 \tag{7.3.3}$$

For loop 2,

$$6i_2 - 2i_1 - 3i_1 = 0 \qquad \Rightarrow \qquad i_2 = \frac{5}{6}i_1 \tag{7.3.4}$$

Substituting Eq. (7.3.4) into Eq. (7.3.3) gives

$$\frac{di_1}{dt} + \frac{2}{3}i_1 = 0$$

Rearranging terms,

$$\frac{di_1}{i_1} = -\frac{2}{3}dt$$

Since i1 = i, we may replace i1 with i and integrate:

$$\ln i \Big|_{i(0)}^{i(t)} = -\frac{2}{3}t \Big|_0^t$$

or

$$\ln \frac{i(t)}{i(0)} = -\frac{2}{3}t$$

Taking the powers of e, we finally obtain

$$i(t) = i(0)e^{-(2/3)t} = 10e^{-(2/3)t} \text{ A}, \qquad t > 0$$

which is the same as by Method 1. The voltage across the inductor is

$$v = L \frac{di}{dt} = 0.5(10) \left( -\frac{2}{3} \right) e^{-(2/3)t} = -\frac{10}{3} e^{-(2/3)t}$$

V

Since the inductor and the 2-Ω resistor are in parallel,

$$i_x(t) = \frac{v}{2} = -1.6667e^{-(2/3)t} \text{A}, \qquad t > 0$$

Find i and

$$v_x$$

in the circuit of Fig. 7.15. Let $i(0) = 7$ A.

Answer: 7 e⁻2*^t* A, −7 e⁻2*^t* V, t > 0.

Practice Problem 7.3

The switch in the circuit of Fig. 7.16 has been closed for a long time. At Example 7.4 t = 0, the switch is opened. Calculate i(t) for t > 0.

Solution:

When t < 0, the switch is closed, and the inductor acts as a short circuit to dc. The 16-Ω resistor is short-circuited; the resulting circuit is shown in Fig. 7.17(a). To get i1 in Fig. 7.17(a), we combine the 4- Ω and 12-Ω resistors in parallel to get

$$\frac{4 \times 12}{4 + 12} = 3 \ \Omega$$

Hence,

$$i_1 = \frac{40}{2+3} = 8 \text{ A}$$

We obtain i(t) from i1 in Fig. 7.17(a) using current division, by writing

$$i(t) = \frac{12}{12 + 4} i_1 = 6 \text{ A}, \qquad t < 0$$

Since the current through an inductor cannot change instantaneously,

$$i(0) = i(0^-) = 6\,\mathrm{A}$$

When t > 0, the switch is open and the voltage source is disconnected. We now have the source-free RL circuit in Fig. 7.17(b). Combining the resistors, we have

$$R_{\text{eq}} = (12 + 4) || 16 = 8 \Omega$$

The time constant is

$$\tau = \frac{L}{R_{\text{eq}}} = \frac{2}{8} = \frac{1}{4} \,\text{s}$$

Thus,

$$i(t) = i(0)e^{-t/\tau} = 6e^{-4t} A$$

Figure 7.16 For Example 7.4.

Solving the circuit of Fig. 7.16: (a) for t < 0, (b) for t > 0.

For Practice Prob. 7.4.

Figure 7.19 For Example 7.5.

Figure 7.20

The circuit in Fig. 7.19 for: (a) t < 0, (b) t > 0.

Example 7.5 In the circuit shown in Fig. 7.19, find io, vo, and i for all time, assuming that the switch was open for a long time.

Solution:

It is better to first find the inductor current i and then obtain other quantities from it.

For t < 0, the switch is open. Since the inductor acts like a short circuit to dc, the 6-Ω resistor is short-circuited, so that we ha ve the circuit shown in Fig. 7.20(a). Hence, io = 0, and

$$i(t) = \frac{10}{2+3} = 2 \text{ A}, \qquad t < 0$$ $$v_o(t) = 3i(t) = 6 \text{ V}, \qquad t < 0$$

Thus, i(0) = 2.

For t > 0, the switch is closed, so that the v oltage source is shortcircuited. We now have a source-free RL circuit as shown in Fig. 7.20(b). At the inductor terminals,

$$R_{\text{Th}} = 3 \parallel 6 = 2 \text{ }\Omega$$

so that the time constant is

$$\tau = \frac{L}{R_{\text{Th}}} = 1 \text{ s}$$

Hence,

$$i(t) = i(0)e^{-t/\tau} = 2e^{-t}A, \qquad t > 0$$

Because the inductor is in parallel with the 6- and 3-Ω resistors,

$$v_o(t) = -v_L = -L\frac{di}{dt} = -2(-2e^{-t}) = 4e^{-t} \text{ V}, \qquad t > 0$$

and

$$i_o(t) = \frac{v_L}{6} = -\frac{2}{3}e^{-t}A, \qquad t > 0$$

Thus, for all time,

$$i_o(t) = \begin{cases} 0 \text{ A}, & t < 0 \\ -\frac{2}{3}e^{-t} \text{ A}, & t > 0 \end{cases}, \quad v_o(t) = \begin{cases} 6 \text{ V}, & t < 0 \\ 4e^{-t} \text{ V}, & t > 0 \end{cases}$$ $$i(t) = \begin{cases} 2 \text{ A}, & t < 0 \\ 2e^{-t} \text{ A}, & t \ge 0 \end{cases}$$

^o We notice that the inductor current is continuous at t = 0, while the (t) current through the 6-Ω resistor drops from 0 to −2∕3 at t = 0, and the voltage across the 3-Ω resistor drops from 6 to 4 at t = 0. We also notice that the time constant is the same regardless of what the output is defined to be. Figure 7.21 plots i and io.

Determine i, io, and vo for all t in the circuit shown in Fig. 7.22. Assume that the switch was closed for a long time. It should be noted that opening a switch in series with an ideal current source creates an infinite voltage at the current source terminals. Clearly this is impossible. For the purposes of problem solving, we can place a shunt resistor in parallel with the source (which now makes it a voltage source in series with a resistor). In more practical circuits, devices that act like current sources are, for the most part, electronic circuits. These circuits will allow the source to act like an ideal current source over its operating range but voltage-limit it when the load resistor becomes too large (as in an open circuit).

Answer:

$$i = \begin{cases} 16 \text{ A}, & t < 0 \\ 16e^{-2t} \text{ A}, & t \ge 0 \end{cases}, \quad i_o = \begin{cases} 8 \text{ A}, & t < 0 \\ -5.333e^{-2t} \text{ A}, & t > 0 \end{cases},$$ $$v_o = \begin{cases} 32 \text{ V}, & t < 0 \\ 10.667e^{-2t} \text{ V}, & t > 0 \end{cases}$$