7.1 Introduction

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7.1 Introduction

Now that we ha ve considered the three passi ve elements (resistors, ca pacitors, and inductors) and one active element (the op amp)individually, we are prepared to consider circuits that contain various combinations of two or three of the passi ve elements. In this chapter , we shall e xamine two types of simple circuits: a circuit comprising a resistor and capacitor and a circuit comprising a resistor and an inductor . These are called RC and RL circuits, respectively. As simple as these circuits are, the y find continual applications in electronics, communications, and control sys tems, as we shall see.

We carry out the analysis of RC and RL circuits by applying Kirchhoff’s laws, as we did for resisti ve circuits. The only dif ference is that applying Kirchhoff’s laws to purely resistive circuits results in algebraic equations, while applying the la ws to RC and RL circuits produces differential equations, which are more difficult to solve than algebraic equations. The differential equations resulting from analyzing RC and RL circuits are of the first order. Hence, the circuits are collectively known as first-order circuits.

A first-order circuit is characterized by a first-order differential equation.

In addition to there being tw o types of first-order circuits (RC and RL), there are tw o ways to e xcite the circuits. The first way is by ini tial conditions of the storage elements in the circuits. In these so-called source-free circuits, we assume that ener gy is initially stored in the ca pacitive or inductive element. The energy causes current to flow in the circuit and is gradually dissipated in the resistors. Although source-free circuits are by definition free of independent sources, they may ha ve dependent sources. The second way of exciting first-order circuits is by independent sources. In this chapter, the independent sources we will consider are dc sources. (In later chapters, we shall consider sinusoidal and exponential sources.) The two types of first-order circuits and the two ways of exciting them add up to the four possible situations we will study in this chapter.

Finally, we consider four typical applications of RC and RL circuits: delay and relay circuits, a photoflash unit, and an automobile ignition circuit.

7.2 The Source-Free RC Circuit

A source-free RC circuit occurs when its dc source is suddenly disconnected. The energy already stored in the capacitor is released to the resistors.

Consider a series combination of a resistor and an initially char ged capacitor, as shown in Fig. 7.1. (The resistor and capacitor may be the equivalent resistance and equivalent capacitance of combinations of re sistors and capacitors.) Our objective is to determine the circuit response, which, for pedagogic reasons, we assume to be the voltage v(t) across the capacitor. Since the capacitor is initially charged, we can assume that at time t = 0, the initial voltage is

$$v(0) = V_0 \tag{7.1}$$

with the corresponding value of the energy stored as

$$w(0) = \frac{1}{2}CV_0^2$$

(7.2)

Applying KCL at the top node of the circuit in Fig. 7.1 yields

$$i_C + i_R = 0 \tag{7.3}$$

By definition, iC = C dv∕dt and iR = v∕R. Thus,

$$C\frac{dv}{dt} + \frac{v}{R} = 0$$

\n(7.4a)

or

$$\frac{dv}{dt} + \frac{v}{RC} = 0\tag{7.4b}$$

This is a first-order differential equation, since only the first derivative of v is involved. To solve it, we rearrange the terms as

$$\frac{dv}{v} = -\frac{1}{RC}dt\tag{7.5}$$

Integrating both sides, we get

$$\ln v = -\frac{t}{RC} + \ln A$$

where ln A is the integration constant. Thus,

$$\ln\frac{v}{A} = -\frac{t}{RC}$$

(7.6)

Taking powers of e produces

$$v(t) = Ae^{-t/RC}$$

But from the initial conditions, v(0) = A = V0. Hence,

$$v(t) = V_0 e^{-t/RC}$$

\n

$$(7.7)$$

This shows that the voltage response of the RC circuit is an exponential decay of the initial voltage. Since the response is due to the initial energy stored and the physical characteristics of the circuit and not due to some e xternal voltage or current source, it is called the natural response of the circuit.

Figure 7.1 A source-free RC circuit.

A circuit response is the manner in which the circuit reacts to an excitation.

The natural response of a circuit refers to the behavior (in terms of voltages and currents) of the circuit itself, with no external sources of excitation.

The natural response depends on the nature of the circuit alone, with no external sources. In fact, the circuit has a response only because of the energy initially stored in the capacitor.

(t)∕V0 = e−t∕τ Values of v t τ 2τ 3τ 4τ 5τ	TABLE 7.1
		v(t)∕V0
		0.36788
		0.13534
		0.04979
		0.01832
		0.00674

Graphical determination of the time constant τ from the response curve.

The natural response is illustrated graphically in Fig. 7.2. Note that at t = 0, we have the correct initial condition as in Eq. (7.1). As t increases, the voltage decreases toward zero. The rapidity with which the v oltage decreases is e xpressed in terms of the time constant, denoted by τ, the lowercase Greek letter tau.

The time constant of a circuit is the time required for the response to decay to a factor of 1∕e or 36.8 percent of its initial value.1

This implies that at t = τ, Eq. (7.7) becomes

$$V_0 e^{-\tau/RC} = V_0 e^{-1} = 0.368 V_0$$

or

$$\tau = RC \tag{7.8}$$

In terms of the time constant, Eq. (7.7) can be written as

$$v(t) = V_0 e^{-t/\tau} \tag{7.9}$$

With a calculator it is easy to sho w that the v alue of v(t)∕V0 is as shown in Table 7.1. It is evident from Table 7.1 that the voltage v(t) is less than 1 percent of V0 after 5τ (five time constants). Thus, it is customary to assume that the capacitor is fully dischar ged (or charged) after five time constants. In other words, it takes 5τ for the circuit to reach its final state or steady state when no changes take place with time. Notice that for every time interval of τ, the voltage is reduced by 36.8 percent of its pre vious value, v(t + τ) = v(t)∕e = 0.368v(t), regardless of the value of t.

Observe from Eq. (7.8) that the smaller the time constant, the more rapidly the v oltage decreases, that is, the f aster the response. This is illustrated in Fig. 7.4. A circuit with a small time constant gi ves a f ast response in that it reaches the steady state (or final state) quickly due to quick dissipation of ener gy stored, whereas a circuit with a lar ge time constant gives a slo w response because it tak es longer to reach steady state. At any rate, whether the time constant is small or large, the circuit reaches steady state in five time constants.

With the voltage v(t) in Eq. (7.9), we can find the current iR(t),

$$i_R(t) = \frac{v(t)}{R} = \frac{V_0}{R} e^{-t/\tau}$$

\n(7.10)

1 The time constant may be viewed from another perspective. Evaluating the derivative of v(t) in Eq. (7.7) at t = 0, we obtain

$$\frac{d}{dt}\left(\frac{v}{V_0}\right)\Big|_{t=0} = -\frac{1}{\tau}e^{-t/\tau}\Big|_{t=0} = -\frac{1}{\tau}$$

Thus, the time constant is the initial rate of decay, or the time taken for v∕V0 to decay from unity to zero, assuming a constant rate of decay. This initial slope interpretation of the time constant is often used in the laboratory to find τ graphically from the response curve displayed on an oscilloscope. To find τ from the response curve, draw the tangent to the curve at t = 0, as shown in Fig. 7.3. The tangent intercepts with the time axis at t = τ.

Plot of v∕V0 = e −t∕τ for various values of the time constant.

The power dissipated in the resistor is

$$p(t) = vi_R = \frac{V_0^2}{R}e^{-2t/\tau}$$

\n(7.11)

The energy absorbed by the resistor up to time t is

$$w_R(t) = \int_0^t p(\lambda) d\lambda = \int_0^t \frac{V_0^2}{R} e^{-2\lambda/\tau} d\lambda$$

= $-\frac{\tau V_0^2}{2R} e^{-2\lambda/\tau} \Big|_0^t = \frac{1}{2} C V_0^2 (1 - e^{-2t/\tau}), \qquad \tau = RC$ (7.12)

Notice that as t → ∞, wR (∞) → _1 ²CV ⁰ 2 , which is the same as wC (0), the energy initially stored in the capacitor. The energy that was initially stored in the capacitor is eventually dissipated in the resistor.

In summary:

The Key to Working with a Source-Free RC Circuit Is Finding:

• 1. The initial voltage v(0) = V0 across the capacitor.
• 2. The time constant τ.

With these tw o items, we obtain the response as the capacitor v oltage vC (t) = v(t) = v(0) e⁻t∕^τ . Once the capacitor voltage is first obtained, other variables (capacitor current iC, resistor voltage vR, and resistor current iR) can be determined. In finding the time constant τ = RC, R is often the Thevenin equivalent resistance at the terminals of the capacitor; that is, we take out the capacitor C and find R = RTh at its terminals.

In Fig. 7.5, let vC (0) = 15 V. Find vC, vx, and ix for t > 0.

Solution:

We first need to make the circuit in Fig. 7.5 conform with the standard RC circuit in Fig. 7.1. We find the equivalent resistance or the Thevenin The time constant is the same regardless of what the output is defined to be.

When a circuit contains a single capacitor and several resistors and dependent sources, the Thevenin equivalent can be found at the terminals of the capacitor to form a simple RC circuit. Also, one can use Thevenin’s theorem when several capacitors can be combined to form a single equivalent capacitor.

Example 7.1

Figure 7.5

For Example 7.1.

Figure 7.6 Equivalent circuit for the circuit in Fig. 7.5.

The 8-Ω and 12- Ω resistors in series can be combined to give a 20-Ω resistor. This 20-Ω resistor in parallel with the 5-Ω resistor can be combined so that the equivalent resistance is

$$R_{\text{eq}} = \frac{20 \times 5}{20 + 5} = 4 \ \Omega$$

Hence, the equivalent circuit is as shown in Fig. 7.6, which is analogous to Fig. 7.1. The time constant is

Thus,

$$\tau = R_{\text{eq}}C = 4(0.1) = 0.4 \text{ s}$$ $$^{as,}$$ $$v = v(0)e^{-t/\tau} = 15e^{-t/0.4} \text{ V}, \qquad v_C = v = 15e^{-2.5t} \text{ V}$$

From Fig. 7.5, we can use voltage division to get vx; so

$$v_x = \frac{12}{12 + 8}v = 0.6(15e^{-2.5t}) = 9e^{-2.5t}$$

V

Finally,

$$i_x = \frac{v_x}{12} = 0.75e^{-2.5t} \text{ A}$$

Practice Problem 7.1

Refer to the circuit in Fig. 7.7. Let vC (0) = 60 V. Determine vC, vx, and io for t ≥ 0.

For Practice Prob. 7.1.

Answer: 60e⁻0.25*^t* V, 20e⁻0.25*^tV, −5e*⁻0.25*^t*A.

Figure 7.8 For Example 7.2.

Example 7.2 The switch in the circuit in Fig. 7.8 has been closed for a long time, and it is opened at t = 0. Find v(t) for t ≥ 0. Calculate the initial energy stored in the capacitor.

Solution:

For t < 0, the switch is closed; the capacitor is an open circuit to dc, as represented in Fig. 7.9(a). Using voltage division

$$v_C(t) = \frac{9}{9+3}(20) = 15 \text{ V}, \qquad t < 0$$

Since the voltage across a capacitor cannot change instantaneously, the voltage across the capacitor at t = 0⁻ is the same at t = 0, or

$$v_C(0) = V_0 = 15 \text{ V}$$

For t > 0, the switch is opened, and we have the RC circuit shown in Fig. 7.9(b). [Notice that the RC circuit in Fig. 7.9(b) is source free; the independent source in Fig. 7.8 is needed to provide V0 or the initial energy in the capacitor.] The 1-Ω and 9-Ω resistors in series give

$$R_{\text{eq}} = 1 + 9 = 10 \Omega$$

The time constant is

$$\tau = R_{\text{eq}}C = 10 \times 20 \times 10^{-3} = 0.2 \text{ s}$$

Thus, the voltage across the capacitor for t ≥ 0 is

$$v(t) = v_C(0)e^{-t/\tau} = 15e^{-t/0.2} \text{ V}$$

or

$$v(t) = 15e^{-5t} \,\mathrm{V}$$

The initial energy stored in the capacitor is

$$w_C(0) = \frac{1}{2}Cv_C^2(0) = \frac{1}{2} \times 20 \times 10^{-3} \times 15^2 = 2.25 \text{ J}$$

If the switch in Fig. 7.10 opens at t = 0, find v(t) for t ≥ 0 and wC (0). Practice Problem 7.2

Answer: 8 e⁻2*^t* V, 5.333 J.

9 Ω 1 Ω vC(0) 3 Ω + 20 V (a) 9 Ω 1 Ω (b) + ‒ Vo = 15 V 20 mF + ‒

Figure 7.9 For Example 7.2: (a) t < 0, (b) t > 0.

Figure 7.10 For Practice Prob. 7.2.