8.10 Duality

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8.10 Duality

The concept of duality is a time-sa ving, ef fort-effective measure of solving circuit problems. Consider the similarity between Eqs. (8.4) and (8.29). The two equations are the same, e xcept that we must interchange the following quantities: (1) voltage and current, (2) resistance and conductance, (3) capacitance and inductance. Thus, it sometimes occurs in circuit analysis that two different circuits have the same equations and solutions, except that the roles of certain complementary elements are interchanged. This interchangeability is kno wn as the principle of duality.

The duality principle asserts a parallelism between pairs of characterizing equations and theorems of electric circuits.

Dual pairs are sho wn in Table 8.1. Note that po wer does not appear in Table 8.1, because power has no dual. The reason for this is the principle of linearity; since power is not linear, duality does not apply. Also notice from Table 8.1 that the principle of duality extends to circuit elements, configurations, and theorems.

Two circuits that are described by equations of the same form, b ut in which the variables are interchanged, are said to be dual to each other.

Two circuits are said to be duals of one another if they are described by the same characterizing equations with dual quantities interchanged.

The usefulness of the duality principle is self-e vident. Once we know the solution to one circuit, we automatically ha ve the solution for the dual circuit. It is ob vious that the circuits in Figs. 8.8 and 8.13 are dual. Consequently , the result in Eq. (8.32) is the dual of that in Eq. (8.11). We must k eep in mind that the method described here for finding a dual is limited to planar circuits. Finding a dual for a nonplanar circuit is be yond the scope of this te xtbook because nonplanar circuits cannot be described by a system of mesh equations.

To find the dual of a given circuit, we do not need to write do wn the mesh or node equations. We can use a graphical technique. Gi ven a planar circuit, we construct the dual circuit by taking the following three steps:

• 1. Place a node at the center of each mesh of the given circuit. Place the reference node (the ground) of the dual circuit outside the given circuit.
• 2. Draw lines between the nodes such that each line crosses an ele ment. Replace that element by its dual (see Table 8.1).
• 3. To determine the polarity of voltage sources and direction of current sources, follow this rule: A voltage source that produces a positi ve (clockwise) mesh current has as its dual a current source whose reference direction is from the ground to the nonreference node.

In case of doubt, one may verify the dual circuit by writing the nodal or mesh equations. The mesh (or nodal) equations of the original circuit are similar to the nodal (or mesh) equations of the dual circuit. The duality principle is illustrated with the following two examples.

TABLE 8.1

Dual pairs.

Resistance R	Conductance G
Inductance L	Capacitance C
Voltage v	Current i
Voltage source	Current source
Node	Mesh
Series path	Parallel path
Open circuit	Short circuit
KVL	KCL
Thevenin	Norton

Even when the principle of linearity applies, a circuit element or variable may not have a dual. For example, mutual inductance (to be covered in Chapter 13) has no dual.

Construct the dual of the circuit in Fig. 8.44.

Solution:

As shown in Fig. 8.45(a), we first locate nodes 1 and 2 in the two meshes and also the ground node 0 for the dual circuit. We draw a line between one node and another crossing an element. We replace the line joining the nodes by the duals of the elements which it crosses. For example, a line between nodes 1 and 2 crosses a 2-H inductor, and we place a 2-F capacitor (an inductor’s dual) on the line. A line between nodes 1 and 0 crossing the 6-V voltage source will contain a 6-A current source. By drawing lines crossing all the elements, we construct the dual circuit on the given circuit as in Fig. 8.45(a). The dual circuit is redrawn in Fig. 8.45(b) for clarity.

Figure 8.44 For Example 8.14.

Figure 8.45

(a) Construction of the dual circuit of Fig. 8.44, (b) dual circuit redrawn.

Draw the dual circuit of the one in Fig. 8.46.

Answer: See Fig. 8.47.

Obtain the dual of the circuit in Fig. 8.48. Example 8.15

Solution:

The dual circuit is constructed on the original circuit as in Fig. 8.49(a). We first locate nodes 1 to 3 and the reference node 0. Joining nodes 1 and 2, we cross the 2-F capacitor, which is replaced by a 2-H inductor.

Practice Problem 8.14

For Example 8.15.

Joining nodes 2 and 3, we cross the 20- Ω resistor, which is replaced by a ___1 ²⁰ -Ω resistor. We keep doing this until all the elements are crossed. The result is in Fig. 8.49(a). The dual circuit is redrawn in Fig. 8.49(b).

Figure 8.49

For Example 8.15: (a) construction of the dual circuit of Fig. 8.48, (b) dual circuit redrawn.

To v erify the polarity of the v oltage source and the direction of the current source, we may apply mesh currents i1, i2, and i3 (all in the clockwise direction) in the original circuit in Fig. 8.48. The 10-V voltage source produces positive mesh current i1, so that its dual is a 10-A current source directed from 0 to 1. Also, i3 = −3 A in Fig. 8.48 has as its dual v3 = −3 V in Fig. 8.49(b).

Answer: See Fig. 8.51.

For Practice Prob. 8.15.

Figure 8.51 Dual of the circuit in Fig. 8.50.

8.11 Applications

Practical applications of RLC circuits are found in control and communications circuits such as ringing circuits, peaking circuits, resonant circuits, smoothing circuits, and filters. Most of these circuits cannot be covered until we treat ac sources. For now, we will limit ourselves to two simple applications: automobile ignition and smoothing circuits.

8.11.1 Automobile Ignition System

In Section 7.9.4, we considered the automobile ignition system as a charging system. That was only a part of the system. Here, we consider another part—the voltage generating system. The system is modeled by the circuit shown in Fig. 8.52. The 12-V source is due to the battery and alternator. The 4-Ω resistor represents the resistance of the wiring. The ignition coil is modeled by the 8-mH inductor . The 1-μF capacitor (known as the condenser to automechanics) is in parallel with the switch (known as the breaking points or electronic ignition). In the follo wing example, we determine how the RLC circuit in Fig. 8.52 is used in gen erating high voltage.

Automobile ignition circuit.

Assuming that the switch in Fig. 8.52 is closed prior to t = 0⁻, find the inductor voltage vL for t > 0.

Solution:

If the switch is closed prior to t = 0⁻ and the circuit is in steady state, then

$$i(0^{-}) = \frac{12}{4} = 3 \text{ A}, \qquad v_C(0^{-}) = 0$$

At t = 0⁺, the switch is opened. The continuity conditions require that

$$i(0^+) = 3 \text{ A}, \qquad v_C(0^+) = 0$$

(8.16.1)

We obtain di(0⁺)/dt from vL(0⁺). Applying KVL to the mesh at t = 0⁺ yields

$$-12 + 4i(0^{+}) + v_L(0^{+}) + v_C(0^{+}) = 0$$ $$-12 + 4 \times 3 + v_L(0^{+}) + 0 = 0 \implies v_L(0^{+}) = 0$$

Example 8.16

Hence,

$$\frac{di(0^{+})}{dt} = \frac{v_L(0^{+})}{L} = 0$$

\n(8.16.2)

As t → ∞, the system reaches steady state, so that the capacitor acts like an open circuit. Then

$$i(\infty) = 0 \tag{8.16.3}$$

If we apply KVL to the mesh for t > 0, we obtain

$$12 = Ri + L\frac{di}{dt} + \frac{1}{C} \int_0^t i \, dt + v_C(0)$$

Taking the derivative of each term yields

$$\frac{d^2i}{dt^2} + \frac{R}{L}\frac{di}{dt} + \frac{i}{LC} = 0$$

(8.16.4)

We obtain the form of the transient response by following the procedure in Section 8.3. Substituting R = 4 Ω, L = 8 mH, and C = 1 μF, we get

$$\alpha = \frac{R}{2L} = 250$$

, $\omega_0 = \frac{1}{\sqrt{LC}} = 1.118 \times 10^4$

Since α < ω0, the response is underdamped. The damped natural fre quency is

$$\omega_d = \sqrt{\omega_0^2 - \alpha^2} \simeq \omega_0 = 1.118 \times 10^4$$

The form of the transient response is

$$it(t) = e^{-\alpha} (A \cos \omega_d t + B \sin \omega_d t)$$

(8.16.5)

where A and B are constants. The steady-state response is

$$i_{ss}(t) = i(\infty) = 0 \tag{8.16.6}$$

so that the complete response is

$$i(t) = it(t) + iss(t) = e^{-250t}(A \cos 11,180t + B \sin 11,180t)$$

(8.16.7)

We now determine A and B.

$$i(0) = 3 = A + 0 \qquad \Rightarrow \qquad A = 3$$

Taking the derivative of Eq. (8.16.7),

$$\frac{di}{dt} = -250e^{-250t}(A \cos 11,180t + B \sin 11,180t) \n+ e^{-250t}(-11,180A \sin 11,180t + 11,180B \cos 11,180t)$$

Setting t = 0 and incorporating Eq. (8.16.2),

$$0 = -250A + 11{,}180B \qquad \Rightarrow \qquad B = 0.0671$$

Thus,

$$i(t) = e^{-250t}(3 \cos 11,180t + 0.0671 \sin 11,180t)$$

(8.16.8)

The voltage across the inductor is then

$$v_L(t) = L\frac{di}{dt} = -268e^{-250t}\sin 11{,}180t$$

(8.16.9)

This has a maximum value when sine is unity, that is, at 11,180t0 = π∕2 or t0 = 140.5 μs. At time = t0, the inductor voltage reaches its peak, which is

$$v_L(t_0) = -268e^{-250t_0} = -259 \text{ V} \tag{8.16.10}$$

Although this is far less than the voltage range of 6000 to 10,000 V required to fire the spark plug in a typical automobile, a device known as a transformer (to be discussed in Chapter 13) is used to step up the inductor voltage to the required level.

In Fig. 8.52, find the capacitor voltage vC for t > 0.

Answer: 12 − 12e⁻250*^t* cos 11,180t + 267.7e⁻250*^t* sin 11,180t V.

8.11.2 Smoothing Circuits

In a typical digital communication system, the signal to be transmitted is first sampled. Sampling refers to the procedure of selecting samples of a signal for processing, as opposed to processing the entire signal. Each sample is con verted into a binary number represented by a series of pulses. The pulses are transmitted by a transmission line such as a coaxial cable, twisted pair, or optical fiber. At the receiving end, the signal is applied to a digital-to-analog (D/A) con verter whose output is a “staircase” function, that is, constant at each time interv al. In order to recover the transmitted analog signal, the output is smoothed by letting it pass through a “smoothing” circuit, as illustrated in Fig. 8.53. An RLC circuit may be used as the smoothing circuit.

The output of a D/A converter is shown in Fig. 8.54(a). If the RLC circuit in Fig. 8.54(b) is used as the smoothing circuit, determine the output

Practice Problem 8.16

Figure 8.53

A series of pulses is applied to the digitalto-analog (D/A) converter, whose output is applied to the smoothing circuit.

Example 8.17

Figure 8.54 For Example 8.17: (a) output of a D/A converter, (b) an RLC smoothing circuit.

Solution:

voltage vo(t).

This problem is best solved using PSpice. The schematic is shown in Fig. 8.55(a). The pulse in Fig. 8.54(a) is specified using the piecewise

Figure 8.55 For Example 8.17: (a) schematic, (b) input and output voltages.

linear function. The attributes of V1 are set as T1 = 0, V1 = 0, T2 = 0.001, V2 = 4, T3 = 1, V3 = 4, and so on. To be able to plot both input and output voltages, we insert two voltage markers as shown. We select Analysis/Setup/Transient to open up the Transient Analysis dialog box and set Final Time as 6 s. Once the schematic is saved, we select Analysis/Simulate to run and obtain the plots shown in Fig. 8.55(b).

Practice Problem 8.17 Rework Example 8.17 if the output of the D/A converter is as shown in Fig. 8.56.

Answer: See Fig. 8.57.

8.12 Summary

• 1. The determination of the initial v alues x(0) and dx(0)∕dt and final value x(∞) is crucial to analyzing second-order circuits.
• 2. The RLC circuit is second-order because it is described by a second-order dif ferential equation. Its characteristic equation is

s 2 + 2αs + ω⁰ 2 = 0, where α is the neper frequenc y and ω0 is the undamped natural frequency. For a series circuit, α = R∕2L, for a parallel circuit α = 1∕2RC, and for both cases ω0 = 1∕ √ ___ LC .

• 3. If there are no independent sources in the circuit after switching (or sudden change), we regard the circuit as source-free. The complete solution is the natural response.
• 4. The natural response of an RLC circuit is o verdamped, under damped, or critically damped, depending on the roots of the characteristic equation. The response is critically damped when the roots are equal (s1 = s2 or α = ω0), overdamped when the roots are real and unequal (s1 ≠ s2 or α > ω0), or underdamped when the roots are complex conjugate (s1 = s² * or α < ω0).
• 5. If independent sources are present in the circuit after switching, the complete response is the sum of the transient response and the steady-state response.
• 6. PSpice is used to analyze RLC circuits in the same w ay as for RC or RL circuits.
• 7. Two circuits are dual if the mesh equations that describe one circuit have the same form as the nodal equations that describe the other. The analysis of one circuit gives the analysis of its dual circuit.
• 8. The automobile ignition circuit and the smoothing circuit are typical applications of the material covered in this chapter.

Review Questions

8.1 For the circuit in Fig. 8.58, the capacitor voltage at t = 0⁻ (just before the switch is closed) is:

Figure 8.58

For Review Questions 8.1 and 8.2.

8.2 For the circuit in Fig. 8.58, the initial inductor current (at t = 0) is:

(a) 0 A (b) 2 A (c) 6 A (d) 12 A

• 8.3 When a step input is applied to a second-order circuit, the final values of the circuit variables are found by:
  • (a) Replacing capacitors with closed circuits and inductors with open circuits.
  • (b) Replacing capacitors with open circuits and inductors with closed circuits.
  • (c) Doing neither of the above.

8.4 If the roots of the characteristic equation of an RLC circuit are −2 and −3, the response is:

(a) (A cos 2t + B sin 2t)e⁻3*^t* (b) (A + 2Bt)e⁻3*^t* (c) Ae⁻2*^t* + Bte⁻3*^t* (d) Ae⁻2*^t* + Be⁻3*^t*

where A and B are constants.

• 8.5 In a series RLC circuit, setting R = 0 will produce:
  • (a) an overdamped response
  • (b) a critically damped response
  • (c) an underdamped response
  • (d) an undamped response
  • (e) none of the above
• 8.6 A parallel RLC circuit has L = 2 H and C = 0.25 F. The value of R that will produce a unity neper frequency is:

(a) 0.5 Ω (b) 1 Ω (c) 2 Ω (d) 4 Ω

• 8.7 Refer to the series RLC circuit in Fig. 8.59. What kind of response will it produce?
  • (a) overdamped
  • (b) underdamped
  • (c) critically damped
  • (d) none of the above

C1

i

Figure 8.59

For Review Question 8.7.

• 8.8 Consider the parallel RLC circuit in Fig. 8.60. What type of response will it produce?
  • (a) overdamped
  • (b) underdamped
  • (c) critically damped
  • (d) none of the above

For Review Question 8.8.

• 8.9 Match the circuits in Fig. 8.61 with the following items:
  • (i) first-order circuit
  • (ii) second-order series circuit
  • (iii) second-order parallel circuit
  • (iv) none of the above

vs R (c) i ^s C² C1 L R1 (d) C2 R2 + ‒ R¹ R²

Figure 8.61

For Review Question 8.9.

8.10 In an electric circuit, the dual of resistance is:

(a) conductance	(b) inductance
(c) capacitance	(d) open circuit
(e) short circuit

Answers: 8.1a, 8.2c, 8.3b, 8.4d, 8.5d, 8.6c, 8.7b, 8.8b, 8.9 (i)-c, (ii)-b, e, (iii)-a, (iv)-d, f, 8.10a.

Problems

Section 8.2 Finding Initial and Final Values

8.1 For the circuit in Fig. 8.62, find:

(a) i(0⁺) and v(0⁺), (b) di(0⁺)∕dt and dv(0⁺)∕dt,

(c) i(∞) and v(∞).

Figure 8.62 For Prob. 8.1.

Figure 8.63 For Prob. 8.2.

8.3 Refer to the circuit shown in Fig. 8.64. Calculate:

(a) iL(0⁺), vC(0⁺), and vR(0⁺), (b) diL(0⁺)∕dt, dvC(0⁺)∕dt, and dvR(0⁺)∕dt, (c) iL(∞), vC(∞), and vR(∞).

Figure 8.64

• For Prob. 8.3.
  • 8.4 In the circuit of Fig. 8.65, find: (a) v(0⁺) and i(0⁺), (b) dv(0⁺)∕dt and di(0⁺)∕dt, (c) v(∞) and i(∞).

Figure 8.65

For Prob. 8.4.

• 8.5 Refer to the circuit in Fig. 8.66. Determine:
  • (a) i(0⁺) and v(0⁺), (b) di∕(0⁺)dt and dv(0⁺)∕dt,
  • (c) i(∞) and v(∞).

For Prob. 8.5.

8.6 In the circuit of Fig. 8.67, find:

(a) vR(0⁺) and vL(0⁺), (b) dvR(0⁺)∕dt and dvL(0⁺)∕dt, (c) vR(∞) and vL(∞).

Figure 8.67

For Prob. 8.6.

Section 8.3 Source-Free Series RLC Circuit

• 8.7 A series RLC circuit has R = 20 kΩ, L = 0.2 mH, and C = 5 μF. What type of damping is exhibited by the circuit?
• 8.8 Design a problem to help other students better understand source-free RLC circuits.
• 8.9 The current in an RLC circuit is described by

$$\frac{d^2i}{dt^2} + 10\frac{di}{dt} + 25i = 0$$

If i(0) = 10 A and di(0)∕dt = 0, find i(t) for t > 0.

8.10 The differential equation that describes the current in an RLC network is

$$3\frac{d^2i}{dt^2} + 15\frac{di}{dt} + 12i = 0$$

Given that i(0) = 0, di(0)∕dt = 6 mA/s, obtain i(t).

8.11 The natural response of an RLC circuit is described by the differential equation

$$\frac{d^2v}{dt^2} + 2\frac{dv}{dt} + v = 0$$

for which the initial conditions are v(0) = 10 V and dv(0)∕dt = 0. Solve for v(t).

• 8.12 If R = 50 Ω, L = 1.5 H, what value of C will make an RLC series circuit:
  • (a) overdamped,
  • (b) critically damped,
  • (c) underdamped?
• 8.13 For the circuit in Fig. 8.68, calculate the value of R needed to have a critically damped response.

Figure 8.68 For Prob. 8.13.

8.14 The switch in Fig. 8.69 moves from position A to position B at t = 0 (please note that the switch must connect to point B before it breaks the connection at A, a make-before-break switch). Let v(0) = 0, find v(t) for t > 0.

8.15 The responses of a series RLC circuit are

$$v_C(t) = 30 - 10e^{-20t} + 30e^{-10t}$$

V

$$i_L(t) = 40e^{-20t} - 60e^{-10t}$$

mA

where vC and iL are the capacitor voltage and inductor current, respectively. Determine the values of R, L, and C.

8.16 Find i(t) for t > 0 in the circuit of Fig. 8.70.

Figure 8.70

For Prob. 8.16.

8.17 In the circuit of Fig. 8.71, the switch instantaneously moves from position A to B at t = 0. Find v(t) for all t ≥ 0.

8.18 Find the voltage across the capacitor as a function of time for t > 0 for the circuit in Fig. 8.72. Assume steady-state conditions exist at t = 0⁻.

Figure 8.72

For Prob. 8.18.

8.19 Obtain v(t) for t > 0 in the circuit of Fig. 8.73.

Figure 8.73

For Prob. 8.19.

8.20 The switch in the circuit of Fig. 8.74 has been closed for a long time but is opened at t = 0. Determine i(t) for t > 0.

Figure 8.74

For Prob. 8.20.

*8.21 Calculate v(t) for t > 0 in the circuit of Fig. 8.75.

Figure 8.75 For Prob. 8.21.

* An asterisk indicates a challenging problem.

Section 8.4 Source-Free Parallel RLC Circuit

8.22 Assuming R = 2 kΩ, design a parallel RLC circuit that has the characteristic equation

$$s^2 + 100s + 10^6 = 0.$$

8.23 For the network in Fig. 8.76, what value of C is needed to make the response underdamped with unity neper frequency (α = 1)?

Figure 8.76 For Prob. 8.23.

8.24 The switch in Fig. 8.77 moves from position A to position B at t = 0 (please note that the switch must connect to point B before it breaks the connection at A, a make-before-break switch). Determine i(t) for t > 0.

Figure 8.77 For Prob. 8.24.

8.25 Using Fig. 8.78, design a problem to help other students better understand source-free RLC circuits.

Figure 8.78 For Prob. 8.25.

Section 8.5 Step Response of a Series RLC Circuit

8.26 The step response of an RLC circuit is given by

$$\frac{d^2i}{dt^2} + 2\frac{di}{dt} + 5i = 10$$

Given that i(0) = 2 and di(0)∕dt = 4, solve for i(t).

8.27 A branch voltage in an RLC circuit is described by

$$\frac{d^2v}{dt^2} + 4\frac{dv}{dt} + 8v = 24$$

If the initial conditions are v(0) = 0 = dv(0)∕dt, find v(t).

8.28 A series RLC circuit is described by

$$L\frac{d^2i}{dt^2} + R\frac{di}{dt} + \frac{i}{C} = 10$$

Find the response when L = 0.5 H, R = 4 Ω, and C = 0.2 F. Let i(0) = 1, di(0)∕dt = 0.

8.29 Solve the following differential equations subject to the specified initial conditions

(a)

$$d^2v/dt^2 + 4v = 12$$

, $v(0) = 0$ , $dv(0)/dt = 2$
\n(b) $d^2i/dt^2 + 5 \frac{di}{dt} + 4i = 8$ , $i(0) = -1$ ,
\n $\frac{di(0)}{dt} = 0$
\n(c) $\frac{d^2v}{dt^2} + 2 \frac{dv}{dt} + v = 3$ , $v(0) = 5$ ,
\n $\frac{dv(0)}{dt} = 1$
\n(d) $\frac{d^2i}{dt^2} + 2 \frac{di}{dt} + 5i = 10$ , $i(0) = 4$ ,
\n $\frac{di(0)}{dt} = -2$

8.30 The step responses of a series RLC circuit are

$$v_C = 40 - 10e^{-2000t} - 10e^{-4000t} \text{ V}, \qquad t > 0$$ $$i_L(t) = 3e^{-2000t} + 6e^{-4000t} \text{ mA}, \qquad t > 0$$

• (a) Find C. (b) Determine what type of damping is exhibited by the circuit.
• 8.31 Consider the circuit in Fig. 8.79. Find vL(0⁺) and vC(0⁺).

For Prob. 8.31.

Figure 8.80 For Prob. 8.32.

Figure 8.81

For Prob. 8.33.

8.34 Calculate i(t) for t > 0 in the circuit of Fig. 8.82.

For Prob. 8.34.

8.35 Using Fig. 8.83, design a problem to help other students better understand the step response of series RLC circuits.

Figure 8.83

For Prob. 8.35.

8.36 Obtain v(t) and i(t) for t > 0 in the circuit of Fig. 8.84.

Figure 8.84 For Prob. 8.36.

*8.37 For the network in Fig. 8.85, solve for i(t) for t > 0.

Figure 8.85

For Prob. 8.37.

8.38 Refer to the circuit in Fig. 8.86. Calculate i(t) for t > 0.

Figure 8.86 For Prob. 8.38.

Figure 8.87

For Prob. 8.39.

8.40 The switch in the circuit of Fig. 8.88 is moved from position a to b at t = 0. Assume that the voltage across the capacitor is equal to zero at t = 0 and that the switch is a make before break switch. Determine i(t) for all t > 0.

*8.41 For the network in Fig. 8.89, find i(t) for t > 0.

Figure 8.89 For Prob. 8.41.

*8.42 Given the network in Fig. 8.90, find v(t) for t > 0.

• For Prob. 8.42.
  • 8.43 The switch in Fig. 8.91 is opened at t = 0 after the circuit has reached steady state. Choose R and C such that α = 8 Np/s and ωd = 30 rad/s.

Figure 8.91

For Prob. 8.43.

8.44 A series RLC circuit has the following parameters: R = 1 kΩ, L = 1 H, and C = 10 nF. What type of damping does this circuit exhibit?

Section 8.6 Step Response of a Parallel RLC Circuit

8.45 In the circuit of Fig. 8.92, find v(t) and i(t) for t > 0.

For Prob. 8.45.

8.46 Using Fig. 8.93, design a problem to help other students better understand the step response of a parallel RLC circuit.

Figure 8.93

For Prob. 8.46.

8.47 Find the output voltage vo(t) in the circuit of Fig. 8.94.

Figure 8.94

For Prob. 8.47.

8.48 Given the circuit in Fig. 8.95, find i(t) and v(t) for t > 0.

For Prob. 8.48.

8.49 Determine i(t) for t > 0 in the circuit of Fig. 8.96.

Figure 8.97

For Prob. 8.50.

8.51 Find v(t) for t > 0 in the circuit of Fig. 8.98.

Figure 8.98

For Prob. 8.51.

8.52 The step response of a parallel RLC circuit is v = 10 + 20e⁻300*^t* (cos 400t − 2 sin 400t) V, t ≥ 0 when the inductor is 25 mH. Find R and C.

Section 8.7 General Second-Order Circuits

8.53 After being open for a day, the switch in the circuit of Fig. 8.99 is closed at t = 0. Find the differential equation describing i(t), t > 0.

Figure 8.99

For Prob. 8.53.

8.55 For the circuit in Fig. 8.101, find v(t) for t > 0. Assume that i(0⁺) = 2 A.

Figure 8.101

For Prob. 8.55.

8.56 In the circuit of Fig. 8.102, find i(t) for t > 0.

Figure 8.102

For Prob. 8.56.

8.57 Given the circuit shown in Fig. 8.103, determine the characteristic equation of the circuit and the values for i(t) and v(t) for all t > 0.

Figure 8.103

For Prob. 8.57.

• 8.58 In the circuit of Fig. 8.104, the switch has been in position 1 for a long time but moved to position 2 at t = 0. Find:
  • (a) v(0⁺), dv(0⁺)∕dt, (b) v(t) for t ≥ 0.

Figure 8.104 For Prob. 8.58.

8.59 The switch in Fig. 8.105 has been in position 1 for t < 0. At t = 0, it is moved from position 1 to the top of the capacitor at t = 0. Please note that the switch is a make before break switch, it stays in contact with position 1 until it makes contact with the top of the capacitor and then breaks the contact at position 1. Given that the initial voltage across the capacitor is equal to zero, determine v(t).

Figure 8.105

For Prob. 8.59.

Figure 8.106

For Prob. 8.60.

• 8.61 For the circuit in Prob. 8.5, find i and v for t > 0.
• 8.62 Find the response vR(t) for t > 0 in the circuit of Fig. 8.107. Let R = 8 Ω, L = 2 H, and C = 125 mF.

Figure 8.107

For Prob. 8.62.

Section 8.8 Second-Order Op Amp Circuits

Figure 8.108 For Prob. 8.63.

• 8.64 Using Fig. 8.109, design a problem to help other students better understand second-order op amp circuits.

8.65 Determine the differential equation for the op amp circuit in Fig. 8.110. If v1(0⁺) = 2 V and v2(0⁺) = 0 V, find vo for t > 0. Let R = 100 kΩ and C = 1 μF.

Figure 8.110 For Prob. 8.65.

8.66 Obtain the differential equations for vo(t) in the op amp circuit of Fig. 8.111.

*8.67 In the op amp circuit of Fig. 8.112, determine vo(t) for t > 0. Let vin = u(t) V, R1 = R2 = 10 kΩ, C1 = C² = 100 μF.

Figure 8.112

8.68 For the step function vs = u(t), use PSpice or MultiSim to find the response v(t) for 0 < t < 6 s in the circuit of Fig. 8.113.

Figure 8.113

For Prob. 8.68.

8.69 Given the source-free circuit in Fig. 8.114, use PSpice or MultiSim to get i(t) for 0 < t < 20 s. Take v(0) = 30 V and i(0) = 2 A.

Figure 8.114

For Prob. 8.69.

8.70 For the circuit in Fig. 8.115, use PSpice or MultiSim to obtain v(t) for 0 < t < 4 s. Assume that the capacitor voltage and inductor current at t = 0 are both zero.

For Prob. 8.70.

8.71 Obtain v(t) for 0 < t < 4 s in the circuit of Fig. 8.116 using PSpice or MultiSim.

Figure 8.116

For Prob. 8.71.

8.72 The switch in Fig. 8.117 has been in position 1 for a long time. At t = 0, it is switched to position 2. Use PSpice or MultiSim to find i(t) for 0 < t < 0.2 s.

Figure 8.117

For Prob. 8.72.

8.73 Design a problem, to be solved using PSpice or MultiSim, to help other students better understand source-free RLC circuits.

Section 8.10 Duality

8.74 Draw the dual of the circuit shown in Fig. 8.118.

Figure 8.118 For Prob. 8.74.

8.75 Obtain the dual of the circuit in Fig. 8.119.

8.76 Find the dual of the circuit in Fig. 8.120.

Figure 8.120

For Prob. 8.76.

8.77 Draw the dual of the circuit in Fig. 8.121.

Figure 8.121

For Prob. 8.77.

Comprehensive Problems

• 8.80 A mechanical system is modeled by a series RLC circuit. It is desired to produce an overdamped response with time constants 0.1 and 0.5 ms. If a series 50-kΩ resistor is used, find the values of L and C.
• 8.81 An oscillogram can be adequately modeled by a second-order system in the form of a parallel RLC circuit. It is desired to give an underdamped voltage across a 200-Ω resistor. If the damped frequency is 4 kHz and the time constant of the envelope is 0.25 s, find the necessary values of L and C.
• 8.82 The circuit in Fig. 8.123 is the electrical analog of body functions used in medical schools to study convulsions. The analog is as follows:
  • C1 = Volume of fluid in a drug
  • C2 = Volume of blood stream in a specified region
  • R1 = Resistance in the passage of the drug from the input to the blood stream
  • R2 = Resistance of the excretion mechanism, such as kidney, etc.
  • v0 = Initial concentration of the drug dosage
  • v(t) = Percentage of the drug in the blood stream

Find v(t) for t > 0 given that C1 = 0.5 μF, C2 = 5 μF, R1 = 5 MΩ, R2 = 2.5 MΩ, and v0 = 60u(t) V.