Example 10.9

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$$To find **V**Th, consider the circuit in Fig. 10.23(b). Currents **I**1 and **I**² are obtained as$$

\mathbf{I}_1 = \frac{120/75^{\circ}}{8 - j6} \text{ A}, \qquad \mathbf{I}_2 = \frac{120/75^{\circ}}{4 + j12} \text{ A}

$$Applying KVL around loop *bcdeab* in Fig. 10.23(b) gives$$

\mathbf{V}_{\mathrm{Th}} - 4\mathbf{I}_2 + (-j6)\mathbf{I}_1 = 0

$$or$$

\mathbf{V}_{\text{Th}} = 4\mathbf{I}_2 + j6\mathbf{I}_1 = \frac{480/75^\circ}{4 + j12} + \frac{720/75^\circ + 90^\circ}{8 - j6}

$$$$

= 37.95/3.43^\circ + 72/201.87^\circ

$$$$

= -28.936 - j24.55 = 37.95/220.31^\circ \text{ V}

Practice Problem 10.8 Find the Thevenin equivalent at terminals *a*‑*b* of the circuit in Fig. 10.24. For Practice Prob. 10.8. **Answer: Z**Th = 12.4 − *j*3.2 Ω, **V**Th = 47.43∕−51.57° V. # Example 10.9 Find the Thevenin equivalent of the circuit in Fig. 10.25 as seen from terminals *a*‑*b*. # **Figure 10.25** For Example 10.9. # **Solution:** To find **V**Th, we apply KCL at node 1 in Fig. 10.26(a).

15 = Io + 0.5Io \Rightarrow Io = 10 A

$$Applying KVL to the loop on the right ‑hand side in Fig. 10.26(a), we obtain$$

-I_o(2-j4) + 0.5I_o(4+j3) + V_{Th} = 0

$$or$$

\mathbf{V}_{\text{Th}} = 10(2 - j4) - 5(4 + j3) = -j55

$$Thus, the Thevenin voltage is$$

V_{\text{Th}} = 55 \angle -90^{\circ} \text{ V}

$$**Figure 10.26** Solution of the problem in Fig. 10.25: (a) finding **V**Th, (b) finding **Z**Th. To obtain **Z**Th, we remove the independent source. Due to the presence of the dependent current source, we connect a 3-A current source (3 is an arbitrary value chosen for convenience here, a number divisible by the sum of currents leaving the node) to terminals *a*-*b* as shown in Fig. 10.26(b). At the node, KCL gives$$

3 = I_o + 0.5I_o \qquad \Rightarrow \qquad I_o = 2A

$$Applying KVL to the outer loop in Fig. 10.26(b) gives$$

\mathbf{V}_s = \mathbf{I}_o(4 + j3 + 2 - j4) = 2(6 - j)

$$The Thevenin impedance is$$

\mathbf{Z}_{\text{Th}} = \frac{\mathbf{V}_s}{\mathbf{I}_s} = \frac{2(6-j)}{3} = 4 - j0.6667 \ \Omega

$$Determine the Thevenin equivalent of the circuit in Fig. 10.27 as seen from the terminals *a*-*b*. **Answer:**$$

\mathbb{Z}{\text{Th}} = 4.473 \underline{\smash{\big)}, -7.64^{\circ}} ,\Omega, , \mathbf{V}{\text{Th}} = 11.763 \underline{\smash{\big)}, 72.9^{\circ}} \text{ volts.}

**Figure 10.27** For Practice Prob. 10.9. Obtain current **I***o* in Fig. 10.28 using Norton's theorem. Example 10.10 For Example 10.10. # **Solution:** Our first objective is to find the Norton equivalent at terminals *a*-*b*. **Z***^N* is found in the same way as **Z**Th. We set the sources to zero as shown in Fig. 10.29(a). As evident from the figure, the (8 − *j*2) and (10 + *j*4) impedances are short-circuited, so that

\mathbf{Z}_N = 5 \ \Omega

$$To get **I***N*, we short-circuit terminals *a*-*b* as in Fig. 10.29(b) and apply mesh analysis. Notice that meshes 2 and 3 form a supermesh because of the current source linking them. For mesh 1,$$

-j40 + (18 + j2)\mathbf{I}_1 - (8 - j2)\mathbf{I}_2 - (10 + j4)\mathbf{I}_3 = 0 \tag{10.10.1}

$$Solution of the circuit in Fig. 10.28: (a) finding **Z***N*, (b) finding **V***N*, (c) calculating **I***o*. For the supermesh,$$

(13 - j2)I2 + (10 + j4)I3 - (18 + j2)I1 = 0 \t(10.10.2)

$$At node *a*, due to the current source between meshes 2 and 3,$$

I_3 = I_2 + 3 \tag{10.10.3}

$$Adding Eqs. (10.10.1) and (10.10.2) gives$$

-j40 + 5\mathbf{I}_2 = 0 \qquad \Rightarrow \qquad \mathbf{I}_2 = j8

$$From Eq. (10.10.3),$$

\mathbf{I}_3 = \mathbf{I}_2 + 3 = 3 + j8

$$The Norton current is$$

\mathbf{I}_N = \mathbf{I}_3 = (3 + j8) \text{ A}

$$Figure 10.29(c) shows the Norton equivalent circuit along with the im ‑ pedance at terminals *a*‑*b*. By current division,$$

\mathbf{I}_o = \frac{5}{5 + 20 + j15} \mathbf{I}_N = \frac{3 + j8}{5 + j3} = 1.465 / 38.48^{\circ} \text{ A}

# Practice Problem 10.10 Determine the Norton equivalent of the circuit in Fig. 10.30 as seen from terminals *a*‑*b*. Use the equivalent to find **I***o*. # **Figure 10.30** For Practice Prob. 10.10 and Prob. 10.35. **Answer: Z***N* = 3.176 + *j*0.706 Ω, **I***N* = 8.396⧸−32.68° A, **I***o* = 1.9714⧸−2.10° A. # <span id="page-451-0"></span>**10.7** Op Amp AC Circuits The three steps stated in Section 10.1 also apply to op amp circuits, as long as the op amp is operating in the linear region. As usual, we will assume ideal op amps. (See Section 5.2.) As discussed in Chapter 5, the key to analyzing op amp circuits is to keep two important properties of an ideal op amp in mind: - 1. No current enters either of its input terminals. - 2. The voltage across its input terminals is zero. The following examples will illustrate these ideas. # **Figure 10.31** For Example 10.11: (a) the original circuit in the time domain, (b) its frequency domain equivalent. # **Solution:** We first transform the circuit to the frequency domain, as shown in Fig. 10.31(b), where **V***s* = 3⧸ 0°, *ω* = 1000 rad/s. Applying KCL at node 1, we obtain

\frac{3/0^{\circ} - V_1}{10} = \frac{V_1}{-j5} + \frac{V_1 - 0}{10} + \frac{V_1 - V_o}{20}

$$or$$

6 = (5 + j4)\mathbf{V}_1 - \mathbf{V}_o \tag{10.11.1}

$$At node 2, KCL gives$$

\frac{\mathbf{V}_1 - 0}{10} = \frac{0 - \mathbf{V}_o}{-j10}

$$which leads to$$

\mathbf{V}_1 = -j\mathbf{V}_o \tag{10.11.2}

$$Substituting Eq. (10.11.2) into Eq. (10.11.1) yields$$

6 = -j(5 + j4)\mathbf{V}_o - \mathbf{V}_o = (3 - j5)\mathbf{V}_o

$$$$

\mathbf{V}_o = \frac{6}{3 - j5} = 1.029 \angle 59.04^\circ

$$Hence,$$

v_o(t) = 1.029 \cos(1000t + 59.04^{\circ}) \text{ V}

Practice Problem 10.11 Find *vo* and *io* in the op amp circuit of Fig. 10.32. Let *vs*⁼ 12 cos 5000*t* V. **Answer:** 4 sin 5,000*t* V, 400 sin 5,000*t μ*A. **Figure 10.33** For Example 10.12. Example 10.12 Compute the closed ‑loop g ain and phase shift for the circuit in Fig. 10.33. Assume that *R*1 = *R*2 = 10 kΩ, *C*1 = 2 *μ*F, *C*2 = 1 *μ*F, and *ω* = 200 rad/s. # **Solution:** The feedback and input impedances are calculated as

\mathbf{Z}_f = R_2 \left| \frac{1}{j\omega C_2} = \frac{R_2}{1 + j\omega R_2 C_2}

$$$$

\mathbf{Z}_i = R_1 + \frac{1}{j\omega C_1} = \frac{1 + j\omega R_1 C_1}{j\omega C_1}

$$Since the circuit in Fig. 10.33 is an inverting amplifier, the closed‑loop gain is given by$$

G = \frac{V_o}{V_s} = \frac{Z_f}{Z_i} = \frac{-j\omega C_1 R_2}{(1 + j\omega R_1 C_1)(1 + j\omega R_2 C_2)}

$$Substituting the given values of *R*1, *R*2, *C*1, *C*2, and *ω*, we obtain given values of$$

R_1

, $R_2$ , $C_1$ , $C_2$ , and $\omega$ , \n

G = \frac{-j4}{(1+j4)(1+j2)} = 0.434 / 130.6^{\circ}

Thus, the closed‑loop gain is 0.434 and the phase shift is 130.6°. Practice Problem 10.12 Obtain the closed‑loop gain and phase shift for the circuit in Fig. 10.34. Let *R* = 10 kΩ, *C* = 1 *μ*F, and *ω* = 1000 rad/s. **Answer:** 1.0147, −5.6°. # <span id="page-453-0"></span>**10.8** AC Analysis Using PSpice *PSpice* affords a big relief from the tedious task of manipulating com‑ plex numbers in ac circuit analysis. The procedure for using *PSpice* for ac analysis is quite similar to that required for dc analysis. The reader should read Section D.5 in Appendix D for a review of *PSpice* concepts for ac analysis. AC circuit analysis is done in the phasor or frequency domain, and all sources must have the same frequency. Although ac analysis with *PSpice* involves using AC Sweep, our analysis in this chapter requires a single frequency *f* = *ω*∕2*π*. The out‑ put file of *PSpice* contains voltage and current phasors. If necessary, the impedances can be calculated using the voltages and currents in the output file. Obtain *vo* and *io* in the circuit of Fig. 10.35 using *PSpice*. Example 10.13 # **Solution:** We first convert the sine function to cosine.

8 \sin(1000t + 50^{\circ}) = 8 \cos(1000t + 50^{\circ} - 90^{\circ})

$$$$

= 8 \cos(1000t - 40^{\circ})

$$The frequency *f* is obtained from *ω* as$$

f = \frac{\omega}{2\pi} = \frac{1000}{2\pi} = 159.155

Hz The schematic for the circuit is shown in Fig. 10.36. Notice that the current‑controlled current source F1 is connected such that its current flows from node 0 to node 3 in conformity with the original circuit in Fig. 10.35. Since we only want the magnitude and phase of *vo* and *io*, we set the attributes of IPRINT and VPRINT1 each to *AC* = *yes*, *MAG* = *yes*, *PHASE* = *yes*. As a single ‑frequency analysis, we select **Analysis/ Setup/AC Sweep** and enter *Total Pts* = 1, *Start Freq* = 159.155, and *Final Freq* = 159.155. After saving the schematic, we simulate it by selecting **Analysis/Simulate.** The output file includes the source fre‑ quency in addition to the attributes checked for the pseudocomponents IPRINT and VPRINT1, | FREQ | | IM(V_PRINT3) IP(V_PRINT3) | |-----------|-----------|---------------------------| | 1.592E+02 | 3.264E–03 | –3.743E+01 | | FREQ | VM(3) | VP(3) | | 1.592E+02 | 1.550E+00 | –9.518E+01 | **Figure 10.36** The schematic of the circuit in Fig. 10.35. From this output file, we obtain **V***o* = 1.55⧸−95.18° V, **I***o* = 3.264⧸−37.43° mA which are the phasors for *vo* = 1.55 cos(1000*t* − 95.18°) = 1.55 sin(1000*t* − 5.18°) V and *io* = 3.264 cos(1000*t* − 37.43°) mA For Practice Prob. 10.13. **Answer:** 3.219 cos(3,000*t* − 154.6°) V, 6.527 cos(3,000*t* − 55.12°) mA. Example 10.14 Find **V**1 and **V**2 in the circuit of Fig. 10.38. # **Solution:** 1. **Define.** In its present form, the problem is clearly stated. Again, we must emphasize that time spent here will save lots of time and expense later on! One thing that might have created a problem for you is that, if the reference was missing for this problem, you would then need to ask the individual assigning the problem where it is to be located. If you could not do that, then you would need to assume where it should be and then clearly state what you did and why you did it. - 2. **Present.** The given circuit is a frequency domain circuit and the unknown node voltages **V**1 and **V**2 are also frequency domain values. Clearly, we need a process to solve for these unknowns in the frequency domain. - 3. **Alternative.** We have two direct alternative solution techniques that we can easily use. We can do a straightforward nodal analysis approach or use *PSpice*. Since this example is in a section dedicated to using *PSpice* to solve problems, we will use *PSpice* to find **V**1 and **V**2. We can then use nodal analysis to check the answer. - 4. **Attempt.** The circuit in Fig. 10.35 is in the time domain, whereas the one in Fig. 10.38 is in the frequency domain. Since we are not given a particular frequency and *PSpice* requires one, we select any frequency consistent with the given impedances. For example, if we select *ω* = 1 rad/s, the corresponding frequency is *f* = *ω*∕2*π* = 0.15916 Hz. We obtain the values of the capacitance (*C* = 1∕*ωXC*) and inductances (*L* = *XL*∕*ω*). Making these changes results in the schematic in Fig. 10.39. To ease wiring, we have exchanged the positions of the voltage‑controlled current source # **Figure 10.39** Schematic for the circuit in the Fig. 10.38. G1 and the 2 + *j*2 Ω impedance. Notice that the current of G1 flows from node 1 to node 3, while the controlling voltage is across the capacitor C2, as required in Fig. 10.38. The attributes of pseudo components VPRINT1 are set as shown. As a single‑ frequency analysis, we select **Analysis/Setup/AC Sweep** and enter *Total Pts* = 1, *Start Freq* = 0.15916, and *Final Freq* = 0.15916. After saving the schematic, we select **Analysis/Simulate** to simulate the circuit. When this is done, the output file includes | FREQ | VM(1) | VP(1) | |-----------|-----------|------------| | 1.592E–01 | 2.708E+00 | –5.673E+01 | | | | | | FREQ | VM(3) | VP(3) | | 1.592E-01 | 4.468E+00 | –1.026E+02 | from which we obtain, **V**1 = **2.708**⧸**−56.74° V** and **V**2 = **6.911**⧸**−80.72° V** 5. **Evaluate.** One of the most important lessons to be learned is that when using programs such as *PSpice* you still need to validate the answer. There are many opportunities for making a mistake, including coming across an unknown "bug" in *PSpice* that yields incorrect results. So, how can we validate this solution? Obviously, we can rework the entire problem with nodal analysis, and perhaps using *MATLAB*, to see if we obtain the same results. There is another way we will use here: Write the nodal equations and substitute the answers obtained in the *PSpice* solution, and see if the nodal equations are satisfied. The nodal equations for this circuit are given below. Note we have substituted **V**1 = **V***x* into the dependent source.

-3 + \frac{\mathbf{V}_1 - 0}{1} + \frac{\mathbf{V}_1 - 0}{-j1} + \frac{\mathbf{V}_1 - \mathbf{V}_2}{2 + j2} + 0.2\mathbf{V}_1 + \frac{\mathbf{V}_1 - \mathbf{V}_2}{-j2} = 0

(1 + j + 0.25 - j0.25 + 0.2 + j0.5) $\mathbf{V}_1$ -(0.25 - j0.25 + j0.5) $\mathbf{V}_2$ = 3 (1.45 + j1.25) $\mathbf{V}_1$ - (0.25 + j0.25) $\mathbf{V}_2$ = 3 1.9144/40.76° $\mathbf{V}_1$ - 0.3536/45° $\mathbf{V}_2$ = 3 Now, to check the answer, we substitute the *PSpice* answers into this.

1.9144 \underline{/40.76^{\circ}} \times 2.708 \underline{/-56.74^{\circ}} - 0.3536 \underline{/45^{\circ}} \times 6.911 \underline{/-80.72^{\circ}}

= 5.184 \underline{/-15.98^{\circ}} - 2.444 \underline{/-35.72^{\circ}} = 4.984 - j1.4272 - 1.9842 + j1.4269 = 3 - j0.0003 [Answer checks] 6. **Satisfactory?** Although we used only the equation from node 1 to check the answer, this is more than satisfactory to validate the answer from the *PSpice* solution. We can now present our work as a solution to the problem. <span id="page-457-0"></span>Obtain **V***x* and **I***x* in the circuit depicted in Fig. 10.40. Practice Problem 10.14 For Practice Prob. 10.14. **Answer:** 39.37⧸ 44.78° V, 10.336⧸158° A. # **10.9** Applications The concepts learned in this chapter will be applied in later chapters to calculate electric power and determine frequency response. The con ‑ cepts are also used in analyzing coupled circuits, three‑phase circuits, ac transistor circuits, filters, oscillators, and other ac circuits. In this section, we apply the concepts to develop two practical ac circuits: the capaci ‑ tance multiplier and the sine wave oscillators. # **10.9.1** Capacitance Multiplier The op amp circuit in Fig. 10.41 is known as a *capacitance multiplier*, for reasons that will become obvious. Such a circuit is used in integrated‑ circuit technology to produce a multiple of a small physical capacitance *C* when a large capacitance is needed. The circuit in Fig. 10.41 can be used to multiply capacitance values by a factor up to 1,000. For exam‑ ple, a 10‑pF capacitor can be made to behave like a 100‑nF capacitor. In Fig. 10.41, the first op amp operates as a voltage follower, while the second one is an inverting amplifier. The voltage follower iso‑ lates the capacitance formed by the circuit from the loading imposed by the inverting amplifier. Since no current enters the input terminals of the op amp, the input current **I***i* flows through the feedback capaci‑ tor. Hence, at node 1,

\mathbf{I}{i} = \frac{\mathbf{V}{i} - \mathbf{V}{o}}{1/j\omega C} = j\omega C(\mathbf{V}{i} - \mathbf{V}_{o})

$$\n(10.3) Applying KCL at node 2 gives$$

\frac{\mathbf{V}_i - \mathbf{0}}{R_1} = \frac{\mathbf{0} - \mathbf{V}_o}{R_2}

$$or$$

\mathbf{V}_o = -\frac{R_2}{R_1} \mathbf{V}_i \tag{10.4}

$$Substituting Eq. (10.4) into (10.3) gives$$

\mathbf{I}_i = j\omega C \bigg( 1 + \frac{R_2}{R_1} \bigg) \mathbf{V}_i

$$or$$

\frac{\mathbf{I}_i}{\mathbf{V}_i} = j\omega \left( 1 + \frac{R_2}{R_1} \right) C \tag{10.5}

$$The input impedance is$$

\mathbf{Z}{i} = \frac{\mathbf{V}{i}}{\mathbf{I}{i}} = \frac{1}{j\omega C{\text{eq}}}

$$(10.6) where$$

C_{\text{eq}} = \left(1 + \frac{R_2}{R_1}\right)C\tag{10.7}

Thus, by a proper selection of the values of *R*1 and *R*2, the op amp circuit in Fig. 10.41 can be made to produce an effective capacitance between the input terminal and ground, which is a multiple of the physical capaci‑ tance *C*. The size of the effective capacitance is practically limited by the inverted output voltage limitation. Thus, the larger the capacitance multiplication, the smaller is the allowable input voltage to prevent the op amps from reaching saturation. A similar op amp circuit can be designed to simulate inductance. (See Prob. 10.89.) There is also an op amp circuit configuration to create a resistance multiplier. Example 10.15 Calculate *C*eq in Fig. 10.41 when *R*1 = 10 kΩ, *R*2 = 1 MΩ, and *C* = 1 nF. # **Solution:** From Eq. (10.7) *^C*eq = (1 + \_\_\_ *R*2 *R*1 )*C* = (1 + 1 × ¹⁰⁶ \_\_\_\_\_\_\_\_ 10 × 103 ) 1 nF = 101 nF Determine the equivalent capacitance of the op amp circuit in Fig. 10.41 if *R*1 = 10 kΩ, *R*2 = 10 MΩ, and *C* = 10 nF. **Answer:** 10 *μ*F. # **10.9.2** Oscillators We know that dc is produced by batteries. But how do we produce ac? One way is using *oscillators,* which are circuits that convert dc to ac. An oscillator is a circuit that produces an ac waveform as output when powered by a dc input. The only external source an oscillator needs is the dc power supply. Ironically, the dc power supply is usually obtained by con verting the ac supplied by the electric utility company to dc. Having gone through the trouble of conversion, one may wonder why we need to use the oscillator to convert the dc to ac again. The problem is that the ac supplied by the utility company operates at a preset frequenc y of 60 Hz in the United States (50 Hz in some other nations), whereas man y applications such as electronic circuits, communication systems, and micro wave devices require internally generated frequencies that range from 0 to 10 GHz or higher. Oscillators are used for generating these frequencies. In order for sine w ave oscillators to sustain oscillations, the y must meet the *Barkhausen criteria*: - 1. The overall gain of the oscillator must be unity or greater. Therefore, losses must be compensated for by an amplifying device. - 2. The overall phase shift (from input to output and back to the input) must be zero. Three common types of sine wave oscillators are phase ‑shift, twin *T*, and Wien ‑bridge oscillators. Here we consider only the Wien ‑bridge oscillator. The *Wien-bridge oscillator* is widely used for generating sinusoids in the frequency range below 1 MHz. It is an *RC* op amp circuit with only a few components, easily tunable and easy to design. As shown in Fig. 10.42, the oscillator essentially consists of a noninverting amplifier with two feedback paths: The positive feedback path to the noninverting input creates oscillations, while the ne gative feedback path to the in verting input controls the gain. If we define the impedances of the *RC* series and parallel combinations as **Z***s* and **Z***p*, then

Z_s = R_1 + \frac{1}{j\omega C_1} = R_1 - \frac{j}{\omega C_1}

$$(10.8)$$

Z_p = R_2 \Big| \frac{1}{j\omega C_2} = \frac{R_2}{1 + j\omega R_2 C_2}

$$(10.9) The feedback ratio is$$

\frac{\mathbf{V}_2}{\mathbf{V}_o} = \frac{\mathbf{Z}_p}{\mathbf{Z}_s + \mathbf{Z}_p}

$$Practice Problem 10.15 Substituting Eqs. (10.8) and (10.9) into Eq. (10.10) gives Substituting Eqs. (10.8) and (10.9) into Eq. (10.10) gives \n$$

\frac{\mathbf{V}_2}{\mathbf{V}_o} = \frac{R_2}{R_2 + \left(R_1 - \frac{j}{\omega C_1}\right)(1 + j\omega R_2 C_2)}

$$\n$$

= \frac{\omega R_2 C_1}{\omega (R_2 C_1 + R_1 C_1 + R_2 C_2) + j(\omega^2 R_1 C_1 R_2 C_2 - 1)}

$$\n(10.11) To satisfy the second Barkhausen criterion, **V**2 must be in phase with **V***o*, which implies that the ratio in Eq. (10.11) must be purely real. Hence, the imaginary part must be zero. Setting the imaginary part equal to zero gives the oscillation frequency *ωo* as$$

\omega_o^2 R_1 C_1 R_2 C_2 - 1 = 0

$$or$$

\omega_o = \frac{1}{\sqrt{R_1 R_2 C_1 C_2}}\tag{10.12}

$$In most practical applications, *R*1 = *R*2 = *R* and *C*1 = *C*2 = *C*, so that$$

\omega_o = \frac{1}{RC} = 2\pi f_o \tag{10.13}

$$or$$

f_o = \frac{1}{2\pi RC}

$$(10.14) Substituting Eq. (10.13) and *R*1 = *R*2 = *R*, *C*1 = *C*2 = *C* into Eq. (10.11) yields$$

\frac{V_2}{V_o} = \frac{1}{3}

$$(10.15) Thus, in order to satisfy the first Barkhausen criterion, the op amp must compensate by providing a gain of 3 or greater so that the overall gain is at least 1 or unity. We recall that for a noninverting amplifier,$$

\frac{\mathbf{V}_o}{\mathbf{V}_2} = 1 + \frac{R_f}{R_g} = 3\tag{10.16}

$$or$$

R_f = 2R_g \tag{10.17}

Due to the inherent delay caused by the op amp, Wien‑bridge oscil‑ lators are limited to operating in the frequency range of 1 MHz or less. # Example 10.16 Design a Wien‑bridge circuit to oscillate at 100 kHz. # **Solution:** Using Eq. (10.14), we obtain the time constant of the circuit as 14), we obtain the time constant of the circuit as

RC = \frac{1}{2\pi f_o} = \frac{1}{2\pi \times 100 \times 10^3} = 1.59 \times 10^{-6}

(10.16.1) If we select *R* = 10 k Ω, then we can select *C* = 159 pF to satisfy Eq. (10.16.1). Since the gain must be 3, *Rf*∕*Rg* = 2. We could select *Rf* = 20 kΩ while *Rg* = 10 kΩ. <span id="page-461-0"></span>In the Wien‑bridge oscillator circuit in Fig. 10.42, let *R*1 = *R*2 = 2.5 kΩ, *C*1 = *C*2 = 1 nF. Determine the frequency *fo* of the oscillator. Practice Problem 10.16 **Answer:** 63.66 kHz. # **10.10** Summary - 1. We apply nodal and mesh analysis to ac circuits by applying KCL and KVL to the phasor form of the circuits. - 2. In solving for the steady ‑state response of a circuit that has inde ‑ pendent sources with different frequencies, each independent source *must* be considered separately. The most natural approach to analyz‑ ing such circuits is to apply the superposition theorem. A separate phasor circuit for each frequency *must* be solved independently, and the corresponding response should be obtained in the time domain. The overall response is the sum of the time domain responses of all the individual phasor circuits. - 3. The concept of source transformation is also applicable in the fre ‑ quency domain. - 4. The Thevenin equivalent of an ac circuit consists of a voltage source **V**Th in series with the Thevenin impedance **Z**Th. - 5. The Norton equivalent of an ac circuit consists of a current source **I***^N* in parallel with the Norton impedance **Z***N* (=**Z**Th). - 6. *PSpice* is a simple and powerful tool for solving ac circuit problems. It relieves us of the tedious task of working with the complex num‑ bers involved in steady‑state analysis. - 7. The capacitance multiplier and the ac oscillator provide two typical applications for the concepts presented in this chapter . A capaci‑ tance multiplier is an op amp circuit used in producing a multiple of a physical capacitance. An oscillator is a device that uses a dc input to generate an ac output. ‒ # Review Questions 10 0° V ‒j1 Ω **V**^o + **10.1** The voltage **V***o* across the capacitor in Fig. 10.43 is: ‒ For Review Question 10.1. **10.2** The value of the current **I***o* in the circuit of Fig. 10.44 is: (a)

4\angle 0^{\circ}

A (b) $2.4\angle -90^{\circ}$ A (c) $0.6\angle 0^{\circ}$ A (d) $-1$ A **Figure 10.44** For Review Question 10.2. **10.3** Using nodal analysis, the value of **V***o* in the circuit of Fig. 10.45 is: (a)

-24 \text{ V}

(b) $-8 \text{ V}$

(c) 8 V \t\t (d) 24 V

# **10.6** For the circuit in Fig. 10.48, the Thevenin impedance at terminals *a*‑*b* is: | (a) 1 Ω | (b) 0.5 − j0.5 Ω | |------------------|------------------| | (c) 0.5 + j0.5 Ω | (d) 1 + j2 Ω | | (e) 1 − j2 Ω | | # **Figure 10.48** For Review Questions 10.6 and 10.7. - - (a) 10 cos *t* A (b) 10 sin *t* A (c) 5 cos *t* A **10.4** In the circuit of Fig. 10.46, current *i*(*t*) is: (d) 5 sin *t* A (e) 4.472 cos(*t* − 63.43°) A **Figure 10.46** **Figure 10.45** For Review Question 10.3. For Review Question 10.4. - **10.5** Refer to the circuit in Fig. 10.47 and observe that the two sources do not have the same frequency. The current *ix*(*t*) can be obtained by: - (a) source transformation - (b) the superposition theorem - (c) *PSpice* **Figure 10.47** For Review Question 10.5. **10.7** In the circuit of Fig. 10.48, the Thevenin voltage at terminals *a*‑*b* is: (a)

3.535 \angle -45^{\circ} \text{ V}

(b) $3.535 \angle 45^{\circ} \text{ V}$ (c) $7.071 \angle -45^{\circ} \text{ V}$ (d) $7.071 \angle 45^{\circ} \text{ V}$ **10.8** Refer to the circuit in Fig. 10.49. The Norton equivalent impedance at terminals *a*‑*b* is: | (a) −j4 Ω | (b) −j2 Ω | |-----------|-----------| | (c) j2 Ω | (d) j4 Ω | # **Figure 10.49** For Review Questions 10.8 and 10.9. **10.9** The Norton current at terminals *a*‑*b* in the circuit of Fig. 10.49 is: (a)

1/\underline{0^{\circ}}

A (b) $1.5/\underline{-90^{\circ}}$ A (c) $1.5/90^{\circ}$ A (d) $3/90^{\circ}$ A - **10.10** *PSpice* can handle a circuit with two independent sources of different frequencies. - (a) True (b) False *Answers: 10.1c, 10.2a, 10.3d, 10.4a, 10.5b, 10.6c, 10.7a, 10.8a, 10.9d, 10.10b.* # <span id="page-463-0"></span>Problems # Section 10.2 Nodal Analysis **10.1** Determine *i* in the circuit of Fig. 10.50. # **Figure 10.50** For Prob. 10.1. # **Figure 10.51** For Prob. 10.2. **10.3** Determine *vo* in the circuit of Fig. 10.52. # **Figure 10.52** For Prob. 10.3. # **Figure 10.53** For Prob. 10.4. For Prob. 10.5. **10.6** Determine **V***x* in Fig. 10.55. # **Figure 10.55** For Prob. 10.6. **10.7** Use nodal analysis to find **V** in the circuit of Fig. 10.56. # **Figure 10.56** For Prob. 10.7. **10.8** Use nodal analysis to find current *io* in the circuit of Fig. 10.57. Let *is* = 6 cos(200*t* + 15°) A. # **Figure 10.57** For Prob. 10.8. For Prob. 10.9. ‒ **Figure 10.59** **10.11** Using nodal analysis, find *io*(*t*) in the circuit in Fig. 10.60. For Prob. 10.11. **10.12** Using Fig. 10.61, design a problem to help other students better understand nodal analysis. **Figure 10.61** For Prob. 10.12. **10.13** Determine **V***x* in the circuit of Fig. 10.62 using any method of your choice. For Prob. 10.13. **10.14** Calculate the voltage at nodes 1 and 2 in the circuit of Fig. 10.63 using nodal analysis. # **Figure 10.63** For Prob. 10.14. **10.15** Solve for the current **I** in the circuit of Fig. 10.64 using nodal analysis. **Figure 10.64** For Prob. 10.15. # **Figure 10.65** For Prob. 10.16. **10.17** By nodal analysis, obtain current **I***o* in the circuit of Fig. 10.66. **Figure 10.66** For Prob. 10.17. **10.19** Obtain **V***o* in Fig. 10.68 using nodal analysis. **10.20** Refer to Fig. 10.69. If *vs*(*t*) = *Vm* sin *ωt* and *vo*(*t*) = *A* sin(*ωt* + *ϕ*), derive the expressions for *A* and *ϕ*. **Figure 10.69** For Prob. 10.20. **10.21** For each of the circuits in Fig. 10.70, find **V***o*∕**V***i* for *ω* = 0, *ω* → ∞, and *ω*² = 1∕*LC*. **Figure 10.70** For Prob. 10.21. **10.22** For the circuit in Fig. 10.71, determine **V***o*∕**V***s*. **Figure 10.71** For Prob. 10.22. **10.23** Using nodal analysis obtain **V** in the circuit of Fig. 10.72. **Figure 10.72** For Prob. 10.23. # Section 10.3 Mesh Analysis **10.24** Design a problem to help other students better understand mesh analysis. **10.25** Solve for *io* in Fig. 10.73 using mesh analysis. For Prob. 10.25. **10.26** Use mesh analysis to find current *io* in the circuit of Fig. 10.74. **10.29** Using Fig. 10.77, design a problem to help other students better understand mesh analysis. R3 For Prob. 10.26. **10.27** Using mesh analysis, find **I**1 and **I**2 in the circuit of Fig. 10.75. jXL¹ **Figure 10.77** For Prob. 10.29. For Prob. 10.30. **Figure 10.76** For Prob. 10.28. **10.32** Determine **V***o* and **I***o* in the circuit of Fig. 10.80 using mesh analysis. **Figure 10.80** For Prob. 10.32. **10.33** Compute **I** in Prob. 10.15 using mesh analysis. **10.34** Use mesh analysis to find **I***o* in Fig. 10.28 (for Example 10.10). **10.35** Calculate **I***o* in Fig. 10.30 (for Practice Prob. 10.10) using mesh analysis. **10.36** Compute **V***o* in the circuit of Fig. 10.81 using mesh analysis. **Figure 10.81** For Prob. 10.36. **10.37** Use mesh analysis to find currents **I**1, **I**2, and **I**3 in the circuit of Fig. 10.82. **10.38** Using mesh analysis, obtain **I***o* in the circuit shown in Fig. 10.83. For Prob. 10.38. **10.39** Find **I**1, **I**2, **I**3, and **I***x* in the circuit of Fig. 10.84. **Figure 10.84** For Prob. 10.39. # Section 10.4 Superposition Theorem **10.40** Find *io* in the circuit shown in Fig. 10.85 using superposition. # **Figure 10.85** For Prob. 10.40. **10.41** Find *vo* for the circuit in Fig. 10.86, assuming that *is*(*t*) = 2 sin (2*t*) + 3 cos (4*t*) A. **Figure 10.86** For Prob. 10.41. **10.42** Using Fig. 10.87, design a problem to help other students better understand the superposition theorem. **Figure 10.87** For Prob. 10.42. For Prob. 10.43. **10.43** Using the superposition principle, find *ix* in the circuit of Fig. 10.88. For Prob. 10.46. **10.47** Determine *io* in the circuit of Fig. 10.92, using the superposition principle. For Prob. 10.47. **10.44** Use the superposition principle to obtain *vx* in the circuit of Fig. 10.89. Let *vs* = 50 sin 2*t* V and *is* = 12 cos(6*t* + 10°) A. **10.45** Use superposition to find *i*(*t*) in the circuit of Fig. 10.90. **Figure 10.90** For Prob. 10.45. **10.48** Find *io* in the circuit of Fig. 10.93 using superposition. **Figure 10.93** For Prob. 10.48. # Section 10.5 Source Transformation **10.49** Using source transformation, find *i* in the circuit of Fig. 10.94. For Prob. 10.49. **Figure 10.95** For Prob. 10.50. - **10.51** Use source transformation to find **I***o* in the circuit of Prob. 10.42. - **10.52** Use the method of source transformation to find **I***x* in the circuit of Fig. 10.96. **Figure 10.96** For Prob. 10.52. **10.53** Use the concept of source transformation to find **V***^o* in the circuit of Fig. 10.97. For Prob. 10.53. **10.54** Rework Prob. 10.7 using source transformation. # Section 10.6 Thevenin and Norton Equivalent Circuits **10.55** Find the Thevenin and Norton equivalent circuits at terminals *a*‑*b* for each of the circuits in Fig. 10.98. **Figure 10.98** For Prob. 10.55. **10.56** For each of the circuits in Fig. 10.99, obtain Thevenin and Norton equivalent circuits at terminals *a*‑*b*. # **Figure 10.99** For Prob. 10.56. **10.57** Using Fig. 10.100, design a problem to help other students better understand Thevenin and Norton equivalent circuits. # **Figure 10.100** For Prob. 10.57. **10.58** For the circuit depicted in Fig. 10.101, find the Thevenin equivalent circuit at terminals *a*‑*b*. **Figure 10.101** For Prob. 10.58. **10.59** Calculate the output impedance of the circuit shown in Fig. 10.102. For Prob. 10.59. **10.60** Find the Thevenin equivalent of the circuit in Fig. 10.103 as seen from: **10.61** Find the Thevenin equivalent at terminals *a*-*b* of the circuit in Fig. 10.104. **Figure 10.104** For Prob. 10.61. **10.62** Using Thevenin's theorem, find *vo* in the circuit of Fig. 10.105. For Prob. 10.62. **10.63** Obtain the Norton equivalent of the circuit depicted in Fig. 10.106 at terminals *a*-*b*. For Prob. 10.63. **10.64** For the circuit shown in Fig. 10.107, find the Norton equivalent circuit at terminals *a*-*b*. # **Figure 10.107** For Prob. 10.64. **10.65** Using Fig. 10.108, design a problem to help other students better understand Norton's theorem. Problems **449** For Prob. 10.70. **10.71** Find *vo* in the op amp circuit of Fig. 10.114. **Figure 10.114** For Prob. 10.71. **10.72** Compute *io*(*t*) in the op amp circuit in Fig. 10.115 if *vs* = 4 cos(104 *t*) V. # **Figure 10.115** For Prob. 10.72. **10.73** If the input impedance is defined as **Z**in = **V***s*∕**I***s*, find the input impedance of the op amp circuit in Fig. 10.116 when *R*1 = 10 kΩ, *R*2 = 20 kΩ, *C*1 = 10 nF, *C*2 = 20 nF, and *ω* = 5000 rad/s. **Figure 10.116** For Prob. 10.73. **Figure 10.111** For Prob. 10.68. **Figure 10.110** For Prob. 10.67.