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CHAPTER 3 Methods of Analysis 107

P R A C T I C E P R O B L E M 3 . 1 3

The transistor circuit in Fig. 3.45 hasβ=80 andV_BE =0.7 V. Findv_o andi_o.

1 V

10 V 30 k"

20 k"

+

! i_o

V_BE +

!

v_o +

! +

!

Figure 3.45 For Practice Prob. 3.13.

Answer: −3 V,−150µA.

3.10 SUMMARY

1. Nodal analysis is the application of Kirchhoff’s current law at the nonreference nodes. (It is applicable to both planar and nonplanar circuits.) We express the result in terms of the node voltages.

Solving the simultaneous equations yields the node voltages.

2. A supernode consists of two nonreference nodes connected by a (dependent or independent) voltage source.

3. Mesh analysis is the application of Kirchhoff’s voltage law around meshes in a planar circuit. We express the result in terms of mesh currents. Solving the simultaneous equations yields the mesh currents.

4. A supermesh consists of two meshes that have a (dependent or independent) current source in common.

5. Nodal analysis is normally used when a circuit has fewer node equations than mesh equations. Mesh analysis is normally used when a circuit has fewer mesh equations than node equations.

6. Circuit analysis can be carried out usingPSpice.

7. DC transistor circuits can be analyzed using the techniques cover- ed in this chapter.

R E V I E W Q U E S T I O N S

3.1 At node 1 in the circuit in Fig. 3.46, applying KCL gives:

(a) 2+12−v1

3 = v1

6 +v1−v2

4 (b) 2+v1−12

3 = v1

6 +v2−v1

4

(c) 2+12−v₁

3 = 0−v₁

6 +v₁−v₂ 4 (d) 2+v1−12

3 = 0−v1

6 +v2−v1

4

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The transistor circuit in Fig. 3.45 has β = 80 and V_BE = 0.7 V. Find v_o and i_o.

1 V

10 V 30 k"

20 k"

+

! i_o

V_BE +

!

v_o +

! +

!

Answer: −3 V, −150 µA.

3.10 SUMMARY

6. Circuit analysis can be carried out using PSpice.

(a) 2+ 12−v₁

3 = v₁

6 + v₁ −v₂ 4 (b) 2+ v₁−12

3 = v₁

6 + v₂ −v₁ 4

(c) 2+ 12−v₁

3 = 0−v₁

6 + v₁ −v₂ 4 (d) 2+ v₁ −12

3 = 0−v₁

6 + v₂ −v₁ 4

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The transistor circuit in Fig. 3.45 has β = 80 and V_BE = 0.7 V. Find v_o and i_o.

1 V

10 V 30 k"

20 k"

+

! i_o

V_BE +

!

v_o +

! +

!

Answer: −3 V, −150 µA.

3.10 SUMMARY

6. Circuit analysis can be carried out using PSpice.

(a) 2+ 12−v₁

3 = v₁

6 + v₁ −v₂ 4 (b) 2+ v₁ −12

3 = v₁

6 + v₂ −v₁ 4

(c) 2+ 12−v₁

3 = 0−v₁

6 + v₁ −v₂ 4 (d) 2+ v₁ −12

3 = 0−v₁

6 + v₂ −v₁ 4

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108 PART 1 DC Circuits

2 A

v₁

1 2

v₂ 12 V +

!

3 " 4 "

6 " 6 "

8 "

Figure 3.46 For Review Questions 3.1 and 3.2.

3.2 In the circuit in Fig. 3.46, applying KCL at node 2 gives:

(a) v₂−v₁ 4 + v₂

8 = v₂ 6 (b) v₁−v₂

4 + v₂ 8 = v₂

6 (c) v₁−v₂

4 + 12−v₂ 8 = v₂

6 (d) v2−v1

4 + v2−12 8 = v2

6

3.3 For the circuit in Fig. 3.47,v1andv2are related as:

(a) v1=6i+8+v2 (b) v1=6i−8+v2

(c) v1=−6i+8+v2 (d) v1=−6i−8+v2

12 V +

! 4 "

6 " 8 V v₂ v₁

i

! +

3.4 In the circuit in Fig. 3.47, the voltagev₂is:

(a) −8 V (b) −1.6 V

(c) 1.6 V (d) 8 V

3.5 The currentiin the circuit in Fig. 3.48 is:

(a) −2.667 A (b) −0.667 A

(c) 0.667 A (d) 2.667 A

10 V +! +! 6 V

4 "

i

2 "

3.6 The loop equation for the circuit in Fig. 3.48 is:

(a) −10+4i+6+2i =0 (b) 10+4i+6+2i =0 (c) 10+4i−6+2i =0 (d) −10+4i−6+2i= 0

3.7 In the circuit in Fig. 3.49, currenti₁is:

(a) 4 A (b) 3 A (c) 2 A (d) 1 A

i₁ 2 A i₂

20 V +!

2 " 1 "

3 " 4 "

v⁺

!

3.8 The voltagevacross the current source in the circuit of Fig. 3.49 is:

(a) 20 V (b) 15 V (c) 10 V (d) 5 V 3.9 ThePSpicepart name for a current-controlled

voltage source is:

(a) EX (b) FX (c) HX (d) GX

3.10 Which of the following statements are not true of the pseudocomponent IPROBE:

(a) It must be connected in series.

(b) It plots the branch current.

(c) It displays the current through the branch in which it is connected.

(d) It can be used to display voltage by connecting it in parallel.

(e) It is used only for dc analysis.

(f) It does not correspond to a particular circuit element.

Answers: 3.1a, 3.2c, 3.3b, 3.4d, 3.5c, 3.6a, 3.7d, 3.8b, 3.9c, 3.10b,d.

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2 A

v₁

1 2

v₂ 12 V +

!

3 " 4 "

6 " 6 "

8 "

(a) v₂−v₁ 4 + v₂

8 = v₂ 6 (b) v₁−v₂

4 + v₂ 8 = v₂

6 (c) v1−v2

4 + 12−v2

8 = v2

6 (d) v2−v1

4 + v2−12 8 = v2

6

3.3 For the circuit in Fig. 3.47,v₁andv₂ are related as:

(a) v1 = 6i+8+v2 (b) v1 =6i −8+v2

(c) v₁ = −6i +8+v₂ (d) v₁ =−6i−8+v₂

12 V +! 4 "

6 " 8 V v₂ v₁

i

! +

3.4 In the circuit in Fig. 3.47, the voltagev2 is:

(a) −8 V (b) −1.6 V

(c) 1.6 V (d) 8 V

3.5 The currenti in the circuit in Fig. 3.48 is:

(a) −2.667 A (b) −0.667 A

(c) 0.667 A (d) 2.667 A

10 V +

! + 6 V

! 4 "

i

2 "

(a) −10+4i +6+2i = 0 (b) 10+4i +6+2i =0 (c) 10+4i −6+2i =0 (d) −10+4i −6+2i = 0

3.7 In the circuit in Fig. 3.49, currenti1 is:

(a) 4 A (b) 3 A (c) 2 A (d) 1 A

i₁ 2 A i₂

20 V +

!

2 " 1 "

3 " 4 "

v⁺

!

3.8 The voltagev across the current source in the circuit of Fig. 3.49 is:

voltage source is:

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2 A

v₁

1 2

v₂ 12 V +

!

3 " 4 "

6 " 6 "

8 "

(a) v2−v1

4 + v2

8 = v2

6 (b) v₁−v₂

4 + v₂ 8 = v₂

6 (c) v₁−v₂

4 + 12−v₂ 8 = v₂

6 (d) v₂−v₁

4 + v₂−12 8 = v₂

6

3.3 For the circuit in Fig. 3.47,v₁ andv₂ are related as:

(a) v₁ =6i+8+v₂ (b) v₁ =6i−8+v₂ (c) v₁ =−6i+8+v₂ (d) v₁ =−6i−8+v₂

12 V +

! 4 "

6 " 8 V v₂ v₁

i

! +

3.4 In the circuit in Fig. 3.47, the voltagev₂is:

(a) −8 V (b) −1.6 V

(c) 1.6 V (d) 8 V

3.5 The currentiin the circuit in Fig. 3.48 is:

(a) −2.667 A (b) −0.667 A

(c) 0.667 A (d) 2.667 A

10 V +

! + 6 V

! 4 "

i

2 "

(a) −10+4i+6+2i=0 (b) 10+4i+6+2i =0 (c) 10+4i−6+2i =0 (d) −10+4i−6+2i = 0

3.7 In the circuit in Fig. 3.49, currenti₁is:

(a) 4 A (b) 3 A (c) 2 A (d) 1 A

i₁ 2 A i₂

20 V +

!

2 " 1 "

3 " 4 "

v+

!

3.8 The voltagevacross the current source in the circuit of Fig. 3.49 is:

voltage source is:

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2 A

v₁

1 2

v₂ 12 V +

!

3 " 4 "

6 " 6 "

8 "

(a) v₂ −v₁

4 + v₂

8 = v₂ 6 (b) v₁ −v₂

4 + v₂

8 = v₂ 6 (c) v₁ −v₂

4 + 12−v₂

8 = v₂ 6 (d) v₂ −v₁

4 + v₂ −12

8 = v₂ 6

3.3 For the circuit in Fig. 3.47,v₁ and v₂ are related as:

(a) v₁ = 6i+8+v₂ (b) v₁ = 6i−8+v₂ (c) v₁ = −6i +8+v₂ (d) v₁ = −6i −8+v₂

12 V +

! 4 "

6 " 8 V v₂ v₁

i

! +

3.4 In the circuit in Fig. 3.47, the voltagev₂ is:

(a) −8 V (b) −1.6 V

(c) 1.6 V (d) 8 V

3.5 The current i in the circuit in Fig. 3.48 is:

(a) −2.667 A (b) −0.667 A

(c) 0.667 A (d) 2.667 A

10 V +

! + 6 V

! 4 "

i

2 "

(a) −10+4i +6+2i =0 (b) 10+4i +6+2i = 0 (c) 10+4i −6+2i = 0 (d) −10+4i −6+2i = 0

3.7 In the circuit in Fig. 3.49, currenti₁ is:

(a) 4 A (b) 3 A (c) 2 A (d) 1 A

i₁ 2 A i₂

20 V +

!

2 " 1 "

3 " 4 " v+

!

3.8 The voltagev across the current source in the circuit of Fig. 3.49 is:

voltage source is:

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108 PART 1 DC Circuits

2 A

v₁

1 2

v₂ 12 V +

!

3 " 4 "

6 " 6 "

8 "

Figure 3.46

For Review Questions 3.1 and 3.2.

(a) v₂ − v₁

4 + v₂

8 = v₂ 6 (b) v₁ − v₂

4 + v₂

8 = v₂ 6 (c) v₁ − v₂

4 + 12 − v₂

8 = v₂ 6 (d) v₂ − v₁

4 + v₂ − 12

8 = v₂ 6

3.3 For the circuit in Fig. 3.47,v₁ and v₂ are related as:

(a) v₁ = 6i + 8 + v₂ (b) v₁ = 6i − 8 + v₂ (c) v₁ = −6i + 8 + v₂ (d) v₁ = −6i − 8 + v₂

12 V +

! 4 "

6 " 8 V v₂ v₁

i

! +

Figure 3.47

3.4 In the circuit in Fig. 3.47, the voltage v₂ is:

(a) −8 V (b) −1.6 V

(c) 1.6 V (d) 8 V

(a) −2.667 A (b) −0.667 A

(c) 0.667 A (d) 2.667 A

10 V +

! + 6 V

! 4 "

i

2 "

Figure 3.48

(a) −10 + 4i +6 + 2i = 0 (b) 10+ 4i +6 + 2i = 0 (c) 10+ 4i −6 + 2i = 0 (d) −10 + 4i −6 + 2i = 0

3.7 In the circuit in Fig. 3.49, current i₁ is:

(a) 4 A (b) 3 A (c) 2 A (d) 1 A

i₁ 2 A i₂

20 V +

!

2 " 1 "

3 " 4 "

v⁺

!

Figure 3.49

3.8 The voltage v across the current source in the circuit of Fig. 3.49 is:

(a) 20 V (b) 15 V (c) 10 V (d) 5 V 3.9 The PSpice part name for a current-controlled

voltage source is:

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2 A

v₁

1 2

v₂ 12 V +

!

3 " 4 "

6 " 6 "

8 "

(a) v₂−v₁ 4 + v₂

8 = v₂ 6 (b) v₁−v₂

4 + v₂ 8 = v₂

6 (c) v₁−v₂

4 + 12−v₂ 8 = v₂

6 (d) v₂−v₁

4 + v₂−12 8 = v₂

6

3.3 For the circuit in Fig. 3.47,v₁ andv₂ are related as:

(a) v₁ =6i +8+v₂ (b) v₁ = 6i −8+v₂ (c) v₁ =−6i +8+v₂ (d) v₁ = −6i−8+v₂

12 V +

! 4 "

6 " 8 V v₂

v₁

i

! +

3.4 In the circuit in Fig. 3.47, the voltagev₂ is:

(a) −8 V (b) −1.6 V

(c) 1.6 V (d) 8 V

3.5 The currenti in the circuit in Fig. 3.48 is:

(a) −2.667 A (b) −0.667 A

(c) 0.667 A (d) 2.667 A

10 V +

! + 6 V

! 4 "

i

2 "

(a) −10+4i +6+2i = 0 (b) 10+4i+6+2i = 0 (c) 10+4i−6+2i = 0 (d) −10+4i −6+2i = 0

3.7 In the circuit in Fig. 3.49, current i₁is:

(a) 4 A (b) 3 A (c) 2 A (d) 1 A

i₁ 2 A i₂

20 V +

!

2 " 1 "

3 " 4 "

v⁺

!

(a) 20 V (b) 15 V (c) 10 V (d) 5 V 3.9 The PSpicepart name for a current-controlled

voltage source is:

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108 PART 1 DC Circuits

2 A

v₁

1 2

v₂ 12 V +

!

3 " 4 "

6 " 6 "

8 "

Figure 3.46

(a) v₂ − v₁

4 + v₂

8 = v₂ 6 (b) v₁ − v₂

4 + v₂

8 = v₂ 6 (c) v₁ − v₂

4 + 12 − v₂

8 = v₂ 6 (d) v₂ − v₁

4 + v₂ − 12

8 = v₂ 6

3.3 For the circuit in Fig. 3.47, v₁ and v₂ are related as:

(a) v₁ = 6i + 8 + v₂ (b) v₁ = 6i − 8 + v₂ (c) v₁ = −6i + 8 + v₂ (d) v₁ = −6i − 8 +v₂

12 V +

! 4 "

6 " 8 V v₂ v₁

i

! +

Figure 3.47

3.4 In the circuit in Fig. 3.47, the voltage v₂ is:

(a) −8 V (b) −1.6 V

(c) 1.6 V (d) 8 V

(a) −2.667 A (b) −0.667 A

(c) 0.667 A (d) 2.667 A

10 V +

! + 6 V

! 4 "

i

2 "

Figure 3.48

(a) −10 + 4i + 6 + 2i = 0 (b) 10 + 4i + 6 + 2i = 0 (c) 10 + 4i − 6 + 2i = 0 (d) −10 + 4i − 6 + 2i = 0

3.7 In the circuit in Fig. 3.49, current i₁ is:

(a) 4 A (b) 3 A (c) 2 A (d) 1 A

i₁ 2 A i₂

20 V +

!

2 " 1 "

3 " 4 "

v⁺

!

Figure 3.49

(a) 20 V (b) 15 V (c) 10 V (d) 5 V 3.9 The PSpice part name for a current-controlled

voltage source is:

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P R O B L E M S

Sections 3.2 and 3.3 Nodal Analysis

3.1 Determinev₁,v₂, and the power dissipated in all the resistors in the circuit of Fig. 3.50.

10 A

6 A 8 "

4 "

v₁ v₂

2 "

Figure 3.50 For Prob. 3.1.

3.2 For the circuit in Fig. 3.51, obtainv₁ andv₂.

3 A 6 A

5 "

10 "

2 "

v₁ v₂

4 "

3.3 Find the currentsi₁ throughi₄ and the voltagevo in the circuit in Fig. 3.52.

10 A 10 " 20 " 30 " 2 A 60 "

i₁ i₂ i₃ i₄

v_o

3.4 Given the circuit in Fig. 3.53, calculate the currents i₁ throughi₄.

4 A 5 A

2 A

5 "

10 "

5 "

i₁ i₂ i₃ i₄

3.5 Obtainvo in the circuit of Fig. 3.54.

30 V +

! 2 k"

20 V +

! 5 k"

4 k" ⁺v_o

!

3.6 Use nodal analysis to obtainvo in the circuit in Fig.

3.55.

6 " 2 "

12 V

10 V

+!

+! 4 "

i₃ i₂

v_o i₁

3.7 Using nodal analysis,findvo in the circuit of Fig.

3.56.

3 V

4v_o +!

2 "

v_o⁺

! 1 "

3 " 5 "

+!

3.8 Calculatevo in the circuit in Fig. 3.57.

12 V ⁺_! 8 " 2v_o

6 "

+! 3 "

v_o

+ !

3.9 Findio in the circuit in Fig. 3.58.

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P R O B L E M S

Sections 3.2 and 3.3 Nodal Analysis

3.1 Determinev₁,v₂, and the power dissipated in all the resistors in the circuit of Fig. 3.50.

10 A

6 A 8 "

4 "

v₁ v₂

2 "

3.2 For the circuit in Fig. 3.51, obtain v₁ andv₂.

3 A 6 A

5 "

10 "

2 "

v₁ v₂

4 "

3.3 Find the currentsi₁ throughi₄ and the voltagevo in the circuit in Fig. 3.52.

10 A 10 " 20 " 30 " 2 A 60 "

i₁ i₂ i₃ i₄

v_o

3.4 Given the circuit in Fig. 3.53, calculate the currents i₁ throughi₄.

4 A 5 A

2 A

5 "

10 "

5 "

i₁ i₂ i₃ i₄

3.5 Obtainv_o in the circuit of Fig. 3.54.

30 V +

! 2 k"

20 V +

! 5 k"

4 k" +v_o

!

3.6 Use nodal analysis to obtainvo in the circuit in Fig.

3.55.

6 " 2 "

12 V

10 V

+!

+! 4 "

i₃ i₂

v_o i₁

3.7 Using nodal analysis,findvo in the circuit of Fig.

3.56.

3 V

4v_o +!

2 "

v_o+

! 1 "

3 " 5 "

+!

3.8 Calculatevo in the circuit in Fig. 3.57.

12 V + 2v_o

! 8 "

6 "

+! 3 "

v_o

+ !

3.9 Findio in the circuit in Fig. 3.58.

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112 PART 1 DC Circuits

∗3.25 Obtain the node voltages v₁, v₂, and v₃ in the circuit of Fig. 3.73.

10 k"

4 mA

5 k"

v₁

20 V 10 V v₂

v₃ + 12 V

!

! + + !

Figure 3.73

For Prob. 3.25.

Sections 3.4 and 3.5 Mesh Analysis

3.26 Which of the circuits in Fig. 3.74 is planar? For the planar circuit, redraw the circuits with no crossing branches.

2 "

6 "

5 "

2 A (a) 4 "

3 "

1 "

(b) 12 V +

! 2 "

3 "

5 "

4 "

1 "

Figure 3.74

For Prob. 3.26.

3.27 Determine which of the circuits in Fig. 3.75 is planar and redraw it with no crossing branches.

10 V +

!

3 "

5 "

2 "

7 "

4 "

(a) 1 "

6 "

7 "

6 "

1 " 3 "

4 A (b) 8 "

2 "

5 " 4 "

Figure 3.75

For Prob. 3.27.

3.28 Rework Prob. 3.5 using mesh analysis.

3.29 Rework Prob. 3.6 using mesh analysis.

3.30 Solve Prob. 3.7 using mesh analysis.

3.31 Solve Prob. 3.8 using mesh analysis.

3.32 For the bridge network in Fig. 3.76, find i_o using mesh analysis.

30 V +

!

2 k"

6 k" 6 k"

4 k"

i_o

Figure 3.76

For Prob. 3.32.

3.33 Apply mesh analysis to find i in Fig. 3.77.

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3.10 SUMMARY

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3.10 SUMMARY

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108 PART 1 DC Circuits

Figure 3.46

Figure 3.47

Figure 3.48

Figure 3.49

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108 PART 1 DC Circuits

Figure 3.46

Figure 3.47

Figure 3.48

Figure 3.49

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112 PART 1 DC Circuits

Figure 3.73

Sections 3.4 and 3.5 Mesh Analysis

Figure 3.74

Figure 3.75

Figure 3.76

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