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Problem 2.1 An AWG-14 copper wire has a resistance of 17.1 Ω at 20◦ C. How long is it? Solution: AWG-14 has a diameter of 1.6 mm (Table 2-2), and at 20◦ C, copper’s conductivity is σ = 5.81 × 107 (S/m) [Table 2-1]. ℓ σA ℓ = Rσ A

R=

= Rσ π (d/2)2 1.6 × 10−3 = 17.1 × 5.81 × 10 π × 2 7



2

≃ 2 km.

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Problem 2.2 A 3-km long AWG-6 metallic wire has a resistance of approximately 6 Ω at 20◦ C. What material is it made of? Solution: ℓ , AWG-6 has a diameter of 4.1 mm. σA ℓ σ= RA 3 × 103 ℓ = = 3.79 × 107 = Rπ (d/2)2 6π (4.1 × 10−3 /2)2 R=

(S/m),

which, according to Table 2-1, is approximately the value of the conductivity of aluminum.

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Problem 2.3 A thin-film resistor made of germanium is 2 mm in length and its rectangular cross section is 0.2 mm × 1 mm, as shown in Fig. P2.3. Determine the resistance that an ohmmeter would measure if connected across its: (a) Top and bottom surfaces (b) Front and back surfaces (c) Right and left surfaces z 2 mm y 0.2 mm

1 mm x

Figure P2.3: Film resistor of Problem 2.3.

Solution: (a) R= =

ℓ σA

ℓ = 0.22 mm,

A = 1 mm × 2 mm = 2 × 10−6 m2

2 × 10−4 ≃ 47 Ω. 2.13 × 2 × 10−6

(b) R= =

ℓ σA

ℓ = 1 mm,

A = 2 mm × 0.2 mm = 4 × 10−7 m2

10−3 ≃ 1, 174 Ω. 2.13 × 4 × 10−7

(c) R= =

ℓ σA

ℓ = 2 mm,

A = 1 mm × 0.2 mm = 2 × 10−7 m2

2 × 10−3 ≃ 4, 695 Ω. 2.13 × 4 × 10−7

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Problem 2.4 A resistor of length ℓ consists of a hollow cylinder of radius a surrounded by a layer of carbon that extends from r = a to r = b, as shown in Fig. P2.4. (a) Develop an expression for the resistance R. (b) Calculate R at 20◦ C for a = 2 cm, b = 3 cm, and ℓ = 10 cm. l

Carbon Hollow

2a 2b

Figure P2.4: Carbon resistor of Problem 2.4.

Solution: (a) R = σℓA . The area through which current can flow is the cross section consisting of carbon. Hence, A = π b2 − π a2 . Thus, R=

ℓ

σ π (b2 − a2 )

.

(b) R=

0.1 7.14 × 104 π (0.032 − 0.022 )

= 0.89

(mΩ).

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Problem 2.5 A standard model used to describe the variation of resistance with temperature T is given by R = R0 (1 + α T ), where R is the resistance at temperature T (measured in ◦ C), R0 is the resistance at T = 0◦ C, and α is a temperature coefficient. For copper, α = 4 × 10−3 ◦ C−1 . At what temperature is the resistance greater than R0 by 1%? Solution: R = 1.01R0 . Hence, R = 1.01 = 1 + α T R0

α T = 0.01 T=

0.01 = 2.5◦ C. 4 × 10−3

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Problem 2.6 A light bulb has a filament whose resistance is characterized by a temperature coefficient α = 6 × 10−3 ◦ C−1 (see resistance model given in Problem 2.5). The bulb is connected to a 100-V household voltage source via switch. After turning on the switch, the temperature of the filament increases rapidly from the initial room temperature of 20◦ C to an operating temperature of 1800◦ C. When it reaches its operating temperature, it consumes 80 W of power. (a) Determine the filament resistance at 1800◦ C. (b) Determine the filament resistance at room temperature. (c) Determine the current that the filament draws at room temperature and also at 1800◦ C. (d) If the filament deteriorates when the current through it approaches 10 A, is the damage done to the filament greater when it is first turned on or later on when it arrives at its operating temperature? Solution: (a) R = resistance at 1800◦ C. p=

V2 ; R

R=

V 2 (100)2 = = 125 Ω. p 80

(b) R = R0 (1 + α T ). R0 =

R 125 = = 10.6 Ω @ 0◦ C. 1 + αT 1 + 6 × 10−3 × 1800

At T = 20◦ C, R = R0 (1 + α T ) = 10.6(1 + 6 × 10−3 × 20) = 11.9 Ω @ 20◦ C. (c) At T = 20◦ C, I=

100 V = = 8.43 A. R(20◦ C) 11.9

At T = 1800◦ C,

p 80 = = 0.8 A. V 100 (d) The damage is greater at room temperature because the current is much closer to 10 A. I=

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Problem 2.7 A 110-V heating element in a stove can boil a standard-size pot of water in 1.2 minutes, consuming a total of 136 kJ of energy. Determine the resistance of the heating element and the current flowing through it. Solution: W = p ∆t = υ i ∆t W 136 × 103 = = 17.2 A. υ ∆t 110 × 1.2 × 60 υ 110 R= = = 6.41 Ω. i 17.2 i=

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Problem 2.8 A certain copper wire has a resistance R characterized by the model given in Problem 2.5 with α = 4 × 10−3 ◦ C−1 . If R = 60 Ω at 20◦ C and the wire is used in a circuit that cannot tolerate an increase in the magnitude of R by more than 10 percent over its value at 20◦ C, what would be the highest temperature at which the circuit can be operated within its tolerance limits? Solution: The model is given by R = R0 (1 + α T ). We are given that R = 60 Ω @ T = 20◦ C, and the maximum R that can be tolerated is 66 Ω. At 20◦ C, 60 = R0 (1 + 20α ). At unknown T , 66 = R0 (1 + α T ). The ratio gives 66 1 + αT . = 60 1 + 20α Solving for T , we have T=

6 + 132α 6 + 132 + 4 × 10−3 = = 27.2◦ C. 60α 60 × 4 × 10−3

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Problem 2.9 The circuit shown in Fig. P2.9 includes two identical potentiometers with per-length resistance of 20 Ω/cm. Determine Ia and Ib . Ia

4 cm

7.5 cm

80 mA

10 cm

10 cm

2.5 cm 6 cm

Ib Figure P2.9: Circuit of Problem 2.9.

Solution: Effectively, the circuit looks as shown in Fig. P2.9(a). R1

Ia

80 mA

R2

Ib

Fig. P2.9 (a)

Resistors R1 and R2 are: R1 = (4 + 7.5) × 20 = 230 Ω, R2 = (6 + 2.5) × 20 = 170 Ω. Hence, by current division  R2 Ia = (80 mA R1 + R2 170 = 34 mA, = 8 × 10−2 × 400 230 = 46 mA. Ib = 8 × 10−2 × 400 

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Problem 2.10 Determine VL in the circuit of Fig. P2.10.

5Ω

12 V

+ _

5Ω

+ 6Ω

10 Ω

4Ω

5Ω

6Ω

V

_L

5Ω

Figure P2.10: Circuit of Problem 2.10.

The parallel combination of the 4 Ω and 6 Ω resistors is R=

4×6 = 2.4 Ω. 4+6

By voltage division 

2.4 VL = 12 5 + 2.4 + 5



= 2.32 V.

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Problem 2.11 Select the value of R in the circuit of Fig. P2.11 so that VL = 9 V. 12 V _

I0

R

+

3I0 500 Ω 6 mA

_ VL + 500 Ω Figure P2.11: Circuit of Problem 2.11.

Solution: The voltage across the 500-Ω resistor in the right-hand segment is VL = (3I0 + 6 × 10−3 ) × 500. Setting VL = 9 V leads to 1 I0 = 3



VL − 6 × 10−3 500



1 = 3



9 − 6 × 10−3 500



= 4 mA.

The left-hand loop has to satisfy KVL: −12 + I0 R + 500I0 = 0, which leads to R=

12 − 500I0 12 = − 500 I0 I0 12 = − 500 = 2500 Ω. 4 × 10−3

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Problem 2.12 A high-voltage direct-current generating station delivers 10 MW of power at 250 kV to a city, as depicted in Fig. P2.12. The city is represented by resistance RL and each of the two wires of the transmission line between the generating station and the city is represented by resistance RTL . The distance between the two locations is 2000 km and the transmission lines are made of 10-cm–diameter copper wire. Determine (a) how much power is consumed by the transmission line and (b) what fraction of the power generated by the generating station is used by the city. RTL

V0

+ _

RL

(city)

RTL Station 2000 km Figure P2.12: Diagram for Problem 2.12.

Solution: For copper,

ρ = 1.72 × 10−8 Ω-m. For each wire of the transmission line, Eq. (2.2) leads to RTL = ρ

ℓ 1.72 × 10−8 × 2 × 106 = A π (5 × 10−2 )2 = 4.4 Ω.

Delivering 10 MW at 250 kV to the city means that PL =

VL2 = 107 W, RL

with VL = 2.5 × 105 V.

Hence, VL2 (2.5 × 105 )2 = = 6.25 × 103 Ω. PL 107 Also, the current flowing through RL is RL =

VL 2.5 × 105 = = 40 A. RL 6.25 × 103 (a) The power consumed by transmission lines is I=

PTL = 2I 2 RTL = 2 × (40)2 × 4.4 = 14080 W. Total power generated by the source is Ps = PTL + PL = 107 + 14080 = 1.001408 MW, and the fraction used by the city is Fraction =

107 PL = Ps 1.001408 × 107 = 0.9986 ≈ 99.86%.

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Determine the current I in the circuit of Fig. P2.13 given that I0 = 0.

+ V1 _

+

V5

_

V3

1Ω

+

I1

1Ω

V4 V2 2 Ω I0 = 0

_

+ + _

+

1Ω

_

24 V

I

I2

_

3Ω

a I1

+

Problem 2.13

I2 1Ω

b Figure P2.13: Circuit for Problem 2.13.

Solution: Since I0 = 0, the middle branch in the bridge section is of no consequence. The voltage between nodes a and b is the same following either path between them. Hence, I1 × 1 + I1 × 1 = I2 × 1 + I2 × 1, or I1 = I2 , which is obvious considering that all 4 resistors in the bridge section are the same. For the left loop that includes the 24-V source, −24 +V1 +V2 +V3 = 0 V1 = 3I V2 = I1 V3 = I1 and I = I1 + I2 = 2I1 . Hence, −24 + 3I + 2I1 = 0 −24 + 3I + I = 0 or I=

24 = 6 A. 4

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Problem 2.14 Determine currents I1 to I3 in the circuit of Fig. P2.14. 1A 2Ω

3A a I2

I1 + 18 V _

12 Ω

8Ω 4Ω

I3

7Ω b

Figure P2.14: Circuit for Problem 2.14.

Solution: For the loop containing the 18-V source, −18 + 3 × 2 + 8I1 = 0. Hence, I1 = 1.5 A. KCL at node a gives 3 − 1 − I1 − I2 = 0 I2 = 2 − I1 = 2 − 1.5 = 0.5 A. KCL at node b gives 1 + I2 − I3 = 0 I3 = 1 + I2 = 1 + 0.5 = 1.5 A.

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Problem 2.15 Determine Ix in the circuit of Fig. P2.15. I

12 V

+ _

Ix

5Ω

1A

2Ω

Figure P2.15: Circuit for Problem 2.15.

Solution: KVL gives: −12 + 5I + 2Ix = 0. KCL gives: I + 1 − Ix = 0. Solution of the two equations yields Ix =

17 7

= 2.43 A.

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Problem 2.16 Determine currents I1 to I4 in the circuit of Fig. P2.16. 4A

1Ω

8Ω

I1 12 V

+ + _

I3

I2

4Ω

+ + _

6Ω 1V +_ +_

5V

I4

Figure P2.16: Circuit for Problem 2.16.

Solution: Application of KVL to the outer-parameter loop gives −12 + 4 × 1 + 8I3 + 5 − 1 = 0, which gives I3 = 0.5 A. KVL for the left-most loop is −12 + 4 × 1 + 4I1 = 0, which leads to I1 = 2 A. KCL at the top center node gives 4 − I1 − I2 − I3 = 0 or I2 = 4 − I1 − I3 = 4 − 2 − 0.5 = 1.5 A. KCL at the bottom left node gives I4 + I1 − 4 = 0, or I4 = 4 − I1 = 4 − 2 = 2 A.

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Problem 2.17 Determine currents I1 to I4 in the circuit of Fig. P2.17. I1

I2

2Ω

4Ω

I3

I4

2Ω

4Ω

6A

Figure P2.17: Circuit for Problem 2.17.

Solution: The same voltage exists across all four resistors. Hence, 2I1 = 4I2 = 2I3 = 4I4 . Also, KCL mandates that I1 + I2 + I3 + I4 = 6 It follows that I1 = 2 A, I2 = 1 A, I3 = 2 A, and I4 = 1 A.

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Problem 2.18 Determine the amount of power dissipated in the 3-kΩ resistor in the circuit of Fig. P2.18. + V0 _

10 mA

2 kΩ

3 kΩ

10−3V0

Figure P2.42: Circuit for Problem 2.18.

Solution: In the left loop, V0 = 10 × 10−3 × 2 × 103 = 20 V. The dependent current source is I0 = 10−3V0 = 20 mA. The power dissipated in the 3-kΩ resistor is p = I02 R = (20 × 10−3 )2 × 3 × 103 = 1.2 W.

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Problem 2.19 Determine Ix and Iy in the circuit of Fig. P2.19. Ix

2Ω 10 V

+ _

6Ω

Iy

I 4Ω

_ +

4Ix

Figure P2.19: Circuit for Problem 2.19.

Solution: Application of KVL to the two loops gives −10 + 2Ix + 4I = 0 −4I + 6Iy − 4Ix = 0. Additionally, I = Ix − Iy . Solution of the three equations yields Ix = 3.57 A,

Iy = 2.86 A.

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Problem 2.20 Find Vab in the circuit in Fig. P2.20. 2Ω 2Ω + _

6V

a

I

2Ω

+ Vab _

12 V

+ _

b Figure P2.20: Circuit for Problem 2.20.

Solution: For the lower loop, KVL gives −6 + 4I + 12 = 0, or I = −1.5 A. Moving from a to b via the 12-V supply, Vab = (−1.5) × 2 + 12 = 9 V.

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Problem 2.21 Find I1 to I3 in the circuit of Fig. P2.21. 3 kΩ I1

I3

8V _ +

+ _ 16 V

I2 2 kΩ 4 kΩ

+ _ 12 V

Figure P2.21: Circuit for Problem 2.21.

Solution: Application of KVL to the outer-perimeter loop gives −16 + 3 × 103 I1 − 8 + 12 = 0, which leads to I1 = 4 mA. For the left loop, − 16 + 3 × 103 I1 + 4 × 103 I2 = 0, I2 =

16 − 3 × 103 I1 16 − 3 × 103 × 4 × 10−3 = = 1 mA. 4 × 103 4 × 103

Consequently, I3 = I1 − I2 = 4 − 1 = 3 mA.

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Problem 2.22 Find I in the circuit of Fig. P2.22. I 10 V

2I +_

+ _

3Ω

Figure P2.22: Circuit for Problem 2.22.

Solution: −10 + 2I + 3I = 0. Hence, I=

10 = 2 A. 5

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Problem 2.23 Determine the amount of power supplied by the independent current source in the circuit of Fig. P2.23. + V1 _

2Ω

I

2Ω

0.2 A

V1 4

Figure P2.23: Circuit for Problem 2.23.

Solution: KCL at top node gives 0.2 +

V1 −I = 0 4

Also I = V1 /2 (for the 2-Ω resistor). Hence, 0.2 +

V1 V1 − =0 4 2 V1 = 0.8 V.

The voltage across the 0.2-A source is 2V1 = 1.6 V. Power dissipated is P = V I = 1.6 × 0.2 = 0.32 W.

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Problem 2.24 Given that in the circuit of Fig. P2.24, I1 = 4 A, I2 = 1 A, and I3 = 1 A, determine node voltages V1 , V2 , and V3 . I2 I1

1 Ω V1

R1 = 18 Ω 6Ω

+ 40 V _

V2

6Ω V3

6Ω

18 Ω I3

Figure P2.24: Circuit of Problem 2.24.

Solution: R1 = 18 Ω 1 Ω V1

I4

6 Ω V2

+ 40 V _

6Ω

6Ω

V3 18 Ω

Fig. P2.24 (a)

V1 = 40 − (1 Ω) × I1 = 40 − 4 = 36 V. KCL at node V1 leads to I4 = I1 − I2 = 4 − 1 = 3 A. Hence, V2 = V1 + 6I4 = 36 − 6 × 3 = 18 V. The voltage across R1 is 18I2 . Hence, V3 = V1 − 18I2 = 36 − 18 × 1 = 18 V.

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Problem 2.25 After assigning node V4 in the circuit of Fig. P2.25 as the ground node, determine node voltages V1 , V2 , and V3 . 12 V _

3A

+

3Ω

V2

V1 6Ω

3Ω V3 6Ω

6Ω V4

1A

1A

Figure P2.25: Circuit of Problem 2.25.

Solution: 12 V _

3A

+

3Ω V1 6Ω

I1

V2

3Ω V3 6Ω

6Ω 1A

V4

1A

Fig. P2.25 (a)

From KCL at node V1 , the sum of currents leaving the node is 3 + I1 − 1 = 0, or I1 = −3 + 1 = −2 A. Node voltages (relative to V4 ): V1 = −6 × 1 = −6 V, V2 = V1 − 3I1 = −6 − 3(−2) = 0, V3 = 6 × 1 = 6 V.

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Problem 2.26 After assigning node V1 in the circuit of Fig. P2.25 as the ground node, determine node voltages V2 , V3 , and V4 . Solution: 12 V _

3A

+

V1

I1

3Ω

3Ω

V2

6Ω

V3 6Ω

6Ω V4

1A

1A

Fig. P2.26 (a)

From KCL at node V1 , the sum of currents leaving the node is 3 + I1 − 1 = 0, or I1 = −3 + 1 = −2 A. Hence, relative to node V1 : V2 = −3I1 = −3(−2) = 6 V. V3 = 12 V (because (−) terminal of voltage source is at V1 ) V4 = 6(1) = 6 V.

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Problem 2.27 In the circuit of Fig. P2.27, I1 = 42/81 A, I2 = 42/81 A, and I3 = 24/81 A. Determine node voltages V2 , V3 , and V4 after assigning node V1 as the ground node. I3

9Ω V2

6Ω

6Ω

V3

V4 I2

6Ω V1

6V _

6V _

9Ω I1

+

9Ω

+

Figure P2.27: Circuit of Problem 2.27.

Solution: At node V3 , KCL gives I2 + I4 − I3 = 0, or

24 42 18 − = − A. 81 81 81

I4 = I3 − I2 =

I3

9Ω I4

V2

6Ω

6Ω

V3

V4 I2

6Ω V1

6V _

6V _

9Ω I1

+

9Ω

+

V1 − 6 Fig. P2.27 (a)

V3 = V1 − 6 − 9I2 = 0−6+9×

42 −4 = 1.33 = V. 81 3

V2 = V3 − 6I4   18 4 = 0, = − −6 − 3 81 V4 = V1 + 6 − 9I1   4 42 = V. = 0+6−9 81 3

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Problem 2.28 The independent source in Fig. P2.28 supplies 48 W of power. Determine I2 . I1

+ 12 V _

I3

R

R

0.25I1

I2

R

R

Figure P2.28: Circuit of Problem 2.28.

Solution: From P = V0 I1 , I1 =

P 48 = = 4 A. V0 12

Current of dependent current source is the same as I3 . Hence, I3 = 0.25I1 = 0.25 × 4 = 1 A. By KCL, I2 = I1 − I3 = 4 − 1 = 3 A.

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Problem 2.29 Given that I1 = 1 A in the circuit of Fig. P2.29, determine I0 . I5 I0

1Ω

I4

I3

2Ω

4Ω

I2

I1 = 1 A

8Ω

16 Ω

Figure P2.29: Circuit for Problem 2.29.

Solution: Since the 16-Ω and 8-Ω resistors are connected in parallel, they have the same voltage across them, namely V = 16 × I1 = 16 × 1 = 16 V. By KCL, I0 equals the sum of the currents flowing in all five resistors: 16 16 16 16 16 + + + + 1 2 4 8 16 = 31 A.

I0 =

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Problem 2.30 What should R be in the circuit of Fig. P2.30 so that Req = 4 Ω? 1Ω

a Req

6Ω

2Ω

R

5Ω

b

Figure P2.30: Circuit for Problem 2.30.

Solution: The parallel combination of R and 2-Ω resistor is R1 =

2R . 2+R

R1 is in series with 5-Ω resistor. Hence R2 = R1 + 5 =

2R + 5. 2+R

R2 is in parallel with 6-Ω resistor: 

 2R 6× +5 2+R R3 = , 2R 6+ +5 2+R and

 2R +5 6× 2+R = 4. Req = 1 + R3 = 1 + 2R 11 + 2+R Solving for R leads to R = 2 Ω. 

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Problem 2.31 Find I0 in the circuit of Fig. P2.31. I0

4Ω

18 A

6Ω

3Ω

12 Ω Figure P2.31: Circuit for Problem 2.31.

Solution: Combining the 3-Ω and 6-Ω resistors in parallel gives R=

3 × 6 18 = = 2 Ω. 3+6 9

The new circuit becomes I0 18 A

6Ω 12 Ω

Current division leads to I0 =



Req 12



18 =

6 × 12 × 18 = 6 A. 6 + 12

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Problem 2.32 For the circuit in Fig. P2.32, find Ix for t < 0 and t > 0. Ix + _

t=0 1

2Ω

15 V 2 Ω

2

4Ω 3Ω

4Ω

4Ω

Figure P2.32: Circuit with SPDT switch for Problem 2.32.

Solution: For t < 0: Ix + _

15 V 2 Ω

Ix + _

+ _

3Ω

4Ω

4Ω

2Ω 3Ω

4Ω

2Ω

15 V 2 Ω

Ix

4Ω

2Ω

15 V 2 Ω

Ix + _

2Ω

2Ω

6 3 6 || 3 = 6 + 3 = 2 Ω

2Ω

15 V 1 Ω

Ix =

15 =5A 2+1

For t > 0: Ix 15 V

2Ω

+ _

Ix =

15 = 7.5 A. 2

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Problem 2.33 Determine Req at terminals (a, b) in the circuit of Fig. P2.33. a

4Ω

Req

32 Ω 16 Ω

8Ω 8Ω

b

Figure P2.33: Circuit for Problem 2.33.

Solution: a 32 Ω Req

4Ω

8Ω 16 Ω

8Ω

b

Terminals (a, b) are connected together through a short circuit. Hence, Req = 0.

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Problem 2.34 Select R in the circuit of Fig. P2.34 so that VL = 5 V. Solution: Multiple application of the source-transformation method leads to the final circuit below. 1 kΩ

R

+ 5 kΩ

5 mA

2 kΩ

_

Figure P2.34: Circuit for Problem 2.34.

1 kΩ

R

+

5 kΩ 25 V

VL

+ _

2 kΩ

VL

_

1 kΩ R1 = R + 5k 25 Is = R1

+ Is

R1

2 kΩ

VL

_

R2

Vs

1 kΩ 123 R3

+ _

_

R2 = R1 k 2 kΩ = Vs = Is R2 =

+ VL

2R1 × 103 2 × 103 (R + 5 × 103 ) = R1 + 2 × 103 R + 7 × 103

25R2 50 × 103 = R1 R + 7 × 103

Since no current flows through R3 , VL = Vs =

50 × 103 . R + 7 × 103

Setting VL = 5 V leads to R = 3 kΩ.

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Problem 2.35 If R = 12 Ω in the circuit of Fig. P2.35, find I. Solution:

R R

R 20 V _

I

4Ω

+

4Ω

R/2

R

R

R

R/2 20 V _

I

+

R

R/2

R/2

R

Figure P2.35: Circuit for Problem 2.35.

20 V _

I

4Ω

20 V _

I

+

+

4Ω R R

R 12 = =6Ω 2 2 20 I= = 2 A. 4+6

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Problem 2.36 Use resistance reduction and source transformation to find Vx in the circuit of Fig. P2.36. All resistance values are in ohms. Solution: + Vx _

4 Figure P2.36: Circuit for Problem 2.36.

16

16

12

10 A

4

16

16

+ Vx _

4 8

6

10 A

12

6

4

8

+ Vx _ 10 A

6

6

4

8

+ Vx _ 10 A

3

4

8

+ Vx _ 30 V

Vx =

+ _

3

4

8

30 × 4 = 8 V. 3+4+8

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Problem 2.37 Determine A if Vout /Vs = 9 in the circuit of Fig. P2.37. Solution: 3Ω + _ Vs

I1 12 Ω

3Ω + _ Vs

12 Ω

3Ω

AI1

+ 6 Ω Vout _

AI1

2 Ω Vout

Figure P2.37: Circuit for Problem 2.37.

I

+ 6Ω

_

Vs 9 I Vs I1 = = 2 18 I=

Vout = AI1 × 2 =

AVs AVs ×2 = 18 9

Vout A = = 9. Vs 9 Hence A = 81.

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Problem 2.38 For the circuit in Fig. P2.38, find Req at terminals (a, b). Solution: a 32 Ω Req

4Ω

8Ω 16 Ω

8Ω

b

Req = 4 + 5 = 9 Ω.

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Problem 2.39 Find Req at terminals (c, d) in the circuit of Fig. P2.38. Solution: a

5Ω

3Ω 6Ω

b

5Ω 6Ω

3Ω

5Ω

5Ω 12 Ω

6Ω

5Ω

5Ω 4Ω

5Ω

c d

c Req d

c Req d

Req = 4 + 5 + 5 = 14 Ω.

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Problem 2.40 Simplify the circuit to the right of terminals (a, b) in Fig. P2.40 to find Req , and then determine the amount of power supplied by the voltage source. All resistances are in ohms. Solution:

25 V

+ _

a Req

3 4

5 8

6

8 12

6

Figure P2.40: Circuit for Problem 2.40.

12

b

25 V

+ _

a Req

3 4

5 8

6

8 12

4

b

25 V

+ _

a Req

3 4

5 8

6

6

12 || 12 = 6

8

5 + 6 || 6 = 5 + 3 = 8

b

25 V

+ _

a Req

3 4

8

b

25 V

+ _

a Req

3 2

8 || 8 || 4 = 2

b

Req = 3 + 2 = 5 Ω P=

V2 (25)2 = = 125 W. Req 5

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Problem 2.41 For the circuit in Fig. P2.41, determine Req at (a) Terminals (a, b) (b) Terminals (a, c) (c) Terminals (a, d) (d) Terminals (a, f ) e 2Ω

2Ω

d

2Ω

2Ω

c

f 2Ω

2Ω

2Ω

2Ω

2Ω

2Ω

a

b

Figure P2.41: Circuit for Problem 2.41.

Solution: All resistances are in ohms. (a) 2 2

6 2 6 + 2 = 1.5

2

2

2 a

2

2

b

a

2 b

Req

Req

Req = 1.5 + 2 + 2 = 5.5 Ω.

(b) 2 2

2

2

2

2

c

8

c 2 a

Req

2 a

Req

Req = 2 + 2 = 4 Ω.

(c) 2

2

2 d

d 2

6 2 6 + 2 = 1.5

2 2

Req

Req 2 a

2 a

Req = 2 + 2 + 1.5 = 5.5 Ω.

(d) f 2 2

Req

f 2 2

2

2 4 4 4+4 =2

Req

2 a

2 a

Req = 2 + 2 + 2 = 6 Ω.

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Problem 2.42 Find Req for the circuit in Fig. P2.42. All resistances are in ohms. Solution: 5 Req

10 10

10

10

Figure P2.42: Circuit for Problem 2.42.

10

5

5 Req

5

10

20

20

5 Req

5

5

Req = 15 Ω.

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Problem 2.43 Apply voltage and current division to determine V0 in the circuit of Fig. P2.43 given that Vout = 0.2 V. Solution: I5 + V5 8 I _ 3

V0

+ _

I4

+ V3 4 I _ 1

+ V4 4 _

I2 + V2 2 _

+ V1 _

Figure P2.43: Circuit for Problem 2.43.

2 1

+ Vout = 0.2 V _

0.2 = 0.2 A 1 V2 I1 I2 = = (2 + 1) = 0.3 A 2 2 I3 = I1 + I2 = 0.5 A I1 =

V4 V3 +V2 4I3 + 2I2 = = A 4 4 4 = 0.65 I5 = I3 + I4 = 1.15 A

I4 =

V0 = V4 +V5 = 4I4 + 8I5 = 11.8 V.

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Problem 2.44 Apply source transformations and resistance reductions to simplify the circuit to the left of nodes (a, b) in Fig. P2.44 into a single voltage source and a resistor. Then, determine I. 3A

10 Ω

a I

5A

2Ω

12 Ω

4Ω

b Figure P2.44: Circuit of Problem 2.44.

Solution:

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3A

10 Ω 5A

a 12 Ω

2Ω

b 2Ω

30 V 10 Ω _

a

+

10 V

+ _

12 Ω b 12 Ω

40 V

a

+ _

12 Ω b a

40 12

12 Ω

12 Ω b a

40 12

6Ω b 6Ω

20 V

a I

+ _

4Ω b Fig. P2.44 (a)

I=

20 = 2 A. 6+4

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Problem 2.45 Determine the open-circuit voltage Voc across terminals (a, b) in Fig. P2.45. Solution: 5Ω

6Ω

a

+ 30 V

+ _

3Ω

2A

Voc

_ b 6Ω

a

+ 6A

5Ω

3Ω

2A

Voc

_ b 6Ω

a

+ 8A

R=

3 5 15 = Ω 3+5 8

Voc

_ b 15 Ω 8

6Ω

a

+ 15 V

+ _

Voc

_ b Fig. P2.45 (a)

Hence, Voc = 15 V.

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Problem 2.46 Use circuit transformations to determine I in the circuit of Fig. P2.46. Solution: 3A 6Ω

4Ω

3Ω

+_

I 30 V 2A

4Ω

2Ω

6Ω 10 A

12 V _

4Ω

+

4Ω 8V

3Ω I

+ _

2Ω

6

3 9

=2Ω

8Ω

20 V

+ _

10 A 2 Ω

20 V + _

8Ω

20 V

I

2Ω I

+ _

2Ω

Fig. P2.46 (a)

I=

20 − 20 = 0. 8+2+2

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Problem 2.47 Determine currents I1 to I4 in the circuit of Fig. P2.47. 12 Ω

I1

6Ω

I2

+ _

I3

3Ω

I4

6Ω

12 V

Figure P2.47: Circuit of Problems 2.47 and 2.48.

Solution: 12 12 12 I2 = 6 12 I3 = 3 12 I4 = 6 I1 =

= 1 A, = 2 A, = 4 A, = 2 A.

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Problem 2.48 Replace the 12-V source in the circuit of Fig. P2.47 with a 4-A current source pointing upwards. Then, determine currents I1 to I4 . Solution: 12 Ω

I1

I3

3Ω

6Ω

I2

I4

6Ω

4A

Req = I1 = I2 = I3 = I4 =



1 1 1 1 + + + 12 6 3 6 Req 4 × 4 = A, 12 9 Req 8 × 4 = A, 6 9 Req 16 ×4 = A, 3 9 Req 8 × 4 = A. 6 9

−1

=

12 Ω. 9

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Problem 2.49 Determine current I in the circuit of Fig. P2.49. Solution: Resistance combining leads to 10 Ω

40 Ω

25 Ω I 5Ω

30 Ω

60 Ω + _

50 V

40 Ω

10 Ω

10 Ω

25 Ω I 5Ω

40 60 = 24 Ω 40 + 60

5Ω + _

50 V

I 5Ω 64 Ω

+ _

30 Ω 50 V

I 5Ω + _

30 64 = 20.43 Ω 94

50 V

Fig. P2.49 (a)

I=

50 = 1.97 A. 5 + 20.43

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Problem 2.50 Determine the equivalent resistance Req at terminals (a, b) in the circuit of Fig. P2.50. Solution:

5Ω 5Ω

4Ω

a

4Ω

Req 4Ω b 6Ω

5Ω 5Ω

4Ω

a

10 Ω 4Ω b a R = 4 + 4 + (5 || 5 || 10) = 10 Ω b Fig. P2.50 (a)

R = 4 + 4 + (5 k 5 k 10) = 10 Ω.

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*2.51 Determine current I in the circuit of Fig. P2.51. Solution: + _

16 V 2 kΩ

6 mA 2 kΩ

2 kΩ

+ _

5 mA

2 mA

0.5 kΩ

2 kΩ

1 kΩ I

_ +

8V

+ _

16 V

16 V

+ _

4V

_ +

8V

5 mA

I

2 kΩ 8 mA 2 kΩ

5 mA

2 kΩ

2 kΩ

0.5 kΩ

I1

24 mA

5 mA

I1 1 kΩ I _ +

8V

+ _

16 V

2 kΩ

0.5 kΩ

I1 2 kΩ

5 mA

8 mA (2k || 2k || 1k) = 0.5 kΩ

I _ +

8V

Fig. P2.51 (a)

I1 = (19 mA)



0.5k 2.5k



= 3.8 mA.

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19 mA

Problem 2.52 Determine voltage Va in the circuit of Fig. P2.52. Solution: 4Ω

2Ω 2A

_

V

a

+

2A

5A

2.5 A 4Ω

2Ω

2Ω

4Ω

+

2A

V

a

_

_

+

+

_ 4Ω 10 V

4V

+

_

2Ω 10 V

4Ω

V

a

+ _

2A

8Ω _

+

16 V

4Ω

+ a

2A

V

_

2A

8Ω

I 4Ω

V

_

a

+

4A

8Ω

Fig. P2.52 (a)

By current division, I=

4×8 = 2.67 A 4+8

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and V = 4I = 10.67 V.

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Problem 2.53 Convert the circuit in Fig. P2.53(a) from a ∆ to a Y configuration. Solution: a

c

6Ω 3Ω

a

R2

c

R3

1Ω

b

R1

d

b

d

Figure P2.53(a)

6×3 = 1.8 Ω 6+3+1 6×1 = 0.6 Ω R2 = 10 3×1 = 0.3 Ω. R3 = 10 R1 =

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Problem 2.54 Convert the circuit in Fig. P2.53(b) from a T to a Π configuration. Solution: 2Ω

Rb

8Ω Ra

4Ω

Rc

Figure P2.53(b)

2 × 8 + 2 × 4 + 4 × 8 56 = =7Ω 8 8 2 × 8 + 2 × 4 + 4 × 8 56 Rb = = = 14 Ω 4 4 2 × 8 + 2 × 4 + 4 × 8 56 = = 28 Ω. Rc = 2 2 Ra =

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Problem 2.55 Find the power supplied by the generator in Fig. P2.55. Solution: R1 = 18 Ω 1Ω

6Ω

6Ω

+ 20 V _

6Ω

Y

18 Ω

∆

18 Ω 1Ω

18 Ω

+ 20 V _

18 Ω

1Ω + 20 V _

18 Ω

18 Ω

9Ω 18 Ω

9Ω

1Ω I 20 V

+ _

9Ω

20 =2A 10 P = V I = 20 × 2 = 40 W. I=

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Problem 2.56 Repeat Problem 2.55 after replacing R1 with a short circuit. Solution: 1Ω 20 V

6Ω

+ _

6Ω 6Ω

Y

1Ω + 20 V _

18 Ω

∆

18 Ω 18 Ω

18 Ω

18 Ω

9Ω

18 Ω

1Ω + 20 V _

1Ω I 20 V

+ _

9 18 9 + 18 = 6 Ω

20 20 = 6+1 7 20 P = V I = 20 × = 57.1 W. 7 I=

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Problem 2.57 Find I in the circuit of Fig. P2.57. Solution:

9Ω

3V _

6Ω

9Ω 6Ω

9Ω

3V _

I

+

6Ω

+

T

Π Transformation

9Ω 18 Ω

9Ω 3V

9Ω

18 Ω

_ +

18 Ω I

+ _

6Ω 1/3 A

9Ω

18 Ω

3V

9Ω 18 Ω I

+ _

3V 9 || 8 = 6

6Ω 2V

6Ω

_ +

9Ω 18 Ω I

+ _

3V

9Ω 1/6 A

12 Ω

18 Ω I

+ _

3V 36 18 || 12 = 5

36/5 Ω 6/5 V

9Ω

_ + I

+ _

3V

I=

3 + 65 9 + 36 5

≃ 0.26 A.

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Problem 2.58 Find the power supplied by the voltage source in Fig. P2.58. Solution: 4V _ +

3Ω

3Ω

R=6Ω

_

6Ω

4V +

6Ω

Ra = 7.5 Ω Rb = 15 Ω

6Ω

Rc = 15 Ω

6Ω

3×3+3×6+3×6 = 7.5 Ω 6 3×3+3×6+3×6 = 15 Ω Rb = 3 Rc = Rb = 15 Ω. Ra =

7.5 Ω

_

4V

4.3 Ω

I

+

4.3 Ω

4V +

_

4.006 Ω

(4.3 + 4.3) × 7.5 = 4.006 Ω 4.3 + 4.3 + 7.5 4 I= = 0.998 A 4.006 P = I 2 R = (0.998)2 × 4.006 = 3.99 ≃ 4 W.

R=

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Problem 2.59 Repeat Problem 2.58 after replacing R with a short circuit. Solution: 4V _ +

3Ω

3Ω 6Ω

6Ω

4V _ +

6Ω

3Ω

3Ω

4V _

6Ω

I

+

6 3 6+3 =2Ω

2Ω

4 =1A 2+2 P = V I = 4 × 1 = 4 W. I=

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Problem 2.60 Find I in the circuit of Fig. P2.60. All resistances are in ohms. Solution: 1 12 V

2

I

+ _

2 ∆ 1 12 V

R1 = 0.5

R2 = 1 2

1 12 V

2

Y I

+ _

2 4

I

R3 = 1 2

2×2 = 0.5 Ω 2+2+4 2×4 R2 = =1Ω 2+2+4 R3 = R2 = 1 Ω.

R1 =

0.5

+ _ 3

1

3

I 0.5

12 V

+ _ 1.5

I=

12 = 4 A. 1 + 0.5 + 1.5

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Problem 2.61 Find Req for the circuit in Fig. P2.61. Solution: 18 Ω 6Ω 18 Ω 6Ω

6Ω

1Ω

9Ω

18 Ω

Req

18 Ω

18 Ω

18 Ω

1Ω

18 Ω 18 Ω

18 Ω

Req

9Ω

9Ω 1Ω 9Ω

18 Ω

18 Ω

Req

9Ω 9Ω

1Ω 18 Ω

Req

9Ω 1Ω 9Ω

Req

Req = 9 + 1 = 10 Ω.

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Problem 2.62 Find Req at terminals (a, b) in Fig. P2.62 if (a) Terminal c is connected to terminal d by a short circuit (b) Terminal e is connected to terminal f by a short circuit (c) Terminal c is connected to terminal e by a short circuit All resistance values are in ohms. d 3 3 e

3 Req

3 f

a b 3

3 c

Figure P2.62: Circuit for Problem 2.62.

Solution: (a) d 3

3

a b

3

3

3 3

c

d 9

9 4.5

a b

9 9

9

a b 4.5

Req = 9 Ω

9 c

(b) 3

d

3

3

a b

3

e 3

3 f

3

a b

3

Req = 6 Ω

c

(c)

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3

d

3

3

a b

3

e 3

3 c

3

d

3

3

a b

3 3

6 3 6+3 =2 3

a b

3

Req = 3 + 3 + 2 = 8 Ω

c

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Problem 2.63 For the Wheatstone bridge circuit of Fig. 2-30, solve the following problems: (a) If R1 = 1 Ω, R2 = 2 Ω, and Rx = 3 Ω, to what value should R3 be adjusted so as to achieve a balanced condition? (b) If V0 = 6 V, Ra = 0.1 Ω, and Rx were then to deviate by a small amount to Rx = 3.01 Ω, what would be the reading on the ammeter? Solution: (a) From Eq. (2.45), Rx R3 = , R1 R2

R1 Rx 1 × 3 = = 1.5 Ω. R2 2

R3 =

(b) V0 I1

I2

R1

V0

+ −

R2 Ra

Ia

V1

V2 I3

I4

R3 Rx

KCL equations at nodes V1 and V2 are I1 = I3 + Ia I4 = I2 + Ia KVL for the left loop and outside perimeter loop are −V0 + I1 R1 + I3 R3 = 0, −V0 + I2 R2 + I4 Rx = 0. Also, for the upper triangle of the bridge, I1 R1 + Ia Ra − I2 R2 = 0. Simultaneous solution of the five equations leads to Ia =

(R3 R2 − Rx R1 )V0 . R2 Rx (R1 + R3 ) + (R2 + Rx )[R1 R3 + Ra (R1 + R3 )]

For R1 = 1 Ω, R2 = 2 Ω, R3 = 1.5 Ω, Rx = 3.01 Ω, and V0 = 6 V, Ia = −2.5 mA.

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Problem 2.64 If V0 = 10 V in the Wheatstone-bridge circuit of Fig. 2-31 and the minimum voltage Vout that a voltmeter can read is 1 mV, what is the smallest resistance fraction (∆R/R) that can be measured by the circuit? Solution: V0 Vout = 4



∆R R



∆R 4Vout 4 × 10−3 = = = 4 × 10−4 , R V0 10 or 4 parts in 10,000.

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Problem 2.65 Suppose the cantilever system shown in Fig. 2-38 is used in the Wheatstone-bridge sensor of Fig. 2-31 with V0 = 2 V, α = −1 × 10−9 m2 /N, L = 0.5 cm, W = 0.2 cm, and H = 0.2 mm. If the measured voltage is Vout = −2 V, what is the force applied to the cantilever? Solution: From Example 2-16, Vout =

V0 FL α . 4 W H2

Solving for F, we have  Vout W H 2 V0 αL   −2 0.2 × 10−2 × (0.2 × 10−3 )2 =4 2 (−1 × 10−9 ) × 0.5 × 10−2

F =4



= 64 N.

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Problem 2.66 A touch sensor based on a piezoresistor built into a micromechanical cantilever made of silicon is connected in a Wheatstone-bridge configuration with a V0 = 1 V. If L = 1.44 cm and W = 1 cm, what should the thickness H be so that the touch sensor registers a voltage magnitude of 10 mV when the touch pressure is 10 N? Solution: From Example 2-16: Vout =

V0 FL . α 4 W H2

Solving for H, we have H=



V0 4Vout



FL α W

1/2

1 (−1 × 10−9 ) × 10 × 1.44 × 10−2 = × 4 × (−10 × 10−3 ) 10−2 

1/2

= 0.6 mm.

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Problem 2.67 Determine I1 and I2 in the circuit of Fig. P2.67. Assume VF = 0.7 V for both diodes. I1

53 Ω

_ 6V

0.7 V

+

I2

53 Ω

+ _

+ _0.7 V

Figure P2.67: Circuit for Problem 2.67.

Solution: The diode in the left-hand loop is reverse biased, so I1 = 0. In the right-hand loop, the diode is forward biased. Hence, I2 =

6 − 0.7 = 0.1 A. 53

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Problem 2.68 diodes.

Determine V1 in the circuit of Fig. P2.68. Assume VF = 0.7 V for all 50 Ω

9V

+ _

100 Ω

+ _

V1

25 Ω Figure P2.68: Circuit for Problem 2.68.

Solution: 9 − 3(0.7) ≃ 0.04 A, 50 + 100 + 25 V1 = 100I = 4 V. I=

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Problem 2.69 If the voltage source in the circuit of Fig. P2.69 generates a single square wave with an amplitude of 2 V, generate a plot for υout for the same time period.

υs(t)

+ _

100 Ω

+ υout

_

υs(t) 2V T

t

-2 V Figure P2.69: Circuit and voltage waveform for Problem 2.69.

Solution: Current will flow through the loop only when υs (t) is greater than 0.7 V. Hence, ( 2 − 0.7 = 1.3 V for 0 ≤ t ≤ T /2 υout = 0 for T /2 ≤ t ≤ T. vout 1.3 V t T/2

T

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Problem 2.70 If the voltage source in the circuit of Fig. P2.70(a) generates the single square waveform shown in Fig. P2.70(b), generate plots for i1 (t) and i2 (t). Solution: i1

i2

73 Ω

146 Ω

υs(t)

+ _

(a)

υs(t) 8V

2

t (s)

4

−8 V (b) Square wave

0.05 A

8 − 0.7 146 = 0.05 A i2 t (s) i1

−0.1 A

−8 + 0.7 73 = −0.1 A

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Problem 2.71 If the voltage source in the circuit of Fig. P2.70(a) generates the single triangular waveform shown in Fig. P2.70(c), generate plots for i1 (t) and i2 (t). Solution: i1

i2

73 Ω

146 Ω

υs(t)

+ _

(a)

υs(t) 8V

4

2

t (s)

−8 V (c) Triangular wave

0.05 A

i2 t (s) 1

2

3

4 i1

−0.1 A

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Problem 2.72 Use the DC Operating Point Analysis in Multisim to solve for voltage Vout in the circuit of Fig. P2.72. Solve for Vout by hand and compare with the value generated by Multisim. See the solution for Exercise 2-14 (on ) for how to incorporate circuit variables into algebraic expressions. C

M

O DR

+ + 2.5 V _

10 Ω

10 Ω

Vout

_ 15 Ω

25 Ω

Circuit for Problem 2.72.

Solution: A. By-Hand Solution By voltage division: Vout =



 10 × 2.5 = 0.714 V. 10 + 25

+ + 2.5 V _

10 Ω

Vout

_

25 Ω 25 Ω

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B. By Multisim

Circuit in MultiSIM Schematic Capture

DC Operating Point Solution. The value in the last row, V(1)-V(3), is the specific solution to the problem.

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Problem 2.73 Find the ratio Vout /Vin for the circuit in Fig. P2.73 using DC Operating Point Analysis in Multisim. See the Multisim Tutorial included on the CD on how to reference currents in ABM sources (you should not just type in I(V1)). 1 kΩ Iin Vin

+ _

+

10 kΩ

100Iin

Vout

1 kΩ

_ Figure P2.73: Circuit for Problem 2.73.

Solution: Vout = (100Iin ) × 1000 = 105 Hence,

Vin = 10Vin . 104

Vout = 10. Vin

Circuit in MultiSIM Schematic Capture

DC Operating Point solution

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Problem 2.74 Use DC Operating Point Analysis in Multisim to solve for all six labeled resistor currents in the circuit of Fig. P2.74. I2

I1 1Ω

1Ω +_ I3

1A

I4

2V

1Ω

1Ω +_ I5

1Ω

I6

3V 1Ω

Figure P2.74: Circuit for Problem 2.74.

Solution:

Circuit in MultiSIM Schematic Capture

DC Operating Point solution

V5 −V3 = 1.5 − 2 = −0.5 A, 1 V5 −V4 = 1.5 − 0 = 1.5 A, I2 = 1 V3 −V1 I3 = = 2 − 2 = 0, 1 V4 −V2 I4 = = 0 + 1 = 1 A, 1 V1 = 2 A, I5 = 1 V2 = −1 A. I6 = 1 I1 =

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Problem 2.75 Find the voltages across R1 , R2 , and R3 in the circuit of Fig. P2.75 using the DC Operating Point Analysis tool in Multisim. R1 V1

10 Ω + _ 15 V

R3 I R2

15 Ω

30 Ω

1.5I

Figure P2.75: Circuit for Problem 2.75.

Solution: When built, the circuit should look like that shown below:

To set the current source up so that it refers to the proper current, double-click on the current source. This will bring up the window shown below:

In ABM syntax, this becomes: 1.5*V(2)/30. Enter this as shown in the figure above. Using DC Operating Analysis, select the following outputs: V(1) V(2) V(3) V(1) − V(2) V(2) − V(3) The numerical results are shown in the grapher window: c All rights reserved. Do not reproduce or distribute. 2013 National Technology and Science Press

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Problem 2.76 Find the equivalent resistance looking into the terminals of the circuit in Fig. P2.76 using a test voltage source and current probes in the Interactive Simulation in Multisim. Compare the answer you get to what you obtain from series and parallel combining of resistors carried out by hand.

Figure P2.76: Circuit for Problem 2.76.

Solution: The two 12-kΩ resistors in parallel are equal to 6 kΩ because for any parallel combination of identically valued resistors, R Req = , N where Req is the total equivalent resistance of the combination, R is the resistance of the elements, and N is the number of resistors in the parallel network. On the right side there is now a series combination of a 6-kΩ resistor, a 10-kΩ resistor, and a 3.9-kΩ resistor. Together these equal 19.9 kΩ. This new value is in parallel with a 1-kΩ resistor. Therefore 19.9 k 1 = 952.153 Ω. This is then in series with a 1-kΩ resistor, so 1.952153 kΩ. This is then in parallel with another 1-kΩ resistor, so 1.952 k 1 = 661.25 Ω. Finally, this value is in series with a 4.7-kΩ resistor, so Req = 4.7 kΩ + 661.25 Ω = 5.361 kΩ. When you construct the circuit, add a voltage source across the two pins, and add the current probe right above the test source. Run the Interactive Simulation to get the following screen shot:

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Note that you can set the test voltage to anything you want. Unity test voltage (and current) sources usually just make things easier to solve. Take the values from this simulation, to get Req =

1.00 = 5.348 kΩ, 187 × 10−6

which is pretty close (0.24% off) from what we obtained by hand. The differences can be explained by rounding.

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m2.1 Kirchhoff’s Laws: Determine currents I1 to I3 and the voltage V1 in the circuit of Fig. m2.1 with component values Isrc = 1.8 mA, Vsrc = 9.0 V, R1 = 2.2 kΩ, R2 = 3.3 kΩ, and R3 = 1.0 kΩ.

Vsrc

R1

I1

R2

I2

R3

I3

+ _

_ Isrc

V1

+

Figure m2.1: Circuit for Problem m2.1.

Solution:

Summary Comparison: Results: Analysis Simulaon Measurement Rela ve Differences: Simulaon -- Analysis Measurement -- Analysis

I1 (mA) I2 (mA) I3 (mA) -0.56 2.36 -1.80 -0.56 2.36 -1.80 -0.51 2.43 -1.92

0.0% -8.9%

0.0% 3.0%

0.0% 6.7%

V1 (V) 9.59 9.58 9.80

-0.1% 2.2%

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Analytical Solution:

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Multisim Results:

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myDAQ Results:

Further Exploration: No change observed in I1, I2, and I3, but V1 varied from 7.88 V to 9.88 V. Explana!on: The current source maintains a constant current in the lower branch and therefore the remaining circuit sees no difference. However, as the voltage changes across R3 the current source voltage adapts its own voltage V1.

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m2.2 Equivalent Resistance: Find the equivalent resistance between the following terminal pairs in the circuit of Fig. m2.2 under the stated conditions: (a) a-b with the other terminals unconnected, (b) a-d with the other terminals unconnected, (c) b-c with a wire connecting terminals a and d, and (d) a-d with a wire connecting terminals b and c. Use these component values: R1 = 10 kΩ, R2 = 33 kΩ, R3 = 15 kΩ, R4 = 47 kΩ, and R5 = 2 kΩ. a

R1

c

R2

R3

R4

b

R5

d

Figure m2.2: Circuit for Problem m2.2.

Solution:

Summary Comparison: Results: Analysis Simulaon Measurement Rela ve Differences: Simulaon -- Analysis Measurement -- Analysis

R_AB (kohms) R_AD (kohms) R_BC (kohms) R_AD (kohms) 16.72 9.77 21.45 9.025 16.72 9.766 21.45 9.025 16.64 9.79 21.4 9.04

0.0% -0.5%

0.0% 0.2%

0.0% -0.2%

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0.0% 0.2%

Analytical Solution:

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Multisim Results:

c All rights reserved. Do not reproduce or distribute. 2013 National Technology and Science Press

c All rights reserved. Do not reproduce or distribute. 2013 National Technology and Science Press

c All rights reserved. Do not reproduce or distribute. 2013 National Technology and Science Press

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myDAQ Results: See summary comparison table for DMM ohmmeter measurements.

Further Exploration Results: Measurement R_AB R_AD R_BC with wire R_AD with wire

Voltage (V) 4.89 4.88 4.91 4.88

Current (mA) 0.29 0.50 0.23 0.54

Calculated resistance (kΩ) 16.9 9.76 21.4 9.04

These calculated resistance values agree with the myDAQ ohmmeter measurements within 2%.

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m2.3 Current and Voltage Dividers: Apply the concepts of voltage dividers, current dividers, and equivalent resistance to find the currents I1 to I3 and the voltages V1 to V3 in the circuit of Fig. m2.3. Use these component values: Vsrc = 12 V, R1 = 1.0 kΩ, R2 = 10 kΩ, R3 = 1.5 kΩ, R4 = 2.2 kΩ, R5 = 4.7 kΩ, and R6 = 3.3 kΩ. I1

R1

R3

Vsrc

I2

+

+ _

V1

_

R2

+ V2 R4

_

R5 I3

R6

+ V _ 3 Figure m2.3: Circuit for Problem m2.3.

Solution:

Summary Comparison: Results: Analysis Simulaon Measurement Rela ve Differences: Simulaon -- Analysis Measurement -- Analysis

I1 (mA) 1.82 1.82 1.83

I2 (mA) 1.40 1.40 1.41

I3 (mA) 0.45 0.45 0.45

0.2% 0.8%

0.2% 0.9%

0.0% 1.1%

V1 (V) 4.19 4.19 4.21

V2 (V) 2.09 2.09 2.10

V3 (V) -5.99 -5.99 -5.96

0.0% 0.5%

0.0% 0.3%

0.0% -0.6%

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Analytical Solution:

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Multisim Results:

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myDAQ Results:

See summary comparison table for DMM voltmeter and ammeter measurements.

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m2.4 Wye-Delta Transformation: Find (a) the currents I1 and I2 in the circuit of Fig. m2.4 and (b) the power delivered by each of the two voltage sources. Use these component values: V1 = 15 V, V2 = 15 V, R1 = 3.3 kΩ, R2 = 1.5 kΩ, R3 = 4.7 kΩ, R4 = 5.6 kΩ, R5 = 1.0 kΩ, and R6 = 2.2 kΩ. R3 R2

R1

R5

R4 V1 _

R6 V2 _

+

+

I1

I2

Figure m2.4: Circuit for Problem m2.4.

Solution:

Summary Comparison: Results: Analysis Simulaon Measurement Rela ve Differences: Simulaon -- Analysis Measurement -- Analysis

I1 (mA) 4.05 4.05 4.05

I2 (mA) 4.45 4.45 4.45

P1 (mW) 60.7 60.7 60.6

P2 (mW) 66.7 66.7 66.5

0.0% 0.0%

0.0% 0.0%

0.0% -0.2%

0.0% -0.3%

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Analytical Solution:

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Multisim Results:

Interac ve simula on results: I1 = 4.05 mA and I2 = 4.4 5mA

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Interac ve simula on results: PL = 60.703 mW delivered, PR = 66.687 mW delivered

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myDAQ Results:

DMM ammeter on 20mA range se!ng measures I1 = 4.06 mA and I2 = 4.45 mA. DMM voltmeter on 20V range se!ng measures 14.91 V for the le" source and 14.85 V for the right source. Calculate power as product of voltage and current: PL = 60.5 mW and PR = 66.1 mW.

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