(6)(2) (-6)(-4) (-4)(6) + (-2)(-3) + (4)(3) + (2)(-3) = = 0

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Chapter 3 Homework Solution P3.2-2, 4, 6, 10, 13, 17, 21 P3.3-2, 4, 6, 11 P3.4-1, 3, 6, 9, 12 P3.5-2 P3.6-1, 4, 9, 14, 21, 31, 40 ---

P 3.2-2 Determine the values of i2, i4, v2, v3, and v6 in Figure P 3.2-2.

Solution:

Apply KCL at node a to get 2 = i2 + 6 = 0  i2 = 4 A Apply KCL at node b to get 3 = i4 + 6  i4 = -3 A Apply KVL to the loop consisting of elements A and B to get

-v2 – 6 = 0  v2 = -6 V Apply KVL to the loop consisting of elements C, D, and A to get

-v3 – (-2) – 6 = 0  v4 = -4 V Apply KVL to the loop consisting of elements E, F and D to get

4 – v6 + (-2) = 0  v6 = 2 V Check: The sum of the power supplied by all branches is

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P 3.2-4 Determine the power absorbed by each of the resistors in the circuit shown in Figure P 3.2-4.

solution

2

Power absorbed by the 4 resistor = 4 = 100 W 2

2

Power absorbed by the 6 resistor = 6 = 24 W 1

2

Power absorbed by the 8 resistor = 8 = 72 W 4 i i i       12 2 A 1 6 20 5 A 2 4 3 2 A 3 2 3 A 4 2 3 i i i i i i i           

(checked using LNAP 8/16/02)

P 3.2-6 Determine the power supplied by each voltage source in the circuit of Figure P 3.2-6.

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Figure P 3.2-6 Solution:

3

3 2 mA 3 2 10 6 10 6 mW P           

3

3 1 mA 7 1 10 7 10 7 mW P          

(checked using LNAP 8/16/02)

P 3.2-10 The circuit shown in Figure P 3.2-10 consists of five voltage sources and four current sources. Express the power supplied by each source in terms of the voltage source voltages and the current source currents.

Figure P 3.2-10 Solution:

The subscripts suggest a numbering of the sources. Apply KVL to get

1 2 5 9 6

vvvvv

1

i and v1 do not adhere to the passive convention, so

1 1 1 1 2 5 9 6

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is the power supplied by source 1. Next, apply KCL to get

2 1 4

i   ii

2

i and v2 do not adhere to the passive convention, so

2 2 2 1 4 2

pi v   ii v is the power supplied by source 2. Next, apply KVL to get

3 6 5 9

vvvv

3

i and v3 adhere to the passive convention, so

3 3 3 3 6 5 9

p  i v  i vvv

is the power supplied by source 3. Next, apply KVL to get

4 2 5 8

vv  v v

4

i and v4 do not adhere to the passive convention, so

4 4 4 4 2 5 8

pi vi vvv is the power supplied by source 4. Next, apply KCL to get

5 3 2 3 1 4 1 3 4

i     i i i ii   i i i

5

i and v5 adhere to the passive convention, so

5 5 5 1 3 4 5

p  i v   i  i i v is the power supplied by source 5. Next, apply KCL to get

6 7 1 3

i  i ii

6

i and v6 adhere to the passive convention, so

6 6 6 7 1 3 6

p  i v   iii v

is the power supplied by source 6. Next, apply KVL to get

7 6

v  v

7

i and v7 adhere to the passive convention, so

 

7 7 7 7 6 7 6

p  i v  ivi v is the power supplied by source 7. Next, apply KCL to get

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8 4

i  i

8

i and v8 do not adhere to the passive convention, so

 

8 8 8 4 8 4 8

pi v  i v  i v is the power supplied by source 8. Finally, apply KCL to get

9 1 3

i  i i

9

i and v9 adhere to the passive convention, so

9 9 9 1 3 9

p  i v   ii v is the power supplied by source 9.

(Check: 9 1 0 n n p  

.)

P 3.2-13 Determine the value of the current that is measured by the meter in Figure P 3.2-13.

Figure P 3.2-13 Solution:

We can label the circuit as shown.

The subscripts suggest a numbering of the circuit elements. Apply KVL to node the left mesh to get 1 1 1 20 15 25 20 0 0.5 A 40 ii    i  

Apply KVL to node the left mesh to get

 

2 25 1 0 2 25 1 25 0.5 12.5 V

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Apply KCL to get im i2. Finally, apply Ohm’s law to the 50  resistor to get 2 m 2 12.5 0.25 A 50 50 v i  i   (Checked: LNAPDC 9/1/04) P 3.2-17 Determine the current i in Figure P 3.3-17.

Answer: i = 4 A

Figure P 3.3-17 Solution:

Apply KCL at node a to determine the current in the horizontal resistor as shown.

Apply KVL to the loop consisting of the voltages source and the two resistors to get -4(2-i) + 4(i) - 24 = 0  i = 4 A

P3.2-21 Determine the value of the voltage v5 for the circuit shown in Figure P3.2-21.

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Solution:

Apply KVL to the left mesh:

2 18 12 0 2 6 V

v     v

Use the element equation of the dependent source:

 

6 0.10 2 0.10 6 0.6 A

iv  

Apply KCL at the right node

5 6 0.25 5 25 0.25 6 25 0.25 0.6 8.75 V 25 v i v i         

P 3.3-2 Consider the circuits shown in Figure P 3.3-2.

(a) Determine the value of the resistance R in Figure P 3.3-2b that makes the circuit in Figure P 3.3-2b equivalent to the circuit in Figure P 3.3-2a.

(b) Determine the current i in Figure P 3.3-2b. Because the circuits are equivalent, the current i in Figure P 3.3-2a is equal to the current i in Figure P 3.3-2b.

(c) Determine the power supplied by the voltage source. Solution:

 

( ) 6 3 2 4 15 28 28 ( ) 1.867 A 15 28 =28(1.867)=52.27 W (28 V and do not adhere to the passive convention.)

a R b i R c p i i           

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P 3.3-4 Determine the voltage v in the circuit shown in Figure P 3.3-4. Figure P 3.3-4 Solution: 1 3 3 1 Voltage division 16 12 8 V 16 8 4 12 4 V 4 8 KVL: 0 4 V v v v v v v           

(checked using LNAP 8/16/02) P 3.3-6 The input to the circuit shown in Figure P 3.3-6 is the voltage of the voltage source, va. The output of this circuit is the voltage measured by the voltmeter, vb. This circuit produces an output that is proportional to the input, that is

vb = k va where k is the constant of proportionality.

(a) Determine the value of the output, vb, when R = 240 Ω and va = 18 V.

(b) Determine the value of the power supplied by the voltage source when R = 240 Ω and va = 18 V.

(c) Determine the value of the resistance, R, required to cause the output to be vb = 2 V when the input is va = 18 V.

(d) Determine the value of the resistance, R, required to cause vb = 0.2va (that is, the value of the constant of proportionality is 2

10).

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Figure P 3.3-6 Solution:

 

  

180 a.) 18 10.8 V 120 180 18 b.) 18 1.08 W 120 180 c.) 18 2 18 2 2 120 15 120 d.) 0.2 0.2 120 0.8 30 120 R R R R R R R R R                       

P 3.3-11 For the circuit of Figure P 3.3-11, find the voltage v3 and the current i and show that the power delivered to the three resistors is equal to that supplied by the source.

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Solution:

3 3

3

From voltage division 12 3 V

3 9 = = 1 A 3 then v v i       

The power absorbed by the resistors is:

 

12

 

6 

 

12

 

3 

 

12

 

3 12 W The power supplied by the source is (12)(1) = 12 W.

P 3.4-1 Use current division to determine the currents i1, i2, i3, and i4 in the circuit shown in Figure P 3.4-1. Figure P 3.4-1. Solution: 1 1 1 6 4 4 A 1 1 1 1 1 1 2 3 6 3 6 3 2 1 1 2 3 4 A; 2 1 1 1 1 3 6 3 2 1 1 2 4 1 A 3 1 1 1 1 6 3 2 1 1 4 2 A 4 1 1 1 1 6 3 2 i i i i                        

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Figure P 3.4-3

(a) Suppose R2 = 12 Ω. Determine the value of R1 and of the current i.

(b) Suppose, instead, R1 = 12 Ω. Determine the value of R2 and of the current i. (c) Instead, choose R1 and R2 to minimize the power absorbed by any one resistor. Solution: 1 1 2 2 2 8 8 or 8 8 8 (2 ) 2 or 2 i R R i R i i R R i         

 

 

1 2 8 2 8 2 A ; 12 2 6 3 3 8 4 8 A ; 12 4 6 3 2 3 a i R b i R            

 

1 2 1 2 1 2 1 2 1 1 1 2 1 2 1

will cause i= 2 1 A. The current in both and will be 1 A. 2 1 2 8 ; 2 8 8 8 2 c R R R R R R R R R R R R R R               

P 3.4-6 Figure P 3.4-6 shows a transistor amplifier. The values of R1 and R2 are to be selected. Resistances R1 and R2 are used to bias the transistor, that is, to create useful operating conditions. In this problem, we want to select R1 and R2 so that vb = 5 V. We expect the value of ib to be approximately 10 μA. When i1 ≥ 10ib, it is customary to treat ib as negligible, that is, to assume ib = 0. In that case R1 and R2 comprise a voltage divider.

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Figure P 3.4-6

(a) Select values for R1 and R2 so that vb = 5 V and the total power absorbed by R1 and R2 is no more than 5 mW.

(b) An inferior transistor could cause ib to be larger than expected. Using the values of R1 and R2 from part (a), determine the value of vb that would result from ib = 15 μA.

Solution:

(a) To insure that ib is negligible we require

so R1R2 150 k

To insure that the total power absorbed by R1 and R2 is no more than 5 mW we require

2 3 1 2 1 2 15 5 10 R R 45 k R R         Next to cause vb = 5 V we require

2 b 1 2 1 2 5 v R 15 R 2R R R     

For example, R140 k ,  R2 80 k , satisfy all three requirements.

(b) KVL gives

80 10 3

i1vb150 KCL gives 1 b 3 15 10 6 40 10 v i      Therefore

80 103

b 3 15 10 6 b 15 40 10 v v            Finally b b 13.8 3 1.2 15 4.6 V 3 v    v  

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P 3.4-9 Determine the power supplied by the dependent source in Figure P 3.4-9.

Figure P 3.4-9 Solution:

Use current division to get

3

a 75 30 10 22.5 mA 25 75 i        so vb 50

22.5 10 3

 1.125 V

The power supplied by the dependent source is given by

3

30 10 1.125 33.75 mW

p     

P 3.4-12 Determine the value of the current measured by the meter in Figure P 3.4-12.

Figure P 3.4-12 Solution:

Replace the (ideal) ammeter with the equivalent short circuit. Label the current measured by the meter.

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Apply KCL at the left node of the VCCS to get a a a a 1.2 1.2 0.2 0.3 4 V 10 0.3 v v v v      

Use current division to get

 

m a 30 30 0.2 0.2 4 0.6 A 30 10 30 10 iv    

P 3.5-2 Determine the power supplied by each source in the circuit shown in Figure P 3.5-2.

Figure P 3.5-2 Solution:

The 20- and 5- resistors are connected in parallel. The equivalent resistance is 20 5 4 20 5

 

 .

The 7- resistor is connected in parallel with a short circuit, a 0- resistor. The equivalent resistance is 0 7 0

0 7 

 

 , a short circuit.

The voltage sources are connected in series and can be replaced by a single equivalent voltage source.

After doing so, and labeling the resistor currents, we have the circuit shown.

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The parallel current sources can be replaced by an equivalent current source.

Apply KVL to get

 

1 1

5 v 4 3.5 0 v 19 V

     

The power supplied by each sources is:

Source Power delivered

8-V voltage source 2 3.5

 

 7 W 3-V voltage source 3 3.5

 

 10.5 W 3-A current source 3 19 57 W 0.5-A current source 0.5 19 9.5 W

(Checked using LNAP, 9/15/04)

P 3.6-1 The circuit shown in Figure P 3.6-1a has been divided into two parts. In Figure P 3.6-1b, the right-hand part has been replaced with an equivalent circuit. The left-hand part of the circuit has not been changed.

(a) Determine the value of the

resistance R in Figure P 3.6-1b that makes the circuit in Figure P 3.6-1b equivalent to the circuit in Figure P 3.6-1a.

(b) Find the current i and the voltage v shown in Figure P 3.6-1b. Because of the equivalence, the current i and the voltage v shown in Figure P 3.6-1a are equal to the current i and the voltage v shown in Figure P 3.6-1b.

(c) Find the current i2 shown in Figure

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P 3.6-1a using current division. Solution:

 

 

 

2 48 24 16 32 48 24 32 32 32 32 24 16 V ; 32 32 8 32 32 16 1 A 32 2 48 1 1 A 48 24 2 3 a R b v i c i                   P 3.6-4

(a) Determine values of R1 and R2 in Figure P 3.6-4b that make the circuit in Figure P 3.6-4b equivalent to the circuit in Figure P 3.6-4a.

(b) Analyze the circuit in Figure P 3.6-4b to determine the values of the currents ia and ib (c) Because the circuits are equivalent, the currents ia and ib shown in Figure P 3.6-4b are

equal to the currents ia and ib shown in Figure P 3.6-4a. Use this fact to determine values of the voltage v1 and current i2 shown in Figure P 3.6-4a.

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(a) 1 1 24 1 12 1 8 4 2 2 R     R   and (b)

First, apply KVL to the left mesh to get  27 6ia3ia 0  ia 3A. Next, apply KVL to the left mesh to get 4ib 3ia 0  ib 2 25. A.

(c)

P 3.6-9 Determine the value of the current i in Figure 3.6-9.

Answer: i = 0.5 mA

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P 3.6-14 All of the resistances in the circuit shown in Figure P 3.6-14 are multiples of R. Determine the value of R.

Figure P 3.6-14 Solution:

 

4 6 4 2 3 2 5 5 R RR RRRR

2 2 2

2 2

2 RR RR R  R R RR

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Then

 

120.1 R 2R 2R 0.1 2RR60 

P 3.6-21 Determine the value of the resistance R in the circuit shown in Figure P 3.6-22, given that Req = 9 Ω.

Answer: R = 15 Ω

Figure P 3.6-22 Solution:

Replace parallel resistors by an equivalent resistor:

8 || 24 = 6 

A short circuit in parallel with a resistor is equivalent to a short circuit.

Replace series resistors by an equivalent resistor:

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Now

eq 9R  5 12 ||R||10 so 60 11 4 15 60 11 R R R      

P 3.6-31 The voltmeter in Figure P 3.6-31 measures the voltage across the current source.

Figure P 3.6-31

(a) Determine the value of the voltage measured by the meter. (b) Determine the power supplied by each circuit element. Solution:

Replace the ideal voltmeter with the equivalent open circuit and label the voltage measured by the meter. Label the element voltages and currents as shown in (b).

Using units of V, A, and W:

a.) Determine the value of the voltage measured by the meter.

Kirchhoff’s laws give

Using units of V, mA, k and mW: a.) Determine the value of the voltage measured by the meter.

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R m

12vv and iR    is 2 10 A3 Ohm’s law gives

3

R 25 10 R v    i Then

3

3



3

R 25 10 R 25 10 2 10 50 V v    i        m 12 R 12 50 62 V v  v   

b.) Determine the power supplied by each element.

voltage source

 

3

s 3 12 12 2 10 24 10 W i          current source

3

3 62 2 10  124 10  W resistor

3

R R 3 50 2 10 100 10 W v i         total 0 R m 12vv and iR   is 2 mA Ohm’s law gives

R 25 R v   i Then

 

R 25 R 25 2 50 V v   i     m 12 R 12 50 62 V v  v   

b.) Determine the power supplied by each element. voltage source

 

 

s 12 12 2 24 mW i      current source 62 2

 

124 mW resistor R R 50

 

2 100 mW v i     total 0

P3.6-40 Consider the circuit shown in Figure P3.6-40. Given that the voltage of the dependent voltage source is va 8 V, determine the values of R1and vo.

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Solution: First, o 20 8 3.2 V 20 30 v      Next,

c b 1 1 1 1 1 1 1 1 1 8 40 40 10 40 10 400 400 40 20 40 40 12 40 || 40 12 40 40 480 52 12 40 i i R R R R R R R R R                        then

 

1 1 400 20 8 400 1000 480 480 52 1000 10 20 480 52R R 8 52          

Figure

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