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Electric Circuits, 9th Edition P16 potx

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b Test your solution for the branch currents by showing that the total power dissipated equals the total power developed.. Figure P4.13 4.14 Use the node-voltage method to find the tota

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126 Techniques of Circuit Analysis

dv 2 = R 3 R 4 {R y R,I, 2 - [R 2 (R 3 + R 4 ) + R 3 R 4 ]I gl }

dR, " [(R l + R 2 )(R 3 + R 4 ) + R 3 R 4 ] 2 (4.98)

Figure 4.69 A Circuit used to introduce sensitivity analysis

We now consider an example with actual component values to illustrate the use of Eqs 4.97 and 4.98

EXAMPLE

Assume the nominal values of the components in the circuit in Fig 4.69 are:

R l = 25 H ; R 2 = 5 ft; R 3 = 50 ft; R 4 = 75 ft; I g] = 12 A and

I g2 = 16 A Use sensitivity analysis to predict the values of v x and v 2 if

the value of R^ is different by 10% from its nominal value

Solution

From Eqs 4.95 and 4.96 we find the nominal values of v v and v 2 Thus

25(3750(16) - [5(125) + 3750112}

and

30(125) + 3750

3750[30(16) - 25(12)]

30(125) + 3750 (4.100)

Now from Eqs 4.97 and 4.98 we can find the sensitivity of V\ and v 2 to

changes in R t Hence

dv { [3750 + 5(125)] - (3750(16) - [3750 + 5(125)]12}

dRi [(30)(125) + 3750]'

and

dv 2 _ 3750(3750(16) - [5(125) + 3750J12}]

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Ai?! = - 2 5 ft and Eq 4.101 predicts Av x will be

A vi = ( ^ V2-5) = -1-4583 V

Therefore, if R^ is 10% less than its nominal value, our analysis predicts

that v\ will be

Vi = 25 - 1.4583 = 23.5417 V (4.103)

Similarly for Eq 4.102 we have

Av 2 = 0.5(-2.5) = -1.25 V,

v 2 = 90 - 1.25 = 88.75 V (4.104)

We attempt to confirm the results in Eqs 4.103 and 4.104 by substituting

the value R^ = 22.5 ft into Eqs 4.95 and 4.96 When we do, the results are

vx = 23.4780 V, (4.105)

Why is there a difference between the values predicted from the sensitivity

analysis and the exact values computed by substituting for R^ in the

equa-tions for V\ and v 2l We can see from Eqs 4.97 and 4.98 that the sensitivity

of Vi and v 2 with respect to R^ is a function of R lt because R^ appears in

the denominator of both Eqs 4.97 and 4.98 This means that as Ri

changes, the sensitivities change and hence we cannot expect Eqs 4.97 and

4.98 to give exact results for large changes in /¾ Note that for a 10%

change in R u the percent error between the predicted and exact values of

vx and v 2 is small Specifically, the percent error in v { = 0.2713% and the

percent error in v 2 = 0.0676%

From this example, we can see that a tremendous amount of work is

involved if we are to determine the sensitivity of v^ and v 2 to changes in

the remaining component values, namely R 2f R 3t R 4f I gU and I g2

Fortunately, PSpice has a sensitivity function that will perform sensitivity

analysis for us The sensitivity function in PSpice calculates two types of

sensitivity The first is known as the one-unit sensitivity, and the second

is known as the 1% sensitivity In the example circuit, a one-unit change

in a resistor would change its value by 1 ft and a one-unit change in a

current source would change its value by 1 A In contrast, 1%

sensitiv-ity analysis determines the effect of changing resistors or sources by

1% of their nominal values

The result of PSpice sensitivity analysis of the circuit in Fig 4.69 is

shown in Table 4.2 Because we are analyzing a linear circuit, we can use

superposition to predict values of v\ and v 2 if more than one component's

value changes For example, let us assume /?, decreases to 24 ft and R 2

decreases to 4 ft From Table 4.2 we can combine the unit sensitivity of V\

to changes in R { and R 2 to get

Auj Av\

A#[ + 1R~ 2 = 0.5833 - 5.417 = -4.8337 V/ft,

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Techniques of Circuit Analysis

Similarly,

At?? Av 2

+

AS, ' AR 2 0.5+ 6.5 = 7.0 V / a

Thus if both R { and R 2 decreased by 1II we would predict

Vi = 25 + 4.8227 = 29.8337 V,

v 2 = 90 - 7 = 83 V

TABLE 4.2 PSpice Sensitivity Analysis Results

Element Element Name Value

Element! Sensitivity (Volts/Unit)

(a) DC Sensitivities of Node Voltage VI

Rl 25 R2 5 R3 50 R4 75

IG1 12

IG2 16

(b) Sensitivities of Output V2

Rl 25

R2 5

R3 50

R4 75

IG1 12 IG2 16

0.5833 -5.417 0.45 0.2 -14.58 12.5

0.5 6.5 0.54 0.24 -12.5

15

Normalized Sensitivity (Volts/Percent)

0.1458 -0.2708 0.225 0.15 -1.75

2

0.125 0.325 0.27 0.18 -1.5 2.4

If we substitute R { = 24 fl and R 2 = 4 ft into Eqs 4.95 and 4.96 we get

vi = 29.793 V,

v 2 = 82.759 V

In both cases our predictions are within a fraction of a volt of the actual node voltage values

Circuit designers use the results of sensitivity analysis to determine which component value variation has the greatest impact on the output of the circuit As we can see from the PSpice sensitivity analysis in Table 4.2,

the node voltages Dj and v 2 are much more sensitive to changes in R 2 than

to changes in R u Specifically, V\ is (5.417/0.5833) or approximately

9 times more sensitive to changes in R 2 than to changes in R x and v 2 is

(6.5/0.5) or 13 times more sensitive to changes in R 2 than to changes in

Ri, Hence in the example circuit, the tolerance on R 2 must be more

strin-gent than the tolerance on R^ if i t is important to keep V\ and v 2 close to their nominal values

NOTE: Assess your understanding of this Practical Perspective by trying Chapter Problems 4.105-4.107

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Summary

• For the topics in this chapter, mastery of some basic terms,

and the concepts they represent, is necessary Those terms

are node, essential node, path, branch, essential branch,

mesh, and planar circuit Table 4.1 provides definitions

and examples of these terms (See page 91.)

• Two new circuit analysis techniques were introduced in

this chapter:

• The node-voltage method works with both planar

and nonplanar circuits A reference node is chosen

from among the essential nodes Voltage variables

are assigned at the remaining essential nodes, and

Kirchhoff s current law is used to write one equation

per voltage variable The number of equations is

n e — 1, where n e is the number of essential nodes

(See page 93.)

• The mesh-current method works only with planar

circuits Mesh currents are assigned to each mesh,

and Kirchhoff's voltage law is used to write one

equation per mesh The number of equations is

b — (n — 1), where b is the number of branches in

which the current is unknown, and n is the number of

nodes The mesh currents are used to find the branch

currents (See page 99.)

• Several new circuit simplification techniques were

introduced in this chapter:

• Source transformations allow us to exchange a

volt-age source (v s) and a series resistor (R) for a current

source (i s) and a parallel resistor (R) and vice versa

The combinations must be equivalent in terms of

their terminal voltage and current Terminal

equiva-lence holds provided that

(See page 109.)

• Thevenin equivalents and Norton equivalents allow

us to simplify a circuit comprised of sources and resis-tors into an equivalent circuit consisting of a voltage source and a series resistor (Thevenin) or a current source and a parallel resistor (Norton) The simplified circuit and the original circuit must be equivalent in terms of their terminal voltage and current Thus keep in mind that (1) the Thevenin voltage (Kiii) is the open-circuit voltage across the terminals of the original circuit, (2) the Thevenin resistance (i?Th) is the ratio of the Thevenin voltage to the short-circuit current across the terminals of the original circuit; and (3) the Norton equivalent is obtained by per-forming a source transformation on a Thevenin equivalent (See page 113.)

• Maximum power transfer is a technique for calculating

the maximum value of p that can be delivered to a load,

RL Maximum power transfer occurs when Ri = Rjh,

the Thevenin resistance as seen from the resistor R L

The equation for the maximum power transferred is

(See page 120.)

• In a circuit with multiple independent sources,

superposition allows us to activate one source at a time

and sum the resulting voltages and currents to deter-mine the voltages and currents that exist when all inde-pendent sources are active Deinde-pendent sources are never deactivated when applying superposition (See page 122.)

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130 Techniques of Circuit Analysis

Problems

Section 4.1

4.1 For the circuit shown in Fig P4.1, state the numerical

value of the number of (a) branches, (b) branches

where the current is unknown, (c) essential branches,

(d) essential branches where the current is unknown,

(e) nodes, (f) essential nodes, and (g) meshes

Figure P4.1

4.4 Assume the current i g in the circuit in Fig P4.4 is

known The resistors R^ - R 5 are also known a) How many unknown currents are there?

b) How many independent equations can be writ-ten using Kirchhoff s current law (KCL)?

c) Write an independent set of KCL equations d) How many independent equations can be derived from Kirchhoff s voltage law (KVL)? e) Write a set of independent KVL equations

Figure P4.4

V L

4.2 a) If only the essential nodes and branches are

identified in the circuit in Fig P4.1, how many

simultaneous equations are needed to describe

the circuit?

b) How many of these equations can be derived

using Kirchhoff s current law?

c) How many must be derived using Kirchhoffs

voltage law?

d) What two meshes should be avoided in applying

the voltage law?

4.3 a) How many separate parts does the circuit in

Fig P4.3 have?

b) How many nodes?

c) How many branches are there?

d) Assume that the lower node in each part of the

circuit is joined by a single conductor Repeat

the calculations in (a)-(c)

Figure P4.3

4.5 A current leaving a node is defined as positive a) Sum the currents at each node in the circuit shown in Fig P4.4

b) Show that any one of the equations in (a) can be derived from the remaining three equations

Section 4.2

4.6 Use the node-voltage method to find Uj and v 2 in PSPICE the circuit in Fig P4.6

Figure P4.6

144 V

son

4.7 Use the node-voltage method to find how much

PSPICE power the 2 A source extracts from the circuit in

HULTISIM p j g p 4 J

Figure P4.7

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4.8 Use the node-voltage method to find v x and v 2 in

PSPICE the circuit shown in Fig P4.8

MULTISIM

Figure P4.8

8 f t

-VvV-<>i|40ft kmVL u 2f 120ft C\ J 1 A

4.9 Use the node-voltage method to find v () in the

cir-PSPICE cuit in Fig P4.9

Figure P4.9

2011

24 V

8012

40 mA

4.10 a) Find the power developed by the 40 mA current

PSPKE source in the circuit in Fig P4.9

b) Find the power developed by the 24 V voltage

source in the circuit in Fig P4.9

c) Verify that the total power developed equals the

total power dissipated

4.11 A 50 O resistor is connected in series with the

PSPICE 40 mA current source in the circuit in Fig P4.9

MULTISIM N ^ ,

a) Find v a

b) Find the power developed by the 40 mA current

source

c) Find the power developed by the 24 V voltage

source

d) Verify that the total power developed equals the

total power dissipated

e) What effect will any finite resistance connected

in series with the 40 mA current source have on

the value of v a l

4.12 The circuit shown in Fig P4.12 is a dc model of a

PSPICE residential power distribution circuit

a) Use the node-voltage method to find the branch

currents i { — /6

b) Test your solution for the branch currents by showing that the total power dissipated equals the total power developed

Figure P4.12

125 V

125 V

':,

4.13 a) Use the node-voltage method to find the

PSPKE branch currents /a - /e in the circuit shown in

MULTISIM F i g p 4 > 1 3

b) Find the total power developed in the circuit

Figure P4.13

4.14 Use the node-voltage method to find the total power

PSPICE dissipated in the circuit in Fig P4.14

MULTISIM

Figure P4.14

40 V

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132 Techniques of Circuit Analysis

4.15 a) Use the node-voltage method to find v h v 2 , and

! ™ v 3 in the circuit in Fig P4.15

MULTISIM

b) How much power does the 40 V voltage source

deliver to the circuit?

Figure P4.15

40 V

4.16 a) Use the node-voltage method to show that the

PSPICE output voltage v 0 in the circuit in Fig P4.16 is

equal to the average value of the source voltages

b) Find v 0 if v { = 100 V, v 2 = 80 V, and

v 3 = - 6 0 V

Figure P4.16

4.19 Use the node-voltage method to calculate the

PSPICE power delivered by the dependent voltage source in

WLTISIM , ^ „ , -. n

the circuit in Fig P4.19

Figure P4.19

160 V

('„ I loo a 150 L

20 a

^vw-Section 4.3

4.17 a) Use the node-voltage method to find the total

power developed in the circuit in Fig P4.17

MULTISIM

b) Check your answer by finding the total power

absorbed in the circuit

Figure P4.17

84/A

PSPICE

MULTISIM

4.18 a) Use the node-voltage method to find v„ in the

circuit in Fig P4.18

b) Find the power absorbed by the dependent source

c) Find the total power developed by the

independ-ent sources

Figure P4.18

20 a

4.20 a) Find the node voltages V\, «2, and v$ in the

cir-PSPICE c ui t in Fig P4.20

•IULTISIM

b) Find the total power dissipated in the circuit

Figure P4.20

5 n 10 a

m V A f

VvV-f \ 5 /(, iv*20ft y2?40O y y k (^/) 96 V

Section 4.4

4.21 Use the node-voltage method to find i 0 in the

cir-PSPICE cuit in Fig P4.21

MULTISIM

Figure P4.21

20 V 6

30kll lkft

PSPICE MULTISIM

4.22 a) Use the node-voltage method to find the

branch currents ij, i 2, and /3 in the circuit in

Fig P4.22

b) Check your solution for ij, i 2, and i3 by showing that the power dissipated in the circuit equals the power developed

Figure P4.22

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4.23 a) Use the node-voltage method to find the power

dissipated in the 2 (2 resistor in the circuit in

Fig P4.23

b) Find the power supplied by the 230 V source

Figure P4.23

230 vC-y

i n

i n

i n :

la-i n

l n

i n

s n 2n :5 n

4.24 Use the node-voltage method to find the value of v 0

KM« in the circuit in Fig P4.24

MUITISIM

Figure P4.26

25 V

55 n

4.27 Use the node-voltage method to find v 0 in the cir-PSPICE cuit in Fig P4.27

MULTISIM

Figure P4.27

15 V

Figure P4.24

50 V

4.25 Use the node-voltage method to find the value of v a

PSPICE in the circuit in Fig P4.25

«!ULTISIM

Figure P4.25

4.26 a) Use the node-voltage method to find v 0 and

in the circuit in Fig P4.26 Use node a as the

reference node

b) Repeat part (a), but use node b as the

refer-ence node

c) Compare the choice of reference node in (a)

and (b) Which is better, and why?

4.28 Use the node-voltage method to find the power

devel-oped by the 20 V source in the circuit in Fig P4.28

PSPICE MULTISIM

Figure P4.28

8 0 n (1)3.125 v A

4.29 Assume you are a project engineer and one of your PSPICE

MULTISIM staff is assigned to analyze the circuit shown in Fig P4.29 The reference node and node numbers given on the figure were assigned by the analyst

Her solution gives the values of v 3 and v 4 as 108 V and 81.6 V, respectively

Test these values by checking the total power developed in the circuit against the total power dis-sipated Do you agree with the solution submitted

by the analyst?

Figure P4.29

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134 Techniques of Circuit Analysis

4.30 Show that when Eqs 4.16,4.17, and 4.19 are solved

for i B, the result is identical to Eq 2.25

Section 4.5

4.31 Solve Problem 4.12 using the mesh-current method

4.32 Solve Problem 4.13 using the mesh-current method

4.33 a) Use the mesh-current method to find the branch

currents L, //„ and L in the circuit in Fig P4.33

b) Repeat (a) if the polarity of the 60 V source is

reversed

Figure P4.33

4.34 a) Use the mesh-current method to find the total

power developed in the circuit in Fig P4.34

MULTISIM

b) Check your answer by showing that the total

power developed equals the total power

dissipated

Figure P4.34

460 V

4.35 Solve Problem 4.21 using the mesh-current method

4.36 Solve Problem 4.23 using the mesh-current method

Section 4.6

4.37 a) Use the mesh-current method to find v 0 in the

PSPICE circuit in Fig P4.37

MULTISIM

b) Find the power delivered by the dependent source

Figure P4.37

io a

7 a

4.38 Use the mesh-current method to find the power

dissi-PSPICE p a t e (j in the 20 fl resistor in the circuit in Fig P4.38

MULTISIM

Figure P4.38

5 0

3 a

• A W

135 V ©

4 f t

2 a l a

-"VW

4.39 Use the mesh-current method to find the power delivered by the dependent voltage source in the

circuit seen in Fig P4.39

Figure P4.39

660 V

25 a

20 /,,

4.40 Use the mesh-current method to find the power

>SPICE developed in the dependent voltage source in the

JLTISIM _ _ _

circuit in Fig P4.40

Figure P4.40

30 V

Section 4.7 4.41 Solve Problem 4.8 using the mesh-current method 4.42 a) Use the mesh-currcnt method to find how much

power the 4 A current source delivers to the cir-cuit in Fig P4.42

b) Find the total power delivered to the circuit c) Check your calculations by showing that the total power developed in the circuit equals the total power dissipated

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Figure P4.42

120 V

5 0

80 V

4.43 Solve Problem 4.20 using the mesh-current method

4.44 a) Use the mesh-current method to solve for i± in

PSPICE the circuit in Fig P4.44

MULTISIM

b) Find the power delivered by the independent

current source

c) Find the power delivered by the dependent

volt-age source

4.47 Solve Problem 4.22 using the mesh-current method 4.48 Use the mesh-current method to find the total

PSPICE power dissipated in the circuit in Fig P4.48

MULTISIM

Figure P4.48

4.49 a) Assume the 20 V source in the circuit in

Fig P4.48 is changed to 60 V Find the total power dissipated in the circuit

b) Repeat (a) if the 6 A current source is replaced

by a short circuit

c) Explain why the answers to (a) and (b) are the same

Figure P4.44

i k n

150 k

4.45 Use the mesh-current method to find the total power

PSPICE developed in the circuit in Fig P4.45

MULTISIM

Figure P4.45

4.46 a) Use the mesh-current method to determine

PSPICE which sources in the circuit in Fig P4.46 are

gen-IULTISIM crating power

b) Find the total power dissipated in the circuit

Figure P4.46

2 f t

4.50 a) Find the branch currents /a - i c for the circuit

nna shown in Fig P4.50

MULTISIM

b) Check your answers by showing that the total power generated equals the total power dissipated

Figure P4.50

4/d - ic

19 A

4.51 a) Use the mesh-current method to find the branch

PSPICE MULTISIM currents in i a — ie in the circuit in Fig P4.51 b) Check your solution by showing that the total power developed in the circuit equals the total power dissipated

Figure P4.51

100ft

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