Figure P3.2: Circuit for Problem 3.2.zyang/Teaching/20162017Summer/Downloads/HW3-sol.pdfProblem 3.26...

Problem 3.2 Apply nodal analysis to determine Vx in the circuit of Fig. P3.2.

1 Ω

2 Ω

4 Ω

2 Ω

3 A Vx

V

+

_

Figure P3.2: Circuit for Problem 3.2.

Solution: At node V , application of KCL gives

V2+1

−3+V

2+4= 0,

which leads toV = 6 V.

By voltage division,

Vx =V ×42+4

=6×4

6= 4 V.

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Problem 3.3 Use nodal analysis to determine the currentIx and amount of powersupplied by the voltage source in the circuit of Fig. P3.3.

2 Ω

4 Ω

8 Ω

40 V9 A

Ix

V

+_

I


Solution: At nodeV , application of KCL gives

−9+V2

+V4

+V −40

8= 0

V

(

12

+14

+18

)

= 9+408

7V8

= 9+5

V = 16 V.

The currentIx is then given by

Ix =V4

=164

= 4 A.

To find the power supplied by the 40-V source, we need to first find the current Iflowing into its positive terminal,

I =V −40

8=

16−408

= −3 A.

Hence,P = V I = 40× (−3) = −120 W

(The minus sign confirms that the voltage source is a supplier of power.)

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

Problem 3.4 For the circuit in Fig. P3.4:

(a) Apply nodal analysis to find node voltagesV1 andV2.

(b) Determine the voltageVR and currentI.

1 Ω 1 Ω1 Ω

1 Ω 1 Ω

VR+_

16 V

V1 V2 I

+_+_


Solution: (a) At nodesV1 andV2,

Node 1:V1−16

1+

V1

1+

V1−V2

1= 0 (1)

Node 2:V2−V1

1+

V2

1+

V2

1= 0 (2)

Simplifying Eqs. (1) and (2) gives:

3V1−V2 = 16 (3)

−V1 +3V2 = 0. (4)

Simultaneous solution of Eqs. (3) and (4) leads to:

V1 = 6 V, V2 = 2 V.

(b)

VR = V1−V2 = 6−2 = 4 V

I =V2

1=

21

= 2 A.


Problem 3.10 The circuit in Fig. P3.10 contains a dependent current source.Determine the voltage Vx.

2Vx3 Ω

2 Ω

6 Ω6 V Vx

Vx

+

_

+_+_


Solution: In terms of the node voltage Vx, KCL gives

Vx −62

+Vx

3−2Vx +

Vx

6= 0,

whose solution leads toVx = −3 V.


Problem 3.14 Apply nodal analysis to find the currentIx in the circuit of Fig. P3.14.

0.1 Ω

0.1 Ω

0.5 Ω0.5 Ω

0.2 Ω

0.1 Ω

3 V2 V

4 V

Ix

+_

+_

+_

+_

+ _

V1 V2 V3


Solution: Application of KCL to the designated node voltagesV1, V2, andV3 gives

V1−20.1

+V1−V2

0.5+

V1−V3−40.2

= 0 (1)

V2−V1

0.5+

V2

0.1+

V2−V3

0.5= 0 (2)

V3−V1 +40.2

+V3−V2

0.5+

V3−30.1

= 0 (3)

Simplification, followed with simultaneous solution, leads to

V1 = 2.865 V, V2 = 0.625 V, V3 = 1.51 V,

and

Ix =V2

0.1=

0.6250.1

= 6.25 A.


Problem 3.26 Apply mesh analysis to find the mesh currents in the circuit ofFig. P3.26. Use the information to determine the voltageV .

2 Ω

4 Ω

3 Ω

2 Ω

I1 I216 V

12 V

V

+_ +

_


Solution: Application of KVL to the two loops gives:

Mesh 1: −16+2I1 +3(I1− I2) = 0,

Mesh 2: 3(I2− I1)+(2+4)I2 +12= 0,

which can be simplified to

5I1−3I2 = 16 (1)

−3I1 +9I2 = −12. (2)

Simultaneous solution of (1) and (2) leads to

I1 = 3 A, I2 = −13

A.

Hence,

V = 3(I1− I2) = 3

(

3+13

)

= 10 V.


Problem 3.31 Apply mesh analysis to determine the amount of power supplied bythe voltage source in Fig. P3.31.

3 Ω 6 Ω

2 Ω

2 Ω 4 Ω

4 A

48 V

+_

I1 I2

I3


Solution:

Mesh 1: 2I1 +3(I1− I3)+2(I1− I2)+48= 0

Mesh 2: −48+2(I2− I1)+6(I2− I3)+4I2 = 0

Mesh 3: I3 = −4 A.

Solution is:I1 = −8.4 A, I2 = 0.6 A, I3 = −4 A.

Current entering “+” terminal of voltage source is:

I = I1− I2 = −8.4−0.6 = −9 A.

Hence,P = V I = 48× (−9) = −432 W.


Problem 3.34 Apply mesh analysis to the circuit in Fig. P3.34 to determine Vx.

2Vx3 Ω

2 Ω

6 Ω6 V

+_+_ I1 I2 I3

+

_Vx

+_


Solution:

Mesh 1: −6+2I1 +3(I1 − I2) = 0Supermesh: 3(I2 − I1)+6I3 = 0Auxiliary 1: I3 − I2 = 2Vx

Auxiliary 2: Vx = 6I3

Solution is:

I1 = 4.5 A, I2 = 5.5 A, I3 = −0.5 A.

Vx = 6I3 = 6× (−0.5) = −3 V.


Problem 3.52 Apply the by-inspection method to develop a node-voltage matrixequation for the circuit in Fig. P3.52 and then use MATLAB R© or MathScript softwareto solve for V1 and V2.

6 Ω

12 Ω

6 Ω 3 A4 A2 A

V1 V2


Solution:

Node 1: G11 =

(

16

+1

12

)

= 0.25

G12 = G21 = −112

= −0.083

G22 =

(

16

+1

12

)

= 0.25

It1 = 2+4 = 6 A

It2 = 3−4 = −1 A

Application of Eq. (3.26) gives:[

0.25 −0.083−0.083 0.25

][

V1

V2

]

=

[

6−1

]

Matrix inversion gives

V1 = 25.5 V, V2 = 4.5 V.


Problem 3.53 Use the by-inspection method to establish a node-voltage matrixequation for the circuit in Fig. P3.53. Solve the matrix equation by MATLABR©

or MathScript software to findV1 to V4.

8 Ω9 Ω

5 Ω

3 Ω2 Ω

6 Ω1 Ω 4 Ω

7 Ω3 A

2 A

V1

V4V2V3


Solution:

G11 =1

2+1+

13+4

= 0.476

G12 = G21 = −1

2+1= −0.333

G13 = G31 = 0

G14 = G41 = −1

3+4= −0.143

G22 =1

1+2+

17

+16

= 0.643

G23 = G32 = −16

= −0.167

G24 = G42 = 0

G33 =15

+16

+19

= 0.478

G34 = G43 = −15

= −0.2

G44 =1

3+4+

15

= 0.343

Application of Eq. (3.26) gives:

0.476 −0.333 0 −0.143−0.333 0.643 −0.167 0

0 −0.167 0.478 −0.2−0.143 0 −0.2 0.343

V1

V2

V3

V4

=

20−2−3

Matrix inversion gives:

V1 =−8.1689 V, V2 =−8.4235 V, V3 =−16.155 V, V4 =−21.5748 V.


Problem 3.60 Determine the current Ix in the circuit of Fig. P3.60 by applying thesource-superposition method. Call I′x the component of Ix due to the voltage sourcealone, and I′′x the component due to the current source alone. Show that Ix = I′x + I′′xis the same as the answer to Problem 3.9.

3 Ω 6 Ω

2 Ω

2 Ω 4 Ω

4 A

48 V

+_

Ix


Solution:

A. Voltage source alone:

3 Ω 6 Ω

2 Ω

2 Ω 4 Ω

48 V

Ix

+_

‘

I1 I2‘ ‘

7I′1 −2I′2 = −48

−2I′1 +12I′2 = 48

I′1 = −6 A, I′2 = 3 A.

HenceI′x = I′2 = 3 A.

B. Current source alone:

3 Ω 6 Ω

2 Ω2 Ω 4 Ω

4 A

Ix‘‘‘‘

‘‘I1 ‘‘I2

I3

I′′3 = −4 A

7I′′1 −2I′′2 −3I′′3 = 0

−2I′′1 +12I′′2 −6I′′3 = 0


Hence,

I′′1 = −2.4 A, I′′2 = −2.4 A.

I′′x = I′′2 − I′′3 = −2.4+4 = 1.6 A.

Ix = I′x + I′′x = 3+1.6 = 4.6 A.


Problem 3.61 Apply the source-superposition method to the circuit in Fig. P3.61to determine:

(a) I′x, the component of Ix due to the voltage source alone

(b) I′′x , the component of Ix due to the current source alone

(c) The total current Ix = I′x + I′′x(d) P′, the power dissipated in the 4-Ω resistor due to I′x(e) P′′, the power dissipated in the 4-Ω resistor due to I′′x(f) P, the power dissipated in the 4-Ω resistor due to the total current I. Is P =

P′+P′′? If not, why not?

2 Ω

4 Ω

8 Ω

40 V9 A

Ix +_


Solution:(a) Voltage source alone:

2 Ω

4 Ω

8 Ω

40 V

+_

V1

Ix‘

‘

V ′1

2+

V ′1

4+

V ′1 −40

8= 0.

Hence,

V ′1 =

407

V.

I′x =40

7×4=

107

A.

(b) Current source alone:

2 Ω

4 Ω

8 Ω

9 A

Ix

V1

‘‘

‘‘

V ′′1

2+

V ′′1

4+

V ′′1

8= 9.

V ′′1 =

727

V.

I′′x =V ′′

1

4=

724×7

=187

A.


(c)

Ix = I′x + I′′x =107

+187

=287

= 4 A.

(d)

P′= (I′x)

2R =

(

107

)2

×4 =40049

W = 8.16 W.

(e)

P′′= (I′′x )

2R =

(

187

)2

×4 = 26.45 W.

(f)P = I2

x R = 42 ×4 = 64 W.

P 6= P′+P′′ because the superposition theorem does not apply to power.


Problem 3.64 Find the Thevenin equivalent circuit at terminals(a,b) for the circuitin Fig. P3.64.

1 Ω

2 Ω 3 Ω

4 Ω

2 Ω

3 A

a

b

Voc

V

+

_


Solution:

V3

+V6

= 3

V = 6 V.

Voltage division gives

VTh = Voc =V6×4 = 4 V.

Suppressing the current source:

1 Ω

2 Ω 3 Ω

4 Ω

2 Ωa

b

RTh

RTh

RTh

5 Ω

3 Ω

4 Ω

a

b

3 Ω

Ω

a

b

20

9

RTh = 3+209

=479

= 5.2 Ω.

Thevenin equivalent circuit:

5.2 Ωa

b

4 V

+_


Problem 3.73 Find the Norton equivalent circuit at terminals(a,b) for the circuitin Fig. P3.73.

0.2 Ω

0.2 Ω

0.1 Ω 0.25 Ω

a

b

I0

0.2I0

Iex

Vex

+_I1 I2 I3

+_


Solution: The circuit contains no independent sources. Hence,

VTh = 0.

To determineRTh, we add an external voltage sourceVex and proceed to findIex.

0.1I1 +0.2(I1− I2)−0.2I0 = 0

0.2I0 +0.2(I2− I1)+0.2I2 +0.25(I2− I3) = 0

0.25(I3− I2)+Vex = 0

Additionally, I0 = I1.Solution is:

I1 = −5Vex, I2 = −2.5Vex, I3 = −6.5Vex,

Iex = −I3 = 6.5Vex

RTh =Vex

Iex=

16.5

= 0.15 Ω

Hence,

a

b

0.15 Ω


Problem 3.81 What value of the load resistorRL will extract the maximum amountof power from the circuit in Fig. P3.81, and how much power will that be?

2 Ω

4 Ω 6 Ω

8 Ω

4 Ω

3 A RL

a

b


Solution: We start by obtaining the Thevenin equivalent circuit at terminals(a,b),as if RL were not there. We first findVoc:

V6−3+

V12

= 0

V = 12 V.

Hence,

2 Ω

4 Ω 6 Ω

8 Ω

4 Ω

3 A

a

bVoc

V

+

_

Voltage division gives:

VTh = Voc =

(

84+8

)

V =812

×12= 8 V.

Next, we suppress the current source to findRTh:

2 Ω

4 Ω 6 Ω

8 Ω

4 Ωa

b

RTh

Simplification leads to:RTh = 10.44 Ω.

Equivalent circuit:

RL

10.44 Ω

8 V

I

+_

For maximum power transfer toRL ,

RL = RTh = 10.44 Ω

I =8

2×10.44= 0.38 A

Pmax = I2RL = (0.38)2×10.44= 1.53 W.


Problem 3.82 For the circuit in Fig. P3.82, choose the value ofRL so that the powerdissipated in it is a maximum.

2 kΩ

6 kΩ

4 kΩ

8 kΩ

2 mA

RL

a

b


Solution: We need to find the Thevenin equivalent circuit at terminals(a,b), as ifRL

were not present.

2 kΩ

6 kΩ

4 kΩ

8 kΩ

2 mA

a

b

Voc

I1

I2

+

_

The current source will divide amongI1 andI2 such that

(4+2)I1 = (8+6)I2

Also, I1 + I2 = 2 mA.The solution yields:

I1 = 1.4 mA, I2 = 0.6 mA.

Voc = (−4I1 +8I2)×103

= −4×1.4+8×0.6 = −0.8 V.

To find RTh, we suppress the current source and simplify the circuit:

2 kΩ

6 kΩ

4 kΩ

8 kΩ

a

b

RTh = 8 kΩ || 12 kΩ = 4.8 kΩ

Hence,RL should be 4.8 kΩ for maximum power transfer to it.


Problem 3.83 Determine the maximum power that can be extracted by the loadresistor from the circuit in Fig. P3.83.

2000Ix

6 kΩ

3 kΩ

4 kΩ

RL

Ix

15 V

+_

+_


Solution: To find the Thevenin equivalent circuit, we start by determiningVTh = Voc.

2000Ix

6 kΩ

3 kΩ

4 kΩ

Ix

15 V

+_

+_

Voc

a

b

V1

+

_

Voltage division:

V1 =15

(3+6)k×6k = 10 V

Ix =V1

6k=

106

mA.

The dependent voltage source is:

2000Ix = 2×106×103×10−3

=206

V.

With (a,b) an open circuit, no current flows through the 4-kΩ resistor. Hence, thereis no voltage drop across it.

VTh = Voc = V1−2000Ix = 10−206

=406

= 6.67 V.

2000Ix

6 kΩ

3 kΩ

4 kΩ

15 V

+_

+_

Isc

a

b

I1 I2

Next, we findIsc:

−15+3kI1 +6k(I1− I2) = 0

6k(I2− I1)+4kI2 +2000Ix = 0

Also,Ix = I1− I2


Solution yields:

I1 = 2.5 mA, I2 = 1.25 mA.

Isc = I2 = 1.25 mA.

RTh =Voc

Isc=

6.671.25×10−3 = 5.33 kΩ.

Hence,RL = 5.33 kΩ extracts maximum power.

RL

RTh = 5.33 kΩ

6.67 V

I

+_ 5.33 kΩ

I =6.67

2×5.33= 0.625 mA

Pmax = I2RL = (0.625×10−3)2×5.33×103

= 2.09 (mW).


Problem 3.87 Determine the maximum power extractable from the circuit inFig. P3.87 by the load resistor RL.

200I0

1 kΩ

2 kΩ

2 kΩ

I0

RL

+_


Solution: The circuit has no independent sources. Hence,

VTh = 0.

Consequently, RL cannot extract any power from the circuit.


Problem 3.88 In the circuit of Fig. P3.88, what value ofRs would result inmaximum power transfer to the 10-Ω load resistor?

2 A RLRs 10 Ω


Solution: Maximum power transfer toRL occurs when all of the 2 A flowsthroughRL , requiringRs to be∞.


Figure P3.2: Circuit for Problem 3.2.zyang/Teaching/20162017Summer/Downloads/HW3-sol.pdfProblem 3.26...

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