Solucionario Circuitos Sadiku Cap 3

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Chapter 3, Solution 1

Using Fig. 3.50, design a problem to help other students to better understandnodal analysis.

R R1 2

Figure 3.50For Prob. 3.1 and Prob. 3.39.

Solution

Given R 1 = 4 k Ω , R 2 = 2 k Ω , and R 3 = 2 k Ω , determine the value of I x usingnodal analysis.

Let the node voltage in the top middle of the circuit be designated as V x .

[(V x –12)/4k] + [(V x –0)/2k] + [(V x –9)/2k] = 0 or (multiply this by 4 k)

(1+2+2)V x = 12+18 = 30 or V x = 30/5 = 6 volts and

Ix = 6/(2k) = 3 mA .

12 V+

I x

+9 V R 3




At node 1,

2

vv6

5

v

10

v 2111

60 = - 8v 1 + 5v 2 (1)

At node 2,

2vv

634

v 212

36 = - 2v 1 + 3v 2 (2)

Solving (1) and (2),

v1 = 0 V , v 2 = 12 V




Applying KCL to the upper node,

60

v20

30

v

20

v

10

v8 0oo0 = 0 or v 0 = –60 V

i1 =10v 0 –6 A , i2 =

20v 0 –3 A ,

i3 =30v 0 –2 A , i 4 =

60v 0 1 A .




At node 1,

–6 – 3 + v 1 /(20) + v 1/(10) = 0 or v 1 = 9(200/30) = 60 V

At node 2,3 – 2 + v 2 /(10) + v 2 /(5) = 0 or v 2 = –1(1600/80) = –20 V

i1 = v 1 /(20) = 3 A , i 2 = v 1 /(10) = 6 A ,i3 = v 2 /(40) = –500 mA , i 4 = v 2 /(40) = –500 mA .

i1

10 40 2 A20

3Av v1 2

i i

6 A

2 3 i4

40





Chapter 3, Solution 6.

Solve for V using nodal analysis.1

10

10

Figure 3.55For Prob. 3.6.

Step 1. The first thing to do is to select a reference node and to identify all the

unknown nodes. We select the bottom of the circuit as the reference node. The onlyunknown node is the one connecting all the resistors together and we will call that nodeV 1 . The other two nodes are at the top of each source. Relative to the reference, the oneat the top of the 10-volt source is –10 V. At the top of the 20-volt source is +20 V.

Step 2. Setup the nodal equation (there is only one since there is only oneunknown).

Step 3. Simplify and solve.

orV 1 = –2 V .

The answer can be checked by calculating all the currents and see if they add up to zero.The top two currents on the left flow right to left and are 0.8 A and 1.6 A respectively.The current flowing up through the 10-ohm resistor is 0.2 A. The current flowing right toleft through the 10-ohm resistor is 2.2 A. Summing all the currents flowing out of the

node, V 1 , we get, +0.8+1.6 –0.2–2.2 = 0. The answer checks.

+

V 1

–

1010 V–+

+5 20 V




0V2.020

0V10

0V2 x

xx

0.35V x = 2 or V x = 5.714 V .

Substituting into the original equation for a check we get,

0.5714 + 0.2857 + 1.1428 = 1.9999 checks!



Chapter 3, Solution 820i v i6 1 1 3

i1 + i 2 + i 3 = 0 020

v5v20

0)60v(10v 0111

But10

v5

2v so that 2v 1 + v 1 – 60 + v 1 – 2v 1 = 0

or v 1 = 60/2 = 30 V, therefore v o = 2v 1 /5 = 12 V .

5V 0

+V 0

– – 20

i2

4

60V

–

+




Let V 1 be the unknown node voltage to the right of the 250- Ω resistor. Let the groundreference be placed at the bottom of the 50- Ω resistor. This leads to the following nodalequation:

0I300V5V1572V3

getwegsimplifyin

0150

0I60V50

0V250

24V

b111

b111

But250

V24I 1 b . Substituting this into the nodal equation leads to

or V08.100V2.24 1 1 = 4.165 V.

Thus, I b = (24 – 4.165)/250 = 79.34 mA .




v 1 v 2 v 3

At node 1. [(v 1 –0)/8] + [(v 1 –v 3)/1] + 4 = 0At node 2. –4 + [(v 2 –0)/2] + 2i o = 0

At node 3. –2i o + [(v 3 –0)/4] + [(v 3 –v 1)/1] = 0

Finally, we need a constraint equation, i o = v 1 /8

This produces,1.125v 1 – v 3 = 4 (1)

0.25v1 + 0.5v

2 = 4 (2)

–1.25v 1 + 1.25v 3 = 0 or v 1 = v 3 (3)

Substituting (3) into (1) we get (1.125–1)v 1 = 4 or v 1 = 4/0.125 = 32 volts. This leads to,

io = 32/8 = 4 amps .




Find V o and the power absorbed by all the resistors in the circuit of Fig. 3.60.

12 6


Solution

At the top node, KCL produces 06

)24(V12

0V12

60V ooo

(1/3)V o = 1 or V o = 3 V .

P 12 = (3–60) 2/1 = 293.9 W (this is for the 12 Ω resistor in series with the 60 Vsource)

P 12Ω = (V o)2/12 = 9/12 = 750 mW (this is for the 12 Ω resistor connecting V o toground)

P 4Ω = (3–(–24)) 2/6 = 121.5 W .

+ _60 V

–+ 24 V

V o

12




There are two unknown nodes, as shown in the circuit below.

20

At node 1,

010

VV20

0V20

40V o111 or

(0.05+0.05+.1)V 1 – 0.1V o = 0.2V 1 – 0.1V o = 2 (1)

At node o,

010

0VI4

10VV o

x1o and I x = V 1/20

–0.1V 1 – 0.2V 1 + 0.2V o = –0.3V 1 + 0.2V o = 0 or (2)

V 1 = (2/3)V o (3)

Substituting (3) into (1),

0.2(2/3)V o – 0.1V o = 0.03333V o = 2 or

V o = 60 V .

10

+ _

V

20 10 40 V

V1 o

I x

4 I x




Calculate v 1 and v 2 in the circuit of Fig. 3.62 using nodal analysis.

10 V

15 A


Solution

At node number 2, [((v 2 + 10) – 0)/10] + [(v 2 –0)/4] – 15 = 0 or(0.1+0.25)v 2 = 0.35v 2 = –1+15 = 14 or

v 2 = 40 volts .

Next, I = [(v 2 + 10) – 0]/10 = (40 + 10)/10 = 5 amps and

v 1 = 8x5 = 40 volts .




Using nodal analysis, find v o in the circuit of Fig. 3.63.

12.5 A


Solution

At node 1,

[(v 1 –100)/1] + [(v 1 –v o)/2] + 12.5 = 0 or 3v 1 – v o = 200–25 = 175 (1)

(2)

dding 4x(1) to 3x(2) yields,

At node o,

[(v o –v 1)/2] – 12.5 + [(v o –0)/4] + [(v o+50)/8] = 0 or –4v 1 + 7v o = 50

A

+

v o

12.5 A

v 1 v 0

4

21 – 50 V

8

– +100 V

+

v o 4

2

8

1 –50 V

100 V –



(1) + 3(2) = –4v o + 21v o = 700 + 150 or 17v o = 850 or

v o = 50 V .

hecking, we get v 1 = (175+v o)/3 = 75 V.

At node 1,

75–100)/1] + [(75–50)/2] + 12.5 = –25 + 12.5 + 12.5 = 0!

At node o,

[(50–75)/2] + [(50–0)/4] + [(50+50)/8] – 12.5 = –12.5 + 12.5 + 12.5 – 12.5 = 0!

4

C

[(



Chapter 3, Solution 155 A

8

Nodes 1 and 2 form a supernode so that v 1 = v 2 + 10 (1)

At the supernode, 2 + 6v 1 + 5v 2 = 3 (v 3 - v 2) 2 + 6v 1 + 8v 2 = 3v 3 (2)

At node 3, 2 + 4 = 3 (v 3 - v 2) v 3 = v 2 + 2 (3)

Substituting (1) and (3) into (2),

2 + 6v 2 + 60 + 8v 2 = 3v 2 + 6 v 2 =1156

v 1 = v 2 + 10 =1154

i0 = 6v i = 29.45 A

P 65 =

61154

GvR v 2

21

21 144.6 W

P 55 =

51156

Gv2

22 129.6 W

P 35 = 12 W 3)2(Gvv 223L

–+40 V

–+20 V

0 v

v 1 1 2

4




At the supernode,

2 = v 1 + 2 (v 1 - v 3) + 8(v 2 – v 3) + 4v 2 , which leads to 2 = 3v 1 + 12v 2 - 10v 3 (1)But

v 1 = v 2 + 2v 0 and v 0 = v 2 .

Hencev 1 = 3v 2 (2)v 3 = 13V (3)

Substituting (2) and (3) with (1) gives,

v1 = 18.858 V , v

2 = 6.286 V , v

3 = 13 V

2 A

v 3v 2v

1

8 S

4 S1 S

i0

2 S

+ 13 V –+

v 0 –




At node 1,2

vv8v

4v60 2111

120 = 7v 1 - 4v 2 (1)

At node 2, 3i 0 + 02

vv

10

v60 212

But i 0 = .4

v60 1

Hence

0

2vv

10v60

4v603 2121 1020 = 5v 1 + 12v 2 (2)

Solving (1) and (2) gives v 1 = 53.08 V. Hence i 0 =4

v60 1 1.73 A

60 V

–+60 V

4

10

2

8

3i 0

i0

v 1

v 2




At node 2, in Fig. (a),2

vv

2

vv 3212

–15 = 0 or –0.5v 1 + v 2 – 0.5v 3 = 15 (1)

At the supernode,8v

4v

2vv

2vv 313212 = 0 and (v 1 /4) – 15 + (v 3/8) = 0 or

2v 1+v 3 = 120 (2)

From Fig. (b), – v 1 – 30 + v 3 = 0 or v 3 = v 1 + 30 (3)

Solving (1) to (3), we obtain,v 1 = 30 V , v 2 = 60 V = v 3

(b)

15A

v 3

84

22

v 2v 1

– +

30 V

+v 1

–

+v 3

–

(a)




At node 1,

32112131 4716

48235 V V V

V V V V V (1)

At node 2,

32132221 270

428V V V

V V V V V (2)

At node 3,

32132313 724360

428

123 V V V

V V V V V (3)

From (1) to (3),

B AV

V

V

V

36

0

16

724

271

417

3

2

1

Using MATLAB,

V267.12 V,933.4 V,10

267.12

933.4

10

3211 V V V B AV




Nodes 1 and 2 form a supernode; so do nodes 1 and 3. Hence

040414

321321 V V V

V V V (1)

.V 1 . V 2 2 V 3

4 1 4

Between nodes 1 and 3,12012 1331

V V V V (2)Similarly, between nodes 1 and 2,

(3)iV V 221

But . Combining this with (2) and (3) gives4/3V i

. (4)2/6 12 V V

Solving (1), (2), and (4) leads to

V15V,5.4 V,3 321 V V V




4 k

Let v 3 be the voltage between the 2k resistor and the voltage-controlled voltage source.At node 1,

2000vv

4000vv

10x3 31213

12 = 3v 1 - v 2 - 2v 3 (1)

At node 2,

1v

2vv

4vv 23121 3v 1 - 5v 2 - 2v 3 = 0 (2)

Note that v 0 = v 2 . We now apply KVL in Fig. (b)

- v 3 - 3v 2 + v 2 = 0 v 3 = - 2v 2 (3)

From (1) to (3),

v 1 = 1 V , v 2 = 3 V

(b)

+v 3

–

+v 2

–

3v 0

3 mA 1 k

v 32 k

3v 0v v1 2

+v 0

–

(a)




At node 1,8

vv3

4v

2v12 0110

24 = 7v 1 - v 2 (1)

At node 2, 3 +1

v5v8

vv 2221

But, v 1 = 12 - v 1

Hence, 24 + v 1 - v 2 = 8 (v 2 + 60 + 5v 1) = 4 V

456 = 41v 1 - 9v 2 (2)

Solving (1) and (2),

v 1 = - 10.91 V , v 2 = - 100.36 V




We apply nodal analysis to the circuit shown below.

At node o,

30V25.0V25.104

)VV2(V2

0V1

30V1o

1oooo (1)

At node 1,

48V4V50316

0V4

V)VV2(o1

1o1o (2)

From (1), V 1 = 5V o – 120. Substituting this into (2) yields29V o = 648 or V o = 22.34 V .

1

2

2 V o

+ –

3 A

+ _30 V

4 V Vo 1

+

V 16 o

_




Consider the circuit below.

8

_

4V125.0V125.108

VV4

10V

41411 (1)

4V25.0V75.004

VV2

0V4 32

322 (2)

2V75.0V25.0022

0V4

VV32

323 (3)

2V125.1V125.001

0V8

VV2 41

414 (4)

2

2

4

4

V

125.100125.0

075.025.00

025.075.00

125.000125.1

Now we can use MATLAB to solve for the unknown node voltages.

>> Y=[1.125,0,0,-0.125;0,0.75,-0.25,0;0,-0.25,0.75,0;-0.125,0,0,1.125]

Y =

1

4

2 1

+ V o

4 A 2 A

V V2 3V V1 4

2



1.1250 0 0 -0.12500 0.7500 -0.2500 00 -0.2500 0.7500 0

-0.1250 0 0 1.1250

>> I=[4,-4,-2,2]'

I =

4-4-22

>> V=inv(Y)*I

V =3.8000

-7.0000-5.00002.2000

V o = V 1 – V 4 = 3.8 – 2.2 = 1.6 V .




Consider the circuit shown below.

20

At node 1.1 2 1 4

1 24 80 2

1 20

V V V V V V

41 20 V (1)

At node 2,2 31 2 2

1 20 80 98 8

1 8 10

V V V V V V V 3

V (2)

At node 3,2 3 3 3 4

2 3 40 2 5 2

10 20 10

V V V V V V V V

(3)

At node 4,3 41 4 4

1 30 3 6 11

20 10 30

V V V V V V V V 4 (4)

Putting (1) to (4) in matrix form gives:

1

2

3

4

80 21 20 0 1

0 80 98 8 0

0 0 2 5 2

0 3 0 6 11

V

V

V

V

B = A V V = A -1 B

Using MATLAB leads to

V 1 = 25.52 V , V 2 = 22.05 V , V 3 = 14.842 V , V 4 = 15.055 V

10

4

1

8

2 1 3

4

10

30

20




At node 1,

32121311 24745

5103

2015 V V V V V V V V (1)

At node 2,

554

532221

V V V I V V o

(2)

But10

31 V V

I o . Hence, (2) becomes

321 31570 V V V (3)At node 3,

321

32331 V11V6V37005

VV

15

V10

10

VV3 (4)

Putting (1), (3), and (4) in matrix form produces

BAV

70

0

45

V

V

V

1163

3157

247

3

2

1

Using MATLAB leads to

89.2

78.2

19.7

BAV 1

Thus,V 1 = –7.19V ; V 2 = –2.78V ; V 3 = 2.89V .




At node 1,

2 = 2v 1 + v 1 – v 2 + (v 1 – v 3)4 + 3i 0 , i 0 = 4v 2 . Hence,

2 – v (2)

2

2 = 7v 1 + 11v 2 – 4v 3 (1)At node 2,

v 1 – v 2 = 4v 2 + v 2 – v 3 0 = – v 1 + 6v 3

At node 3,v3 = 4 + v 2 – v 3 + 12v 2 + 4(v 1 – v 3)

or – 4 = 4v 1 + 13v 2 – 7v 3 (3)

In matrix form,

4

0

2

v

v

v

7134

161

4117

3

2

1

,1767134

161

4117

1107134

160

4112

1

,66

744

101

427

2 286

4134

061

2117

3

v 1 = ,V625.01761101 v 2 = V375.0

176662

v 3 = .V625.11762863

v 1 = 625 mV , v 2 = 375 mV , v 3 = 1.625 V .




At node c,

d cbcbccd V V V

V V V V V 21150

5410 (1)

At node b,

c ba b bc ba V2V4V90

8V

4VV

8V90V (2)

At node a,

d ba baad a V4V2V7600

8V90V

16V

4V60V (3)

At node d,

d cacd d d a V8V2V5300

10VV

20V

4V60V (4)

In matrix form, (1) to (4) become

BAV

300

60

90

0

V

V

V

V

8205

4027

0241

21150

d

c

b

a

We use MATLAB to invert A and obtain

75.43389.1

56.20

56.10

BAV 1

Thus,V a = –10.56 V ; V b = 20.56 V ; V c = 1.389 V ; VC d = –43.75 V .




At node 1,

42121141 45025 V V V V V V V V (1)At node 2,

32132221 4700)(42 V V V V V V V V (2)At node 3,

4324332 546)(46 V V V V V V V (3)At node 4,

43144143 5232 V V V V V V V V (4)In matrix form, (1) to (4) become

B AV

V

V

V

V

2

6

0

5

5101

1540

0471

1014

4

3

2

1

Using MATLAB,

7076.0

309.2

209.1

7708.0

1 B AV

i.e.V7076.0 V,309.2 V,209.1 V,7708.0 4321

V V V V




At node 1,

[(v 1 –80)/10]+[(v 1 –4v o)/20]+[(v 1 –(v o –96))/40] = 0 or(0.1+0.05+0.025)v 1 – (0.2+0.025)v o =0.175v 1 – 0.225v o = 8–2.4 = 5.6 (1)

At node 2,

–2 i o + [((v o –96)–v 1)/40] + [(v o –0)/80] = 0 and i o = [(v 1 –(v o –96))/40] –2[(v 1 –(v o –96))/40] + [((v o –96)–v 1)/40] + [(v o –0)/80] = 0 –3[(v 1 –(v o –96))/40] + [(v o –0)/80] = 0 or –0.0.075v 1 + (0.075+0.0125)v o = 7.2 = –0.075v

1 + 0.0875v

o = 7.2 (2)

Using (1) and (2) we get,

2.7

6.5

0015625.0175.0075.0

225.00875.0

2.7

6.5

016875.00153125.0175.0075.0

225.00875.0

2.7

6.5

0875.0075.0

225.0175.0

1

1

o

o

v

v

or v

v

v1 = –313.6–1036.8 = –1350.4

vo = –268.8–806.4 = –1.0752 kV

and i o = [(v 1 –(v o –96))/40] = [(–1350.4–(–1075.2–96))/40] = –4.48 amps .

v 1

10

20

80

40

1

v 0

i 0

2i 0

– +

2

– + +

–

96 V

4v80 V 0




1

At the supernode,

1 + 2v 0 =1

vv

1

v

4

v 3121

(1)

But v o = v 1 – v 3 . Hence (1) becomes,

4 = -3v 1 + 4v 2 +4v 3 (2)

At node 3,

2v o +2

v10vv

4v 3

313

or 20 = 4v 1 + 0v 2 – v 3 (3)

At the supernode, v 2 = v 1 + 4i o . But i o =4

v 3 . Hence,

v 2 = v 1 + v 3 (4)

Solving (2) to (4) leads to,

v 1 = 4.97V , v 2 = 4.85V , v 3 = –0.12V .

i0

4

v 2v 1

1 A – +10 V

+ v – 0

4

22v v0 3

1




v 35 k 10 V 20 V

We have a supernode as shown in figure (a). It is evident that v 2 = 12 V, Applying KVLto loops 1and 2 in figure (b), we obtain,

-v 1 – 10 + 12 = 0 or v 1 = 2 and -12 + 20 + v 3 = 0 or v 3 = -8 V

Thus, v 1 = 2 V , v 2 = 12 V , v 3 = -8V .

(b)

v 1v 2

+ – – +

(a)

4 mA

10 k

+v 1

+

–

v 3 –

– +12 V

loop 1loop 2




(a) This is a planar circuit. It can be redrawn as shown below.

5

13

4

(b) This is a planar circuit. It can be redrawn as shown below.

– +12 V

5

4

3

2

1

2 A

2

6




(a) This is a planar circuit because it can be redrawn as shown below,

7

231

6

(b) This is a non-planar circuit.

510 V – +

4




Assume that i 1 and i 2 are in mA. We apply mesh analysis. For mesh 1,

-30 + 20 + 7i 1 – 5i 2 = 0 or 7i 1 – 5i 2 = 10 (1)For mesh 2,

-20 + 9i 2 – 5i 1 = 0 or -5i 1 + 9i 2 = 20 (2)

Solving (1) and (2), we obtain, i 2 = 5.

v0 = 4i 2 = 20 volts .

5 k

i1

i2+v 0

–

– +30 V 20 V – +

4 k

2 k




10 V 4

+ – i

Applying mesh analysis gives,

10I 1 – 6I 2 = 12 and –6I 1 + 8I 2 = –10

or5

611

53

34

IIor

56

II

4335

56

2

1

2

1

I1 = (24–15)/11 = 0.8182 and I 2 = (18–25)/11 = –0.6364

i1 = –I 1 = –818.2 mA ; i 2 = I 1 – I 2 = 0.8182+0.6364 = 1.4546 A ; andi3 = I 2 = –636.4 mA .

I 1 I 2

– +12 V 6 2

1 i i2 3




Applying mesh analysis to loops 1 and 2, we get,

30i 1 – 20i 2 + 60 = 0 which leads to i 2 = 1.5i 1 + 3 (1)

–20i 1 + 40i 2 – 60 + 5 v 0 = 0 (2)But, v 0 = –4i 1 (3)

Using (1), (2), and (3) we get –20i 1 + 60i 1 + 120 – 60 – 20i 1 = 0 or

20i 1 = –60 or i 1 = –3 amps and i 2 = 7.5 amps.

Therefore, we get,v 0 = –4i 1 = 12 volts .

20

i1

i2+v 0

– + –

– + 60 V

6

5v 0 4

20




Consider the circuit below with the mesh currents.

4 3

+ _

22.5V + _

Io

1

2 2

10 A

I4I31 60 V

I2I11

4

5 A

I1 = –5 A (1)

1(I 2 –I 1) + 2(I 2 –I 4) + 22.5 + 4I 2 = 07I 2 – I 4 = –27.5 (2)

–60 + 4I 3 + 3I 4 + 1I 4 + 2(I 4 –I 2) + 2(I 3 – I 1) = 0 (super mesh) –2I 2 + 6 I 3 + 6I 4 = +60 – 10 = 50 (3)

But, we need one more equation, so we use the constraint equation –I 3 + I 4 = 10. Thisnow gives us three equations with three unknowns.

1050

5.27

II

I

110662

107

4

3

2

We can now use MATLAB to solve the problem.

>> Z=[7,0,-1;-2,6,6;0,-1,0]



Z =

7 0 -1-2 6 60 -1 0

>> V=[–27.5,50,10]'

V = –27.5

5010

>> I=inv(Z)*V

I = –1.3750 –10.0000

17.8750Io = I 1 – I 2 = –5 – 1.375 = –6.375 A .

Check using the super mesh (equation (3)):

–2I 2 + 6 I 3 + 6I 4 = 2.75 – 60 + 107.25 = 50!






Assume all currents are in mA and apply mesh analysis for mesh 1.

–56 + 12i 1 – 6i 2 – 4i 3 = 0 or 6i 1 – 3i 2 – 2i 3 = 28 (1)

for mesh 2, –6i 1 + 14i 2 – 2i 3 = 0 or –3i 1 + 7i 2 – i 3 = 0 (2)

for mesh 3,

–4i 1 – 2i 2 + 10i 3 = 0 or –2i 1 – i 2 + 5i 3 = 0 (3)

Solving (1), (2), and (3) using MATLAB, we obtain,

i o = i 1 = 8 mA .

4 k

– 56V i1

i3

2 k

6 k

4 k

6 k

i 2

2 k




10

i 1 6 V

For loop 1,6 = 12i 1 – 2i 2 3 = 6i 1 – i 2 (1)

For loop 2,-8 = – 2i 1 +7i 2 – i 3 (2)

For loop 3,

-8 + 6 + 6i 3 – i 2 = 0 2 = – i 2 + 6i 3 (3)

We put (1), (2), and (3) in matrix form,

2

8

3

i

i

i

610

172

016

3

2

1

,234610172

016

240620182

036

2

5

i3

i2

i3

+ –

– + 8 V

2

1

4

ii2

0



38

210

872

316

3

At node 0, i + i 2 = i 3 or i = i 3 – i 2 =234

2403823 = 1.188 A




Although there are many ways to work this problem, this is an example based on the same kindof problem asked in the third edition.

Problem

Determine the mesh currents in the circuit of Fig. 3.88.

Figure 3.88

Solution

For mesh 1,(1)2121 3050120305012 I I I I

For mesh 2,(2)321312 40100308040301008 I I I I I I

For mesh 3,

(3)3223 50406040506 I I I I Putting eqs. (1) to (3) in matrix form, we get

B AI

I

I

I

6

8

12

50400

4010030

03050

3

2

1

Using Matlab,

44.0

40.0

48.01 B A I

i.e. I 1 = 480 mA , I 2 = 400 mA , I 3 = 440 mA




20

For loop 1,

80 = 70i 1 – 20i 2 – 30i 3 8 = 7i 1 – 2i 2 – 3i 3 (1)

For loop 2,

80 = 70i 2 – 20i 1 – 30i 3 8 = -2i 1 + 7i 2 – 3i 3 (2)

For loop 3,

0 = -30i1 – 30i

2 + 90i

3 0 = i

1 + i

2 – 3i

3

(3)

olving (1) to (3), we obtain i 3 = 16/9

Io = i 3 = 16/9 = 1.7778 A

V ab = 30i 3 = 53.33 V .

S

a

i1

i2

i3

– +80 V

– 3080 V +

V ab 3020

– 30

b20




90 V

+

Loop 1 and 2 form a supermesh. For the supermesh,

6i 1 + 4i 2 – 5i 3 + 180 = 0 (1)

For loop 3, –i 1 – 4i 2 + 7i 3 + 90 = 0 (2)

Also, i 2 = 45 + i 1 (3)

Solving (1) to (3), i 1 = –46, i 3 = –20; i o = i 1 – i 3 = –26 A

– 180V

45 A

i1

i2i3 4

5

2

1

ii1 2




4 8

i

For loop 1, 30 = 5i 1 – 3i 2 – 2i 3 (1)

For loop 2, 10i 2 - 3i 1 – 6i 4 = 0 (2)

For the supermesh, 6i 3 + 14i 4 – 2i 1 – 6i 2 = 0 (3)

But i 4 – i 3 = 4 which leads to i 4 = i 3 + 4 (4)

Solving (1) to (4) by elimination gives i = i 1 = 8.561 A .

1i1 i2

3 i4

–

62

30V

3




For loop 1,12811081112)(8312 2121211

iiiiiii (1)For loop 2,

0214826)(8 21212 oo viiviii But ,13iv

o

21121 706148 iiiii (2)Substituting (2) into (1),

1739.012877 222 iii A and 217.17 21

ii A




First, transform the current sources as shown below.

- 6V +

2

I3 V 1 8 V 2 4 V 3

4 8

I1 2 I 2 + +

20V 12V- -

For mesh 1,

321321 47100821420 I I I I I I (1)For mesh 2,

321312 2760421412 I I I I I I (2)For mesh 3,

321123 7243084146 I I I I I I (3)Putting (1) to (3) in matrix form, we obtain

B AI

I

I

I

3

6

10

724

271

417

3

2

1

Using MATLAB,

8667.1,0333.0 ,5.2

8667.1

0333.0

2

3211 I I I B A I

But



111 420

420

I V V

I 10 V

)(2 212 I I V 4.933 V

Also,

23

32 812

8

12 I V

V I 12.267 V .




We apply mesh analysis and let the mesh currents be in mA. 3k

I4

4k 2k 5k

Io

1k I 3 -I1 I 2 3 V

+ +6 V + 10k

- 4 V-

For mesh 1,

421421 I4II520I4II586 (1)For mesh 2,

43214312 I2I10I13I40I2I10II134 (2)For mesh 3,

432423 I5I15I1030I5I10I153 (3)

For mesh 4, 014524 4321 I I I I (4)

Putting (1) to (4) in matrix form gives

BAI

0

3

4

2

I

I

I

I

14524

515100

210131

4015

4

3

2

1

Using MATLAB,

3

896.3

044.4

608.3

BAI 1 0.148

The current through the 10k resistor is I o= I 2 – I 3 = 148 mA .




3

i3

For the supermesh in figure (a),

3i 1 + 2i 2 – 3i 3 + 27 = 0 (1)

At node 0, i 2 – i 1 = 2i 0 and i 0 = –i 1 which leads to i 2 = –i 1 (2)

For loop 3, –i 1 –2i 2 + 6i 3 = 0 which leads to 6i 3 = –i 1 (3)

Solving (1) to (3), i 1 = (–54/3)A, i 2 = (54/3)A, i 3 = (27/9)A

i0 = –i 1 = 18 A , from fig. (b), v 0 = i 3 –3i 1 = (27/9) + 54 = 57 V .

–+

1

i1

+v 0 or

+v 0

– i2

227V

2

(b)

i1

2 i1 i2

21

27 V –+

2i 0

i20

(a)








For mesh 1,

2(i 1 – i 2) + 4(i 1 – i 3) – 12 = 0 which leads to 3i 1 – i 2 – 2i 3 = 6 (1)

(2)

or the independent current source, i 3 = 3 + i 2 (3)

olving (1), (2), and (3), we obtain,

i1 = 3.5 A , i 2 = -0.5 A , i 3 = 2.5 A .

For the supermesh, 2(i 2 – i 1) + 8i 2 + 2v 0 + 4(i 3 – i 1) = 0 But v 0 = 2(i 1 – i 2) which leads to -i 1 + 3i 2 + 2i 3 = 0

F S

– +V S

–2V 0

+iv

0 2

– 3A

i1

i3

8

4

2i2

i3




Applying mesh analysis leads to; –12 + 4kI 1 – 3kI 2 – 1kI 3 = 0 (1)

–3kI 1 + 7kI 2 – 4kI 4 = 0

–3kI 1 + 7kI 2 = –12 (2) –1kI 1 + 15kI 3 – 8kI 4 – 6kI 5 = 0 –1kI 1 + 15kI 3 – 6k = –24 (3)

I4 = –3mA (4) –6kI 3 – 8kI 4 + 16kI 5 = 0 –6kI 3 + 16kI 5 = –24 (5)

Putting these in matrix form (having substituted I 4 = 3mA in the above),

24

24

12

12

I

I

I

I

k

16600

61501

0073

0134

5

3

2

1

ZI = V

Using MATLAB,

>> Z = [4,-3,-1,0;-3,7,0,0;-1,0,15,-6;0,0,-6,16]

Z =

4 -3 -1 0-3 7 0 0-1 0 15 -60 0 -6 16

>> V = [12,-12,-24,-24]'

V =

12

-12-24-24

We obtain,

>> I = inv(Z)*V



I =1.6196 mA

–1.0202 mA–2.461 mA

3 mA

–2.423 mA




Let the mesh currents be in mA. For mesh 1,

2121 22021012 I I I I (1)

For mesh 2,(2)321312 3100310 I I I I I I

For mesh 3,

3223 2120212 I I I I (3)Putting (1) to (3) in matrix form leads to

B AI

I

I

I

12

10

2

210

131

012

3

2

1

Using MATLAB,

mA25.10,mA5.8 ,mA25.5

25.10

5.8

25.5

3211 I I I B A I

I 1 = 5.25 mA , I 2 = 8.5 mA , and I 3 = 10.25 mA .



Chapter 3, Solution 55Chapter 3, Solution 5510 V

It is evident that I 1 = 4 (1)It is evident that I For mesh 4, 12(I 4 – I 1) + 4(I 4 – I 3) – 8 = 0For mesh 4, 12(I (2)(2)

For the supermesh 6(I 2 – I 1) + 10 + 2I 3 + 4(I 3 – I 4) = 0For the supermesh 6(I or -3I 1 + 3I 2 + 3I 3 – 2I 4 = -5 (3)or -3I At node c, I 2 = I 3 + 1 (4)At node c, I Solving (1), (2), (3), and (4) yields, I 1 = 4A, I 2 = 3A, I 3 = 2A, and I 4 = 4ASolving (1), (2), (3), and (4) yields, I

At node b, i 1 = I 2 – I 1 = -1A At node b, i

At node a, i 2 = 4 – I 4 = 0A At node a, i

At node 0, i 3 = I 4 – I 3 = 2A At node 0, i

ddI 1I

1 = 4 (1)

4 – I 1) + 4(I 4 – I 3) – 8 = 0

2 – I 1) + 10 + 2I 3 + 4(I 3 – I 4) = 01 + 3I 2 + 3I 3 – 2I 4 = -5 (3)

2 = I 3 + 1 (4)

1 = 4A, I 2 = 3A, I 3 = 2A, and I 4 = 4A

1 = I 2 – I 1 = -1A

2 = 4 – I 4 = 0A

3 = I 4 – I 3 = 2A

1

I

I 3

I 4

4 A

+ –

+I 2b c

1Ai1

61A

2i2

i3

I 34A 412

a 0I 4

8 V



Chapter 3, Solution 56+ v – 1

For loop 1, 12 = 4i 1 – 2i 2 – 2i 3 which leads to 6 = 2i 1 – i 2 – i 3 (1)

For loop 2, 0 = 6i 2 –2i 1 – 2 i 3 which leads to 0 = -i 1 + 3i 2 – i 3 (2)For loop 3, 0 = 6i 3 – 2i 1 – 2i 2 which leads to 0 = -i 1 – i 2 + 3i 3 (3)

In matrix form (1), (2), and (3) become,

0

0

6

i

i

i

311

131

112

3

2

1

= ,8

311

131112

2 = 24

301

131162

3 = 24

011

031

612

, therefore i 2 = i 3 = 24/8 = 3A,

v1 = 2i 2 = 6 volts , v = 2i 3 = 6 volts

– +12 V

i1

i2

i3

2

22

+2 2v 2

–




Assume R is in kilo-ohms.V306090V90V,V60mA15xk 4V 212

Current through R is

R )15(R 3

330R iViR 3

3i R 1,oR

This leads to R = 90/15 = 6 k Ω .




i2

For loop 1, 120 + 40i 1 – 10i 2 = 0, which leads to -12 = 4i 1 – i 2 (1)

For loop 2, 50i 2 – 10i 1 – 10i 3 = 0, which leads to -i 1 + 5i 2 – i 3 = 0 (2)

For loop 3, -120 – 10i 2 + 40i 3 = 0, which leads to 12 = -i 2 + 4i 3 (3)

Solving (1), (2), and (3), we get, i 1 = -3A , i 2 = 0, and i 3 = 3A

– +

120 V

30

i1 i3

10 1030




For loop 1, -80 + 30i 1 – 20i 2 + 4v 0 = 0, where v 0 = 80i 3

(3)

From (1), (2), and (3),

=

or 4 = 1.5i 1 – i 2 + 16i 3 (1)

For the supermesh, 60i 2 – 20i 1 – 96 + 80i 3 – 4 v 0 = 0, where v 0 = 80i 3

or 4.8 = -i 1 + 3i 2 – 12i 3 (2)

Also, 2I 0 = i 3 – i 2 and I 0 = i 2 , hence, 3i 2 = i 3

130

12313223

0

8.48

3

2

1

i

i

i

,5

130

1231

3223

2 = ,4.22

100

128.41

3283

3 = 2.67

030

8.431

823

I0 = i 2 = 2 / = -28/5 = –4.48 A

v 0 = 8i 3 = (-84/5)80 = –1.0752 kvolts

i1

i2

i3

40– +

96 V

4v 0 – –

+80V

+v 0

– 80

20

10

I

i

0

2I 0

i2 3




0.5 i0

At node 1, [(v 1 –0)/1] + [(v 1 –56)/4] + 0.5[(v 1 –0)/1] = 0 or 1.75v 1 = 14 or v 1 = 8 VAt node 2, [(v 2 –56)/8] – 0.5[8/1] + [(v 2 –0)/2] = 0 or 0.625v 2 = 11 or v 2 = 17.6 V

P1 = (v 1)2/1 = 64 watts , P 2 = (v 2)2/2 = 154.88 watts ,

P4 = (56 – v 1)2/4 = 576 watts , P 8 = (56 – v 2)2/8 = 1.84.32 watts .

v 2

– +56 V

856 V4v 1

1 2

i0




20 10

At node 1, i s = (v 1/30) + ((v 1 – v 2)/20) which leads to 60i s = 5v 1 – 3v 2

s

(1)

But v 2 = -5v 0 and v 0 = v 1 which leads to v 2 = -5v 1 Hence, 60i s = 5v 1 + 15v 1 = 20v 1 which leads to v 1 = 3i s, v 2 = -15i

i0 = v 2 /50 = -15i s/50 which leads to i 0 /i s = -15/50 = –0.3

+v 0

–

v 2

–

v 1

i0

i s 30 40

5v 0




4 k

We have a supermesh. Let all R be in k , i in mA, and v in volts.

For the supermesh, -100 +4i 1 + 8i 2 + 2i 3 + 40 = 0 or 30 = 2i 1 + 4i 2 + i 3 (1)

At node A, i 1 + 4 = i 2 (2)

At node B, i 2 = 2i 1 + i 3 (3)

Solving (1), (2), and (3), we get i 1 = 2 mA , i 2 = 6 mA , and i 3 = 2 mA .

i1 i2 i3 – +100V

B8 k 2 k

–

A

40 V




10 A

For the supermesh, -50 + 10i 1 + 5i 2 + 4i x = 0, but i x = i 1 . Hence,

2 (2)

50 = 14i 1 + 5i 2 (1)

At node A, i 1 + 3 + (v x /4) = i 2 , but v x = 2(i 1 – i 2), hence, i 1 + 2 = i Solving (1) and (2) gives i 1 = 2.105 A and i 2 = 4.105 A

v x = 2(i 1 – i 2) = –4 volts and i x = i 2 – 2 = 2.105 amp

– +50 V

4i x–

i1 i2

5




i

For mesh 2, 20i 2 – 10i 1 + 4i 0 = 0 (1)

But at node A, i o = i 1 – i 2 so that (1) becomes i 1 = (16/6)i 2 (2)

or the supermesh, –250 + 50i 1 + 10(i 1 – i 2) – 4i 0 + 40i 3 = 0

r 28i 1 – 3i 2 + 20i 3 = 125

e B,

becomes i 3 = 5 + (2/3)i 2 (5)

Solving (1) to

v 0 = 10i 2 = 2.941 volts , i 0 = i 1 – i 2 = (5/3)i 2 = 490.2mA .

F o (3)

At nod i 3 + 0.2v 0 = 2 + i 1 (4)

But, v 0 = 10i 2 so that (4)

(5), i 2 = 0.2941 A,

40i1

i2

i3

– 250V

–4i 0

50 iA 10

0.2V 0

1 2

10

5 A

B

i0 + v 0

i i1 3




For mesh 1, –12 + 12I 1 – 6I 2 – I 4 = 0 or

421 61212 I I I (1)For mesh 2,

–6I 1 + 16I 2 – 8I 3 – I 4 – I 5 = 0 (2)For mesh 3,

–8I 2 + 15I 3 – I 5 – 9 = 0 or9 = –8I 2 + 15I 3 – I 5 (3)

For mesh 4, –I 1 – I 2 + 7I 4 – 2I 5 – 6 = 0 or

(4)6 = –I 1 – I 2 + 7I 4 – 2I 5 For mesh 5,

–I 2 – I 3 – 2I 4 + 8I 5 – 10 = 0 or 5432 8210 I I I I (5)

Casting (1) to (5) in matrix form gives

BAI

10

6

9

0

12

I

I

I

I

I

82110

27011

101580

118166

010612

5

4

3

2

1

Using MATLAB we input:

Z=[12,-6,0,-1,0;-6,16,-8,-1,-1;0,-8,15,0,-1;-1,-1,0,7,-2;0,-1,-1,-2,8]and V=[12;0;9;6;10]

This leads to>> Z=[12,-6,0,-1,0;-6,16,-8,-1,-1;0,-8,15,0,-1;-1,-1,0,7,-2;0,-1,-1,-2,8]

Z =

12 -6 0 -1 0-6 16 -8 -1 -10 -8 15 0 -1

-1 -1 0 7 -20 -1 -1 -2 8

>> V=[12;0;9;6;10]

V =

12



096

10

>> I=inv(Z)*V

I =

2.17011.99121.81192.09422.2489

Thus,

I = [2.17, 1.9912, 1.8119, 2.094, 2.249] A .




The mesh equations are obtained as follows.

1 2 3 412 24 30 4 6 2 0I I I I

or30I 1 – 4I 2 – 6I 3 – 2I 4 = –12 (1)

1 2 4 524 40 4 30 2 6 0I I I I

or –4I 1 + 30I 2 – 2I 4 – 6I 5 = –16 (2)

–6I 1 + 18I 3 – 4I 4 = 30 (3)

–2I 1 – 2I 2 – 4I 3 + 12I 4 –4I 5 = 0 (4)

–6I 2 – 4I 4 + 18I 5 = –32 (5)

Putting (1) to (5) in matrix form

32

0

30

16

12

I

184060

412422

041806

620304

026430

ZI = V

Using MATLAB,

>> Z = [30,-4,-6,-2,0;-4,30,0,-2,-6;-6,0,18,-4,0;-2,-2,-4,12,-4;

0,-6,0,-4,18]Z =

30 -4 -6 -2 0-4 30 0 -2 -6-6 0 18 -4 0-2 -2 -4 12 -4



0 -6 0 -4 18

>> V = [-12,-16,30,0,-32]'

V =

-12-16300

-32

>> I = inv(Z)*V

I =

-0.2779 A-1.0488 A1.4682 A-0.4761 A-2.2332 A





V 3

+ V o -

15

0

V35

V

5.05.00

5.095.025.0

025.035.0 o

Since we actually have four unknowns and only three equations, we need a constraintequation.

Vo = V

2 – V

3

Substituting this back into the matrix equation, the first equation becomes,

0.35V 1 – 3.25V 2 + 3V 3 = –5

This now results in the following matrix equation,

V 2

10 5 10 A 3 V o

V 1 2

5 A

4



15

0

5

V

5.05.00

5.095.025.0

325.335.0

Now we can use MATLAB to solve for V.

>> Y=[0.35,-3.25,3;-0.25,0.95,-0.5;0,-0.5,0.5]

Y =

0.3500 -3.2500 3.0000-0.2500 0.9500 -0.5000

0 -0.5000 0.5000

>> I=[–5,0,15]'

I =

–5015

>> V=inv(Y)*I

V = –410.5262 –194.7368 –164.7368

V o = V 2 – V 3 = –77.89 + 65.89 = –30 V .

Let us now do a quick check at node 1.

–3(–30) + 0.1(–410.5) + 0.25(–410.5+194.74) + 5 =90–41.05–102.62+48.68+5 = 0.01; essentially zero considering theaccuracy we are using. The answer checks.




Although there are many ways to work this problem, this is an example based on thesame kind of problem asked in the third edition.

Problem

Find the voltage V o in the circuit of Fig. 3.112.

3 A

40 20

_

+

V o

25 10

+ _4A 24 V


Solution

Consider the circuit below. There are two non-reference nodes.

3 A

V 1 V o

40 20

_

+

V o

25 10

+ _4 A 24 V



04.2

7

25/243

34V

19.01.0

1.0125.0

Using MATLAB, we get,

>> Y=[0.125,-0.1;-0.1,0.19]

Y =

0.1250 -0.1000-0.1000 0.1900

>> I=[7,-2.04]'

I =

7.0000-2.0400

>> V=inv(Y)*I

V =

81.890932.3636

Thus, V o = 32.36 V .We can perform a simple check at node V o ,

3 + 0.1(32.36–81.89) + 0.05(32.36) + 0.04(32.36–24) =3 – 4.953 + 1.618 + 0.3344 = – 0.0004; answer checks!






7I4

20I4V

50

03

x

x

With two equations and three unknowns, we need a constraint equation,

Ix = 2V 1 , thus the matrix equation becomes,

7

20V

58

05

This results in V 1 = 20/(–5) = –4 V andV 2 = [–8(–4) – 7]/5 = [32 – 7]/5 = 5 V .




015

30

I915174

549

We can now use MATLAB solve for our currents.

>> R=[9,–4,–5;–4,7,–1;–5,–1,9]

R =

9 –4 –5 –4 7 –1 –5 –1 9

>> V=[30,–15,0]'

V =

30 –15

0

>> I=inv(R)*V

I =

6.255 A1.9599 A3.694 A




R 11 = 5 + 2 = 7, R 22 = 2 + 4 = 6, R 33 = 1 + 4 = 5, R 44 = 1 + 4 = 5,R 12 = -2, R 13 = 0 = R 14 , R 21 = -2, R 23 = -4, R 24 = 0, R 31 = 0,R 32 = -4, R 34 = -1, R 41 = 0 = R 42 , R 43 = -1, we note that R ij = R ji for

all i not equal to j.

v1 = 8, v 2 = 4, v 3 = -10, and v 4 = -4

Hence the mesh-current equations are:

4

10

4

8

5100

1540

0462

0027

4

3

2

1

i

i

i

i




R 11 = 2 + 3 +4 = 9, R 22 = 3 + 5 = 8, R 33 = 1+1 + 4 = 6, R 44 = 1 + 1 = 2,R 12 = -3, R 13 = -4, R 14 = 0, R 23 = 0, R 24 = 0, R 34 = -1

v 1 = 6, v 2 = 4, v 3 = 2, and v 4 = -3 Hence,

3

2

4

6

2100

1604

0083

0439

4

3

2

1

i

i

i

i




R 11 = R 1 + R 4 + R 6 , R 22 = R 2 + R 4 + R 5 , R 33 = R 6 + R 7 + R 8 ,,R 44 = R 3 + R 5 + R 8 , R 12 = -R 4 , R 13 = -R 6 , R 14 = 0, R 23 = 0

R 24 = -R 5 , R 34 = -R 8 , again, we note that R ij = R ji for all i not equal to j.

The input voltage vector is =

4

3

2

1

V

V

V

V

4

3

2

1

4

3

2

1

85385

88766

55424

64641

0

0

0

0R

V

V

V

V

i

i

i

i

R R R R R

R R R R R

R R R R R

R R R R




* Schematics Netlist *

R_R4 $N_0002 $N_0001 30R_R2 $N_0001 $N_0003 10

R_R1 $N_0005 $N_0004 30R_R3 $N_0003 $N_0004 10R_R5 $N_0006 $N_0004 30V_V4 $N_0003 0 120Vv_V3 $N_0005 $N_0001 0v_V2 0 $N_0006 0v_V1 0 $N_0002 0

i3

i2i1

Clearly, i 1 = –3 amps , i 2 = 0 amps , and i 3 = 3 amps , which agrees with the answers inProblem 3.44.





I_I2 0 $N_0001 DC 4AR_R1 $N_0002 $N_0001 0.25

R_R3 $N_0003 $N_0001 1R_R2 $N_0002 $N_0003 1F_F1 $N_0002 $N_0001 VF_F1 3VF_F1 $N_0003 $N_0004 0VR_R4 0 $N_0002 0.5R_R6 0 $N_0001 0.5I_I1 0 $N_0002 DC 2AR_R5 0 $N_0004 0.25

Clearly, v 1 = 625 mVolts , v 2 = 375 mVolts , and v 3 = 1.625 volts , which agrees withthe solution obtained in Problem 3.27.






The schematic is shown below. When the circuit is saved and simulated the nodevoltages are displayed on the pseudocomponents as shown. Thus,

,V15V,5.4 V,3 321

V V V

.




The schematic is shown below. When the circuit is saved and simulated, we obtain thenode voltages as displayed. Thus,

V a = –10.556 volts ; V b = 20.56 volts ; V c = 1.3889 volts ; and V d = –43.75 volts .

R1

20

R2

8

R3

10

R4

5

R5

4

R6

4 R7

16

R8

8

V1

90VdcV2

60Vdc

20.56 V

1.3889 V

–43.75 V

–10.556 V





H_H1 $N_0002 $N_0003 VH_H1 6VH_H1 0 $N_0001 0V

I_I1 $N_0004 $N_0005 DC 8AV_V1 $N_0002 0 20VR_R4 0 $N_0003 4R_R1 $N_0005 $N_0003 10R_R2 $N_0003 $N_0002 12R_R5 0 $N_0004 1R_R3 $N_0004 $N_0001 2

Clearly, v 1 = 84 volts , v 2 = 4 volts , v 3 = 20 volts , and v 4 = -5.333 volts




Clearly, v 1 = 26.67 volts , v 2 = 6.667 volts , v 3 = 173.33 volts , and v 4 = –46.67 volts which agrees with the results of Example 3.4.

This is the netlist for this circuit.


R_R1 0 $N_0001 2R_R2 $N_0003 $N_0002 6R_R3 0 $N_0002 4R_R4 0 $N_0004 1R_R5 $N_0001 $N_0004 3I_I1 0 $N_0003 DC 10AV_V1 $N_0001 $N_0003 20VE_E1 $N_0002 $N_0004 $N_0001 $N_0004 3




This network corresponds to the Netlist.This network corresponds to the Netlist.

+ v 0 – + v

4 4

3 k3 k

0 –

2 k

4 k 8 k

6 k

2i 0

+4A

–+100V

2 30 3v1

0




The circuit is shown below.

20 70

When the circuit is saved and simulated, we obtain v 2 = –12.5 volts

0

2 A 30 – +

1 2 3

20 V 50






The amplifier acts as a source.

R s

+V s R L

-

For maximum power transfer,

9s L R R






v 1 = 500(v s)/(500 + 2000) = v s/5

v 0 = -400(60v 1)/(400 + 2000) = -40v 1 = -40(v s /5) = -8v s,

Therefore, v 0/v s = –8




Let v 1 be the potential at the top end of the 100-ohm resistor.

(v s – v 1)/200 = v 1/100 + (v 1 – 10 -3v 0)/2000 (1)

For the right loop, v 0 = -40i 0(10,000) = -40(v 1 – 10 -3)10,000/2000,

or, v 0 = -200v 1 + 0.2v 0 = -4x10 -3v 0 (2)

Substituting (2) into (1) gives, (v s + 0.004v 1)/2 = -0.004v 0 + (-0.004v 1 – 0.001v 0)/20

This leads to 0.125v 0 = 10v s or (v 0/v s) = 10/0.125 = –80





For the left loop, applying KVL gives

–2.25 – 0.7 + 10 5IB + V BE = 0 but V BE = 0.7 V means 10 5IB = 2.25 or

IB = 22.5 µA .

For the right loop, –V CE + 15 – I Cx10 3 = 0. Addition ally, I C = βIB = 100x22.5x10 –6 =2.25 mA. Therefore,

V CE = 15–2.25x10 –3x10 3 = 12.75 V .

+ _

1 k

+ _

2.25 V

0.7 V 100 k

C 15 V | |

+ I C

_V CE

E




For loop 1, -v s + 10k(I B) + V BE + I E (500) = 0 = -v s + 0.7 + 10,000I B + 500(1 + )I B

which leads to v s + 0.7 = 10,000I B + 500(151)I B = 85,500I B

But, v 0 = 500I E = 500x151I B = 4 which leads to I B = 5.298x10 -5

Therefore, v s = 0.7 + 85,500I B = 5.23 volts

+V 0

–

1 k

500-+

18V -+

v s i1

i2

I10 k B

+V CE +

– V BE –

I E




We first determine the Thevenin equivalent for the input circuit.

R Th = 6||2 = 6x2/8 = 1.5 k and VTh = 2(3)/(2+6) = 0.75 volts

For loop 1, -0.75 + 1.5kIB + VBE + 400IE = 0 = -0.75 + 0.7 + 1500I B + 400(1 + )IB

IB = 0.05/81,900 = 0.61 A

v0 = 400IE = 400(1 + )IB = 49 mV

For loop 2, -400IE – VCE – 5kIC + 9 = 0, but, IC = IB and IE = (1 + )IB

VCE = 9 – 5k IB – 400(1 + )IB = 9 – 0.659 = 8.641 volts

+V 0 –

5 k

400-+

9 V -+

I

i1

i2

C

I1.5 k B +V CE + – V BE

– 0.75 V

I E




Although there are many ways to work this problem, this is an example based on the same kindof problem asked in the third edition.

ProblemFind I B and V C for the circuit in Fig. 3.128. Let = 100, V BE = 0.7V.

Figure 3.128

Solution

I 5 k 1

I1 = I B + I C = (1 + )I B and I E = I B + I C = I 1

+

V 0 –

4 k -+

12V

10 k

V

+V CE

– +

V BE –

I B

I E

C

I C



Applying KVL around the outer loop,

4kI E + V BE + 10kI B + 5kI 1 = 12

12 – 0.7 = 5k(1 + )I B + 10kI B + 4k(1 + )I B = 919kI B

IB = 11.3/919k = 12.296 A

Also, 12 = 5kI 1 + V C which leads to V C = 12 – 5k(101)I B = 5.791 volts




From (b), -v 1 + 2i – 3v 0 + v 2 = 0 which leads to i = (v 1 + 3v 0 – v 2)/2

2

2

At node 1 in (a), ((24 – v 1)/4) = (v 1 /2) + ((v 1 +3v 0 – v 2)/2) + ((v 1 – v 2)/1), where v 0 = v

or 24 = 9v 1 which leads to v 1 = 2.667 volts

At node 2, ((v 1 – v 2)/1) + ((v 1 + 3v 0 – v 2)/2) = (v 2 /8) + v 2/4, v 0 = v

v 2 = 4v 1 = 10.66 volts

Now we can solve for the currents, i 1 = v 1 /2 = 1.333 A , i 2 = 1.333 A , and

i3 = 2.6667 A.

+v 0

–

i3v 1 v 2i

(b) (a)

2 3v 0

– +24V

1

3v4 0

i i 21 i2 + +2 8 v 1 v 2 4

– –

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Solucionario Circuitos Sadiku Cap 3