Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

1. (a) 5 nodes

(b) 6 elements

(c) 6 branches

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

2. (a) 4 nodes;

(b) 7 elements;

(c) 7 branches (we omit the 2  resistor as it is not associated with a path in our
definition)

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

3. (a) 4 nodes

(b) path, yes; loop, no

(c) path, no; loop, no

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

4. (a) 6 elements;

(b) path, yes; loop, no.

(c) path, yes; loop, no.

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

5. (a) 4 nodes

(b) 4 elements

(c) 4 branches

(d) i) neither; ii) path only;

iii) both path and loop; iv) neither (‘c’ encountered twice).

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

6. The parallel-connected option would allow most of the sign to still light even if one or
more individual bulbs burn out. For that reason, it would be more useful to the owner.

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

7. By KCL, iA – iB + iC + iD – iE = 0. Thus,

(a) 1 – iB + 3 – 2 – 0 = 0 so iB = 2 A.

(b) –1 – (–1) – 1 – 1 – iE = 0 so iE = -2 A

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

8. (a) By KCL, 7 = 6 + I so I = 1 A

(b) By KCL, 2 = 3 + I + 3 so I = -4 A.

(c) No net current can flow through the resistor or KCL would be violated.

Hence, I = 0.

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

9. We note that KCL requires that if 7 A flows out of the “+” terminal of the 2 V source, it
flows left to right through R1. Equating the currents into the top node of R2 with the
currents flowing out of the same node, we may write

7A

7 + 3 = i2 + 1

or

i2 = 10 – 1 = 9 A

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

10. By KCL, iA + iB + iC = ibattery + isign. Hence, ibattery = iA + iB + iC – isign = 8.3 A.

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

11. We can determine RA from Ohm’s law if either the voltage across, or the current through
the element is known. The problem statement allows us to add labels to the circuit
diagram:

7.6 A
Iwire

IRA
= 1.5 A

Applying KCL to the common connection at the top of the 6  resistor,

Iwire = 1.5 – (-1.6) = 3.1 A

Applying KCL to the top of RA then results in

IRA = 7.6 – Iwire = 7.6 – 3.1 = 4.5 A.

Since the voltage across RA = 9 V, we find that RA = 9/4.5 = 2 

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

12. By KCL,

IB + IC = IE [1].

Given that IC = 1.5 mA, we are left with one equation in two unknowns.

Noting that IC = 50 IB, we may rewrite Eq. [1] as

IC/50 + IC = IE. Solving, IE = 1.53 mA, and IB = IC/50 = 0.03 mA.

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

13. By inspection, I3 = –5Vx due to the action of the dependent source.

Vx = (2×10-3)(4.7×103) = 9.4 V

Hence, I3 = –47 A

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

14. With finite values of R1 some value of current I will flow out of the source and through
the left-most resistor. If we allow some small fraction of that current kI (k < 1) to flow
through the resistor connected by a single node, to where does the current continue? With
nowhere for the current to go, KCL is violated. At best we must imagine an equal current
flowing the opposite direction, yielding a net current of zero. Consequently, Vx must be
zero.

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

15. The order of the resistors in the schematic below is unimportant.

North East South West

Wall Wall Wall Wall

VAC ~ 400  400  400  400 

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

16. (a) -v1 + v2 – v3 = 0

Hence, v1 = v2 – v3 = 0 + 17 = 17 V

(d) v1 = v2 – v3 = -2 – 2 = -4 V

(e) v2 = v1 + v3 = 7 + 9 = 16 V

(f) v3 = -v1 + v2 = 2.33 – 1.70 = 0.63 V

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

17. (a) By KVL, +9 + 4 + vx = 0

Therefore vx = –13 V

From Ohm’s law, ix = vx/7 = –13/7 = –1.86 A

(b) By KVL, +2 + (-7) + vx = 0. Thus, vx = 5 V.

Using Ohm’s law, ix = vx/8 = 625 mA.

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

18. (a) Combining KVL and Ohm’s law,

-1 + 2 + 2i – 5 + 10i = 0

Hence, 12i = 4 so i = 333 mA

(b) Again combining KVL and Ohm’s law,

10 + 2i – 1.5 + 1.5 + 2i + 2i + 2 + 2i – 1 + 2i = 0

Hence, i = -11/10 so i = –1.1 A

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

19. From KVL, +4 – 23 + vR = 0 so vR = 23 – 4 = 19 V

Also, -vR + 12 + 1.5 – v2 – v3 + v1 = 0

or -19 + 12 + 1.5 – v2 – 1.5 + 3 = 0

Solving, v2 = –4 V

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

20. Vdep = V2 – 1000IC – 1000IC.

With V2 = 15 V, IC = 50IB = 50(20×10-6) = 0.001 A
Therefore,
Vdep = 15 – 2000(0.001) = 13 V.

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

21. We apply KCL/KVL and Ohm’s law alternately, beginning with the far left. Knowing
that 500 mA flows through the 7.3  resistor, we calculate 3.65 V as labelled below.
Then application of KVL yields 2.3 – 7.3 = –1.35 V across the 1  resistor.

This tells us that -1.35/1 = -1.35 A flows downward through the 1  resistor. KCL now
tells us that 0.5 – (–1.35) = 1.85 flows through the top 2  resistor. Ohm’s law dictates a
voltage of 3.7 V across this resistor in this case.

Application of KVL determines that –(–1.35) + 3.7 + v2 = 0 or -5.05 V appears across

the right-most 2  resistor, as labelled below. Since this voltage also appears across the
current source, we know that

Vx = –5.05 V

+ 3.7 V –

1.85 A
+ 3.65 V – +
+
-5.05 V
-1.35 V
–
– -1.35 A

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

22. (a) By KVL, -vs + v1 + v2 = 0 [1]

Define a clockwise current I. Then v1 = IR1 and v2 = IR2.

Hence, Eq. [1] becomes

-vs + IR1 + IR2 = 0
Thus, vs = (R1 + R2)I and I = v1/R1
so vs = (R1 + R2)v1/R1 or v1 = vsR1/(R1 + R2). QED

Similarly, v2 = IR2 so we may also write,

vs = (R1 + R2)v2/R1 or v2 = vsR2/(R1 + R2). QED

(Proof, so final answer was given to begin with.)

(b) R1 > 0.

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

23. (a) By inspection:

v1 = 2 V; v2 = 2 V;

i2 = 2/6 = 333 mA

i3 = 5v1 = 10 A

i1 = i2 + i3 = 10.33 A

v4 = v5 = 5i2 = 5(1/3) = 5/3 V

i5 = (5/3)/5 = 333 mA

i3 = i4 + i5 therefore i4 = i3 – i5 = 10 – 1/3 = 9.67 A

-v2 + v3 + v4 = 0 therefore v3 = v4 + v2 = 6/3 = 333 mV

(b) Using the passive sign convention (assuming all elements absorb power):

v1(-i1) + v2(i2) + v3(i3) + v4(i4) + v5(i5) = 0 (should be true, within rounding

error).

= -62/3 + 2/3 + 10/3 + 145/9 + 5/9 = 0, so true.

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

24. Define the current I flowing out of the “+” reference terminal of the 5 V source. This
current also flows through the 100  and 470  resistors since no current can flow into
the input terminals of the op amp.

Then, –5 + 100I + 470I +vout = 0 [1]

Further, since Vd = 0, we may also write

–5 + 100I = 0 [2]

Solving, Vout = 5 – 570I = 5 -570(5/100) = –23.5 V

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

25. With a clockwise current i, KVL yields

–(–8) + (1)i + 16 +4.7i = 0

Therefore, i = –(16 – 8)/5.7 = –4.21 A

Absorbed power:
Source vs1: +(–8)(+4.21) = –33.68 W

Source vs2: (16)(–4.21) = –67.36 W

R1: (4.21)2(1) = 17.21 W

R2: (4.21)2(4.7) = 83.30 W

Check: The sum of the above ‘absorbed’ powers = -0.02, which is within rounding error
of zero, as expected.

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

26. Define a clockwise current I. Then, KVL yields

-12 + 2i + 3i + 5i + 2 +I – 4 = 0

Hence, I = 14/11 = 1.27 A.

Using the passive sign convention to compute power absorbed:

P12V = (12)(-1.27) = -15.24 W

P2 = 2i2 = 3.23 W

P3 = 3i2 = 4.84 W

P5 = 5i2 = 8.06 W

P2V = 2(i) = 2.54 W

P4V = 4(-i) = -5.8 W

P1 = (1)i2 = 1.61 W

Summing these yields -0.04 W, which is within rounding error of zero, as expected.

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

27. Define a clockwise current I

Then invoking KVL and Ohm’s law we may write

500I – 2 + 1000I + 3vx + 2200I = 0

Since vx = -500I, the above equation can be recast as

500I – 2 + 1000I – 1500I + 2200I = 0

Solving, I = 2/2200 = 909.1 A

ABSORBED POWER
500 : 500(I2) = 413.2 W

2 V source: (2)(-I) = -1.818 mW

1 k: (1000)(I2) = 826.5 W

Dep source: [(3)(-500I)](I) = -1.240 mW

2.2 k: (2200)(I2) = 1.818 mW

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

28. (a) By KVL, -12 + 27ix + 33ix + 13ix + 2 + 19ix = 0

Hence, ix = 10/92 = 108.7 mA
Element Pabsorbed (W)
12 V (12)(-0.1087) = -1.304
27   
33   
13   
19   
2V (2)(0.1087) = 0.2174

(b) By KVL, -12 + 27ix + 33ix + 4v1 + 2 + 19ix = 0

and v1 = 33ix. Solving together, ix = 10/211 = 47.39 mA
Element Pabsorbed (W)
12 V -0.5687
27  
33  
Dep source 
19  
2V 0.09478

(c) By KVL, -12 + 27ix + 33ix + 4ix + 2 + 19ix + 2= 0

Solving, ix = 10/83 = 120.5 mA
Element Pabsorbed (W)
12 V -1.446
27  
33  
Dep source 
19  
2V 0.2410

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

29. A simple KVL equation for this circuit is

-3 + 100ID + VD = 0
Substituting in the equation which related the diode current and voltage,

-3 + (100)(45×10-9)
Solving either by trial and error or using a scientific calculator’s equation solver,

VD = 359 mV

which then allows us to use the diode equation to compute ID = 26.38 mA.

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

30. (a) KCL yields 3 – 7 = i1 + i2

or, –4 = v/4 + v/2

Solving, v = –5.333 V

Thus, i1 = v/4 = –1.333 A and i2 = v/2 = –2.667 A

(b)

Element Absorbed Power

3A (-5.333)(-3) = 16.00 W

7A (-5.333)(7) = -37.33 W

4 (4)(-1.333)2 = 7.108 W

2 (2)(-2.667)2 = 14.23 W

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

31. Consider the currents flowing INTO the top node. KCL requires

-2 – i1 – 3 – i2 = 0
Or i1 + i2 = -5 [1]

Also, i1 = v/10 and i2 = v/6 so Eq. [1] becomes

v/10 + v/6 = –5

Solving, v = –18.75 V

Thus the power supplied by the –2 A source is (–2)( –18.75) = 37.5 W

and the power supplied by the 3 A source is – (3)( –18.75) = 56.25 W

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

32. Define ix as the current through X, pointing downward.

By KCL,

1 – v/5 – ix – v/5 + 2 = 0
or
3 – (2/5)v – ix = 0

(a) The current source sets ix = 2 A.

Thus, v = (3 – ix)(5/2) = 2.5 V

(b) v = 2 V, by inspection.

(c) ix = 2V therefore

3 – (2/5)v – 2v = 0
Solving,
V = 15/12 V = 1.25 V

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

33. Summing the currents flowing into the top node, KCL yields

–v/1 + 3ix + 2 – v/3 = 0 [1]

Also, ix = -v/3. Substituting this into Eq. [1] results in

–v –v + 2 – v/3 = 0

Solving, v = 6/7 V (857 mV)

The dependent source supplies power = (3ix)(v)

= (3)( –v/3)(v) = –36/49 W (-735 mW)

The 2 A source supplies power = (2)(v) = 12/7 W = (1.714 W)

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

34. Define the center node as +v; the other node is then the reference terminal.

KCL yields

Solving, v = –1.274 V

(a) R Pabsorbed
1 k 1.623 mW
4.7 k 345.3 W
2.8 k 579.7 W

(b) Source Pabsorbed

3 mA (v)(3×10-3) = –3.833 mW
5 mA (v)( –5×10 ) = +6.370 mW
-3

(c)

Thus, .

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

35. By KVL, veq = v1 + v2 – v3

(a) veq = 0 – 3 – 3 = –6 V;

(b) veq = 1 + 1 – 1 = 1 V;

(c) veq = -9 + 4.5 – 1 = –5.5 V

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

36. KCL requires that ieq = i1 – i2 + i3

(a) ieq = 0 + 3 + 3 = 6A

(b) ieq = 1 – 1 + 1 =1A

(c) ieq = –9 – 4.5 + 1 = –12.5 A

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

37. The voltage sources are in series, hence they may be replaced with veq = –2 + 2 – 12 + 6
= –6 V. The result is the circuit shown below:

1 k
–6 V

Analyzing the simplifed circuit, i = –6/1000 = –6 mA

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

38. We may first reduce the series connected voltage sources, or simply write a KVL
equation around the loop as it is shown:

+2 + 4 + 7i + v1 + 7i + 1 = 0

Setting i = 0, v1 = –7 V

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

39. The current sources are combined using KCL to obtain ieq = 7 – 5 – 8 = –6 A.
The resulting circuit is shown below.

–6 A 2 3

(a) KCL stipulates that v.

Solving, v = –36/5 V

(b) Psupplied by equivalent source = (v)(-6) = 43.2 W

Source Psupplied (W)
7 A source (7)(-36/5) = –50.4 W
5 A source (-5)(-36/5) = 36 W
8 A source (-8)(-36/5) = 57.6 W
43.2 W so confirmed.

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

40. We apply KCL to the top node to write

1.28 – 2.57 = v/1 + Is + vs/1

Setting v = 0, we may solve our equation to obtain Is = –1.29 A

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

41. (a) Employing KCL, by inspection Ix = 3 A; Vy = 3 V

(b) Yes. Current sources in series must carry the same current. Voltage sources in
parallel must have precisely the same voltage.

(c)

1 4V

(all other sources are irrelevant to determining i and v).

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

42. Left-hand network: 1 + 2 || 2 = 1 + 1 = 2

Right-hand network: 4 + 1||2 + 3 = 4 + 0.6667 + 3 = 7.667 

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

43. For Fig. 3.82(a), Req = 1 + 2 || R = 1 + 2R/(R + 2)

(a) R = 2  so Req = 2 
(b) R = 4  so Req = 2.33 
(c) R = 0  so Req = 1 

For Fig. 3.82(b), 1/Req = 1 + 1/R + 1/3

(a) R = 2  so Req = 0.545 

(b) R = 4  so Req = 0.632 
(c) R = 0  so Req = 0.750 
(d)

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

44. (a) The 4 resistors may be replaced with a single resistor having value

2 + 7 + 5 + 1 = 15 .

(b) I = (3 – 1)/ 15 = 133.3 mA

(c) 3 V

(d) P1 = i2R = 17.78 mW

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

45. KCL yields ieq = -2 + 5 + 1 = 4 A

And Req = 5 || 5 = 2.5 

(a)

4A 2.5 

(b) Ohm’s law yields v = (4)(2.5) = 10 V

(c) Power supplied by 2 A source = (–2)(10) = –20 W

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

46. We can create an effective resistance by making the following combination to appear in
parallel with the 1 A source:

6 || 3 1 + 5 + 3 ||6 = 10 .

Noting that 3 || 10 || 5 = 1.579 , we can compute vx = (1)(1.579) = 1.579 V.

Then, i3 = vx/3 = 536.3 mA.

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

47. At the far right we have the resistor combination 9 + 6 || 6 = 9 + 3 = 12 .

After this, we have three resistors in parallel but should not involve the 15  resistor as it
controls the dependent source. Thus, Req-1 = 1/3 + 1/12 and Req = 2.4 .
The simplified circuit is shown below.
Summing the currents flowing into the top node,

[1]

Since i = vx/15, Eq. [1] becomes

Solving, vx = 2.667 V

2.4 

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

48. We combine the left-hand set of resistors: 6 + 3 || 15 = 8.5 

The independent sources may be combined into a single 4 + 3 – 9 = -2 A source (arrow
pointing up). We leave the 6  resistors; at least one has to remain as it controls the
dependent source. A voltage v is defined across the simplified circuit, with the + terminal
at the top node.

Applying KCL to the top node,

-2 – 2i = v/8.5 + v/6 + v/6 [1]
where i = v/6. Thus, Eq. [1] becomes -2 – 2v/6 = v/8.5 + 2v/6 or v = -2.55 V.

We have lost the 15  resistor temporarily, however. Fortunately, the voltage we just
found appears across the original resistor combination we replaced. Hence, a current -
2.55/8.5 = -0.3 A flows downward through the combination.

Hence, the voltage across the 3  || 15  combination is

v – 6(-0.3) = -0.75 V

Thus, P15 = (-0.75)2/15 = 37.5 mW

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

49. Starting from the far right, we define RA = R7 || [R8 + R10||R11 + R9]
= 10 || [10 + 10||10 + 10] = 10 || 25 = 7.143 .

Next, R4 || [R5 + RA + R6] = 10 || [10 + 7.143 + 10] = 7.308 

Finally, Req = R1 || [R2 + 7.308 + R3] = 7.32 

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

50. Four 100  resistors may be combined as:

(a) 25  = 100  || 100  || 100  || 100 

(b) 60  = [(100  || 100 ) + 100 ] || 100 

(c) 40  = (100  + 100 ) || 100  || 100 

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

51. (a) v2 = v1 – v2 = 9.2 – 3 = 6.2 V

(b) v1 = v – v2 = 2 – 1 = 1 V

(c) v = v1 + v2 = 3 + 6 = 9 V

(d) v1 = vR1/(R1 + R2) = v2R2/(R1 + R2).

Thus, setting v1 = v2 and R1 = R2, R1/R2 = 1

(e) v2 = vR2/(R1 + R2) = vR2/(2R2 + R2) = v/3 = 1.167 V

(f) v1 = vR1/(R1 + R2) = (1.8)(1)/(1 + 4.7) = 315.8 mV

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

52. (a) i1 = i – i2 = 8 – 1 = 7A

(b) v = i(R1 || R2) = i(50×103) = 50 V

(c) i2 = iR1/(R1 + R2) = (20×10-3)(1/5) = 4 mA

(d) i1 = iR2/(R1 + R2) = (10)(9)/18 = 5A

(e) i2 = iR1/(R1 + R2) = (10)(10×106)/(10×106 + 1) = 10 A

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

53. One possible solution: Choose v = 2 V. Then

R1 = v/i1 = 2 

R2 = v/i2 = 1.67 

R3 = v/i3 = 250 m

R4 = v/i4 = 645 m

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

54. First, replace the 2  || 10  combination with 1.66 . Then

900 mV

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

55. Employing voltage division,

V3 = (9)(3)/(1 + 3 + 5 + 7 + 9) = 1.08 V

V7 = (9)(7)/(1 + 3 + 5 + 7 + 9) = 2.52 V

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

56. We begin by simplifying the circuit to determine i1 and i2.

We note the resistor combination (4 + 4) || 4 + 5 = 8 || 4 + 5 = 7.667 .
This appears in parallel with the 1  and 2  resistors, and experiences current i2.

Define the voltage v across the 25 A source with the ‘+’ reference on top. Then,

Solving, v = 15.33 V. Thus, i1 = 25 – v/1 = 9.67 A

i2= i1 – v/2 = 2.005 A

Now, from current division we know i2 is split between the 4  and 4 + 4  = 8 
branches, so we may write

v3 = 4[4i2/(4 + 8)] = 2.673 V

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

57. We may do a little resistor combining:

2 k || 4 k = 8/6 k

3 k || 7 k = 21/10 k

The 4 k resistor is in parallel with the series combination of 8/6 + 3 + 21/10 =

6.433 k. That parallel combination is equivalent to 2.466 k.

Thus, voltage division may be applied to yield

v4k = (3)(2.466)/(1 + 2.466) = 2.134 V

and vx = (2.134)(21/10)/(8/6 + 3 + 21/10) = 697.0 mV

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

58. We could use voltage division for determining those voltages if iB = 0. Since it is nonzero
(presumably – circuit analysis will verify whether this is the case), we do not have equal
currents through the two resistors, hence voltage division is not valid.

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

59. (a) i1k = (-gm v)(10/11) where v = 12×10-3 (15/18) cos1000t

Thus, i1k = -10.9cos1000t V

(b) vout = (1×103) i1k = -10.9 cos1000t mV

(c) |vout|/|vs| = 10.9/12 < 1, so no, it does not amplify (in terms of voltage magnitude).

(d) |vout|/|v| = 10.9/10 > 1, so yes.

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

60. vout = (3.3×103)(-gm v) = (3.3×103)(-322×10-3)v

where v = [6×10-6 cos 2300t] (45/18)/(1 + 45/18) = 4.286×10-6 cos 2300t V.

Thus, vout = 4.55 cos 2300t mV

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

61. (a) Looking to the right of the 10  resistors, we see

40 + 50 || (20 + 47) = 68.63 

Further, the 10  || 10  can be replaced with a 5  resistor without losing the desired
voltage. Hence, v appears across 5 || 68.63 = 4.66 

So v = 20 (4.66)/(20 + 4.66) = 3.78 V

(b) The 4  resistor has been “lost” but we can return to the original circuit and note the
voltage determined in the previous part. By voltage division, the voltage across the 50 
resistor is v(50 || 67)/(40 + 50 || 67) = 1.57 V.

v47 = v50(47)/(20 + 47) = 1.10 V

Hence, the power dissipated by the 47  resistor is (v47)2/4 = 25.7 mW

(d) No, the power consumption (26 mW) is far below the maximum rating of 250 mW.

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

62. (a) Define

Req = 40 + 50 || (20 + 47) = 68.63 

RTOTAL = 20 + 10 || Req = 28.73 

itotal = 20/RTOTAL = 0.696 A

By current division, i40 = itotal (10/(10 + Req) = 8.5 mA

(b) P20V = (20)itotal = (20)(0.696) = 13.92 W

(d) I50 = i40 (20 + 47)/[50 + (20 + 47)] = 0.0507 A

P50 = (i50)2 × 50 = 128 mW

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 Engineering Circuit Analysis 9th Edition Chapter Three Exercise Solutions

63. (a) 5 nodes; 6 loops; 7 branches;

(b) Define all currents as flowing either left to right or downward.

Note the combination Req = 2/3 + 2 + 5 = 7.667 .

By inspection, I2left = –2 A;

I5(left) = (-2)(7.667)/(5 + 7.667) = –1.21 A;

I2right = I5 = (-2)(5)/(5 + 7.667) = –790 mA;

The remaining currents are found by current division:

I1 = –0.7895(2)/3 = –526 mA;

I2middle = –0.7895(1)/3 = –263 mA

(a) V2A = V2left) + V5left) = (-2)(2) + (-1.211)(5) = -10.06 V.

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