Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

1. Assuming the passive sign convention, .

(a) i=0 (dc)

(b) A

(c)

(d)

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

2. where C = 10×10-9 F.

(a)

(b)

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

3. (a) C = 1 F, assume passive sign convention.

For t > 4 s, v = 1 V therefore iC = 0

For all other times,

(a) C = 1 F, assume passive sign convention.

for 0 ≤ t ≤ 4 s
ic (A)

0.5
4
t (s)

–0.5

(b) for 0 ≤ t ≤ 4 s
ic (pA)

4
t (s)

–10

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

4. Plate area = 2.5 mm × 2.5 mm = 6.25×10-6 m2. D = 25×10-6 m.

. Thus,

(a)

(b)

(c)

(d)

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

5.

. Therefore,

(a)

(b) For mylar, r = 3.1, so

(c) For SiO2, r = 3.9, so

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

6. <DESIGN> One possible solution:

A standard AAA battery has a height of approximately 44.5 – 2(0.8) mm = 42.9×10-3 m.

Construct a parallel plate capacitor using two gold layers separated by a 1 m

thick layer of mylar (r = 3.1).

Then, . Solving, we need each gold layer to have

area A = 3.64×10 m . One side of each gold square is constrained by the height of the
-3 2

battery-sized container (42.9×10-3 m). Thus, the other side must be A/42.9×10 -3 = 85
mm. This is much larger than the diameter (10.5 mm) of a AA battery, so we gently roll
the structure along the length after first contacting what will be the inner gold foil.

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

7. <DESIGN> One possible solution:

We construct something as sketched here:

fixed plate moveable plate

stops

Our design has two 1mm thick metal plates facing each other, each with area A. One is
fixed, and the other can move along the x axis. Stops are placed to restrict the maximum
plate spacing to 10 mm. If the left side of the fixed plate is at x = 0, the left stop is at x =
d, and the right stop is at 12 mm.
The dielectric layer is simply air, so r = 1. Then

which we can solve for A = 0.113 m2.

Taking the square root, each square plate must be 336 mm tall and 336 mm long.

The location of the left stop can now be determined:

or d = 5 mm.

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

8. <DESIGN> One possible solution:

[1]

[2]

Divide Eq. [1] by Eq. [2] to yield .

Also, we have the constraint that d1 – d2 = 0.1×10-3 m.

Solving together, d1 = 2×10-4 m and d2 = 1×10-4 m.

Use silicone as the spacer (r = 2.9). Construct the capacitor using two squares of copper
foil, each 1 mm thick, having area A. Coat one square on one side with a 2×10-4 m thick
layer of silicone; place the other foil square directly on top of the silicone layer.

Now we can solve for the needed area of each piece of foil:

or A = 9.74×10-4 m2. Hence each foil piece

must measure 31.2 mm x 31.2 mm.

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

9. Taking care to convert all distance units to meters,

VA W (m) C
-1 V 2.06×10-8 10.2 fF
-3 V 3.07×10-8 6.81 fF
-10 V 5.23×10-8 4.00 aF

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

10. We note that where C = 2 F.

vC (V)
(a) (1/2)(2)(1) = 4/2 V = 2 V.

t (s)
(b) (1/2)(2)(1) = 2 V -1 1 2 3

vC (V)

t (s)
-1 1 2 3

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

11. C = 1 mF. so .

(a) (103)(4)(0.2) = 800 V

(103)(8)(0.2) = 1600 V
Then this repeats, and adds to what has been previously computed.

vC (V)

4800
4000
3200
2400
1600
800
t (s)
0.2 0.4 0.6 0.8 1.0 1.2 1.4

(b) v (0.2) = 800 V

v (0.6) = 2400 V

v (1.2) = 4800 V

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

12. Energy stored is given by .

(a)

(b)

(c)

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

13. C = 150 pF

. Energy = .

(a) t = 0: = 10.8 nJ

(b) t = 0.2 s:

(c) t = 0.5 s:

(d) t = 1 s:

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

14. (a)

(b)

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

15. (a) no current through 5  resistor so

(b) vC = voltage across current source so

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

16. <DESIGN> One possible solution:

From Table 2.4, this wire has a cross-sectional area of 0.0804 mm2.

where here due to air as filler.

Solving for the length s of our helix,

Select a coil radius of 2.5 mm, which is larger than our wire diameter. Select N2 = 20.
Solving,
s = 329 mm (which is a bit over 1 ft long).

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

17. L = 75 mH.

From t = -1 to 0, i = 2(t + 1) so slope = +2. From t = 2 to 2.5, slope = -2.

(a) v (mV)
150

2 2.5
t (s)
-1

-150

(b) t = 1, 2.9 s, 3.1 s

v(1) = 0; v(2.9) = 0; v(3.1) = 0;

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

18. L = 17 nH

Slope = . vL = (17×10-9)(-0.5×106) = -8.5 mV

vL (mV)

5
t (s)

-8.5

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

19.

(a)

(b)

(c)

(d)

(e)

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

20. L = 8 pH, .

(a)

(b)

(c)

(d)

(e)

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

21. is = 1 mA, vs = 2 V

(a) vL = 0; iL = is = 1 mA

(b) 14 k resistor is irrelevant here. vL = 0; iL = is = 1mA

(c) vL = 0; iL = vs/4.7×103 = 426 A

(d) vL = 0; iL = vs/14×103 = 143 mA

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

22. L = 200 nH.

Since the current has a linear slope, .

(a) t = 200 ms. The first pulse then results in = 1 V. The
second pulse results in the same voltage but with opposite sign.
vL (V)
1

2
t (s)

-1

(b) The first pulse results in = 20 V. The second pulse results in

= 4 V.

vL (V)
20

2
t (s)
-4

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

23. L = 1 H.
Between -2 and 1 s, slope = (2 – 0)/(-2 – 0) = -1
Between 1 and 3 s, slope = -1/2
For t > 3 s, slope = -2/-3 = 2/3

(a) v(-1) = (1)(-1) = -1 V

(b) v(0) = (1)(-1) = -1 V
(c) v(1.5) = (1)(-1/2) = -0.5 V
(d) v(2.5) = (1)(-1/2) = -0.5 V
(e) v(4) = (1)(2/3) = 0.67 V
(f) v(5) = 0.67 V

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

24. where L = 6 mH.

(a) Any dc current (including 0 A) will result in a zero voltage across the inductor.

(b)

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

25.

(a) At t = 0,

(b) At t = 1.5 ms,

(c) At t = 45 ms,

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

26. L = 1 nH. .

(a) i = 0 therefore wL = 0 J

(b)

(c)

(d)

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

27. L = 33 mH, t = 1 ms. w = 0.5Li2

(a) 809 mJ

(b) 1.60 mJ

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

28. (a) A source transformation results in a (10/16) mA current source in parallel with 16 k.

Then, by current division,

(b) We can neglect the left-hand portion of the circuit and invoke current division:

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

29. (a)
Solving, V = –41.12 V
By voltage division, Vx = (12/27)( –41.95) = –18.3 V

(b)
Solving, V = –36.12 V
By voltage division, Vx = (12/27)( –36.12) = –16.1 V

(c)
V = –56.87 V
By voltage division, Vx = (12/27)( –56.87) = –25.28 V

(d) Same as case (b): Vx = –25.28 V

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

30. (a) Our equivalent is found as:

(c) The 47 k resistor is shorted out, therefore P47k = 0

and P10k =

(d)

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

31. Left leg: 1 F in series with 1 F = (1)(1)/(1+1) = 0.5 F. This appears in parallel with 1 F,
so the group can be replaced with 1 + 0.5 = 1.5 F. The right leg is identical. Thus, we are
left with 1.5 F in series with (1+1 = 2 F) in series with 1.5 F, or

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

32. Each  reduces to (L + L) || L = (2L)(L)/(2L + L) = 2L/3 H.

This leaves us with (2L/3) + L + (2L/3) = 2.33L.

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

33. <DESIGN> Two possible solutions:

(a) Connect one inductor in series with a parallel connection of 4 inductors.

Then, Leq = 1 + 1||1||1||1 = 1 + 0.25 = 1.25 nH.

(b) Create a string by connecting two inductors in series with a parallel combination of
two inductors. Then connect two such strings in parallel.

Leq = 0.5*(1 + 1 + 1 || 1) = 0.5*(2.5) = 1.25 nH

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

34. We have 2 F in series with 1 F in series with 2 F, which equates to

= 0.5 F

Next we have 8 F in series with 5 F, which equates to (8)(5)/(8+5) = 3.08 F.

These two combinations appear in parallel, and equate to 0.5 + 3.08 = 3.58 F.

Now we have 4 F in series with 3.58 F in series with 1 F, or

Continuing the same process, we now have 5 F in parallel with 0.654 F, or 5.654 F.

Finally, this appears in series with 7 F, so that

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

35. The left leg can be reduced to

(1/7 + 1/22)-1 + 4 = 9.310 F

On the right leg, we neglect the 2 F, and then combine as

5 + (1/12 + 1/1)-1 = 5.923 F

Thus, Ceq = (1/9.310 + 1/5.923)-1 = 3.62 F

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

36. The left network reduced to [12||10 + 7] || 4 = [(12)(10)/(12 + 10) + 7] ||4

or .

On the right network, we ignore the 2 H, and form the combination (12 + 1) || 5

or

Thus, Leq = 3.03 + 3.61 = 6.64 H

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

37. We note that R||R = R/2, C || C = 2C, L ||L = L/2, L + L = 2L, and two capacitors in series
equate to C/2.

Thus, our network can be reduced to:

R/2

2C L/2
C/2

2L

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

38. (a) We start by naming each resistor as “R” so that we can create an expression and
substitute in the value of 10  at the end.

Noting that R||R||R = 3R,

The network can be reduced to

(b) Inductor combinations work the same way as resistors, so Leq = 11.4 H.

(c)

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

39. All inductances are in parallel, so 1/Leq = (1/1 + ½ + 1/1 + 1/7 + ½ + ¼)×109

Thus, Leq = 295 pH

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

40. Two capacitors of value C in series can be replaced by C/2.

Three inductors of value L in series can be replaced by 3L. Thus, our circuit can be
reduced to:

3L
C/2 C/2

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

41. Temporarily assign capacitors value ‘C’ and inductors value ‘L.’ Then at the far right we
can neglect the two inductors in parallel. Both vertical inductors are actually in parallel,
to be replaced by L*L/(L+L) = L/2. We have three sets of two capacitors in series, each
of which can be replaced with C*C/(C+C) = C/2. The middle group of vertical capacitors
can be replaced with
C + C/2 + C/2 + C = 3 C.
Finally, the upper left combination of three capacitors can be replaced with C + C/2 =
1.5C. Our equivalent circuit is then

Since L = 2 nH and C = 2 nF, 1.5C = 3 nF, 3C = 6 nF, and L/2 = 1 nH.

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

42. Leq = L1 + L2||L3 + L4||L5||L6

or

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

43. At the far right, we have 10 F in series with 10 F, which can be replaced with 10/2 = 5 F.
We then have 10 F || 5 F = 15 F. This is in series with 10 F. That combination is
equivalent to (15)(10)/(15+10) = 6 F. This is in parallel with 10 F, for an equivalent of 16
F.

Finally, the equivalent of 16 F, which represents all but the bottom left capacitor, is in
series with that 10 F, for an equivalent of 16(10)/(16+10) = 6.15 F.

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

44. Starting on the far right, we see 10 + 10 = 20 H. This is in parallel with 10 H, so the
combination may be replaced by 20(10)/(20 + 10) = 6.67 H.

This is in series with a 10 H inductor, so the combination may be replaced with 10 + 6.67
= 16.67 H. This is in parallel with 10 H, for an equivalent of (10)(16.67)/(10 + 16.67) =
2.5 H.

Finally, the equivalent inductance of 2.5 H is in series with the bottom left 10 H inductor,
so

Leq = 10 + 2.5 = 12.5 H

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

45.

or [1]

Next, or

[2]

By inspection, v3 = vs.

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

46. Using the mesh currents defined in the figure,

or

or

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

47. is(0) = 60 mA therefore i2 = i3 – i1 = 40 mA.

(a) By KCL,

(b)

(c)

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

48. (a) By KVL, or

Thus,

(b) =

(c) =

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

49. We first open-circuit the cosine source, leaving us with only dc sources. Thus, we may
treat capacitors as open circuits, and inductors as short circuits. Hence,

vL’ = 0 and vC’ = 20(0.03) + 9 = 9.6 V

Returning to the original circuit and zeroing out the dc sources, we’re left with a single
source. However, the 9 V source, being replaced by a short circuit, leads us to vC”(t) = 0.

Consequently, since vL = vL’+vL” = 0 + =

and vC = vC’ + vC” = 9.6 + 0 = 9.6 V.

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

50. We define three clockwise mesh currents such that i1 flows in the bottom left mesh, i2
flows in the top mesh, and i3 flows in the bottom right mesh.

By inspection,

Then,

and

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

51. (a) Assuming an ideal op amp,

. By KVL,

Thus,

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

52. We note that with no initial energy storage,

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

53. . Assuming an ideal op amp,

Thus,

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

54. We note that this is a differentiator, so that

(a) Taking the derivative,

(b) Taking the derivative,

(c)

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

55. <DESIGN> One possible solution:

We feed a differentiator into an inverting amplifier to correct the polarity.

This yields

We need so that 1 oC/s (which corresponds to vs changing by 1 mV per

second) yields

Choose C1 = 1 mF, which dictates that Rf = 1 M. For simplicity, choose both resistors in
the inverting amplifier to also be 1 M each.

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

56. <DESIGN> One possible solution:

We feed a differentiator into an inverting amplifier to correct the polarity.

This yields

We need so that 1 m/s (which corresponds to vs changing by 1 mV per second)

yields

Choose Rf = 1 , which dictates that C1 = 1 F. For simplicity, choose both resistors in the
inverting amplifier to also be 1  each.

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

57. <DESIGN> One possible solution:

We treat the proton detector as a current source, and place it in parallel with a 1 
resistor so that the voltage across the pair is proportional to the number of protons per
second.

We connect this to the input of an integrator stage, the output of which is fed into an
inverting amplifier stage to correct the polarity of the output voltage.

This yields

With we need

Choosing R1 = 10 M, Cf must be 100 mF. We choose the two resistors of the inverting
amplifier to also be 10 M each for simplicity.

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

58. <DESIGN> One possible solution:

(a) We feed a differentiator into an inverting amplifier to correct the polarity.

This yields

We need so that 1 m/s2 (which corresponds to vs changing by 10 mV per

second) yields

Choose Rf = 1 , which dictates that C1 = 1 F. For simplicity, choose both resistors in the
inverting amplifier to also be 1  each.

(b) Three such units would likely be needed, to detect motion in all three orthogonal
directions.

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

59. <DESIGN> One possible solution:

(a) A simple voltage divider will provide us the needed voltage output, if constructed as:

vsense

Then, when R = 0, vsense = 0 and when R = 10 , vsense = (2)(1/2) = 1 V.

(b) We connect the positive terminal of vsense to the input of a differentiator, the output of
which is fed into an inverting amplifier to correct the output polarity.

This yields
We need so that 1 l/s (which corresponds to vsense changing by 10 mV per
second) yields
Choose Rf = 10 M, which dictates that C1 = 1 F. For simplicity, choose both
resistors in the inverting amplifier to also be 10 M each.

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

60. (a) We begin by overlaying our dual directly on the original circuit:

(b) Redrawing and labelling:

+ v1 - + v2 -

a. Original circuit, nodal analysis:

Dual circuit:

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

61. (a) We begin by drawing the dual circuit directly over the original circuit.

2sint A

2sint V

(b) We first redraw the dual circuit for clarity.

v1 v2
→i1 →i2

(c)

Original circuit, mesh:

Nodal: v1 = - 2sint V

New, mesh: i1 = -2sint A

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

62. (a) With four elements in parallel, we expect our dual to contain four (dual) elements in
series. Thus,

+ i1 –

b. Original circuit: define as top node, and bottom node as reference. Then,

Dual circuit:

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

63. (a) We construct the dual by noting that elements in series will be represented by their
duals in parallel, and vice-versa. Thus,

vC

(b) The current dual is a voltage, as labelled.

(c)

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

64. We begin by overlaying our dual circuit directly on top of the original circuit, using it as
a guide.

Finally, we redraw and label each dual element:

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

65. (a) DC. Thus, iL = 7/80×103 = 87.5 A

(b) Pdiss = 0

(c) We see from the dc simulation that the inductor current is 87.5 mA as computed, and
there is essentially zero current flowing through the 46 k resistor, hence it dissipates
0 W as expected.

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

66. (a) At DC, the capacitor is an open circuit. Hence, the current through each resistor is

given by . Thus

P80k = (55.6×10-6)2(80×103) = 247 mW

P46k = (55.6×10-6)2(46×103) = 142 mW

(b) Thus,

(c) Our simulation confirms both the resistor currents and capacitor voltage:

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

67. (a) At DC, the inductor is a short circuit, and the capacitor is an open circuit.

By current division, .
Thus, vx = (440(iL) = -5.069 V

(b) .
(c) Our simulation confirms both the inductor current and capacitor voltage (the sign of
the inductor current in the simulation is an artifact of how the component was rotated
before placing):

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

68. (a)

(b) We set up the transient simulation as:

Using the cursor, we complete the table below:

t (ns) iL(A)
0 999.99
130 372.96
260 136.81
500 21.40

(c) so
using the table above,

at t = 130 ns,
and

at t = 500 ns,

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

69. (a)
(b) no - the resistor will slowly dissipate the energy stored in the capacitor

(c) We perform the following simulation:

Using the cursor, we complete the table below:

t (ms) vC (V)
0 9
460 3.31
920 1.22
2300 60.4×10-3

(d) Fraction of energy at selected times:

so using the table above,

at t = 460 ms, which is 100(110/(405) = 27%

and

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

at t = 2.3 s, which is 100(0.0182)/405 =

0.004%

70. (a) Define three clockwise mesh currents: i1 in the left bottom mesh, i3 in the top mesh,
and i2 in the bottom right mesh (noting the open circuited capacitor means we can treat
the 2  resistor and the current source as having the same mesh current).

Solving, vx = -1.00 V and iL = i3 = 5.502 A.

Then,

(b) We see that our simulation agrees well with the hand calculations (5.50 A, -1.00 V).

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

71. (a)

(b) Within a reasonable amount of rounding error, we see that our simulation agrees well
with our hand calculation in terms of amplitude.

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

72. (a)

(b) Within rounding error, our simulation agrees well with our hand calculation.

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

73. The circuit in Fig. 7.72 is an integrator: If we assume an ideal op amp with no initial
energy stored in the inductor, then

and equating currents leads to

and, . With Rf = 47 , L1 = 0.1 H and we find

(b) Our simulation agrees well in terms of magnitude (74.5 mV = 148.9/2) but contains
an unexpected dc offset.

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 Engineering Circuit Analysis 9th Edition Chapter Seven Exercise Solutions

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