Download Circuit Variables-Circuit And Network Analysis-Solution Manual and more Exercises Electrical Circuit Analysis in PDF only on Docsity!
Circuit Variables
Assessment Problems
AP 1.1 To solve this problem we use a product of ratios to change units from dollars/year to dollars/millisecond. We begin by expressing $10 billion in scientific notation:
$ 100 billion = $ 100 × 109
Now we determine the number of milliseconds in one year, again using a product of ratios:
1 year
- 25 days
1 day 24 hours
1 hour 60 mins
1 min 60 secs
1 sec 1000 ms
1 year
- 5576 × 109 ms
Now we can convert from dollars/year to dollars/millisecond, again with a product of ratios:
$ 100 × 109 1 year
1 year
- 5576 × 109 ms
= $ 3. 17 /ms
AP 1.2 First, we recognize that 1 ns = 10 −^9 s. The question then asks how far a signal will travel in 10 −^9 s if it is traveling at 80% of the speed of light. Remember that the speed of light c = 3 × 108 m/s. Therefore, 80% of c is (0.8)(3 × 108 ) = 2. 4 × 108 m/s. Now, we use a product of ratios to convert from meters/second to inches/nanosecond:
- 4 × 108 m 1 s
1 s 109 ns
100 cm 1 m
1 in
- 54 cm
(2. 4 × 108 )(100)
(10^9 )(2.54)
- 45 in 1 ns
Thus, a signal traveling at 80% of the speed of light will travel 9. 45 ′′^ in a nanosecond.
1–2 CHAPTER 1. Circuit Variables
AP 1.3 Remember from Eq. (1.2), current is the time rate of change of charge, or i = dqdt In this problem, we are given the current and asked to find the total charge. To do this, we must integrate Eq. (1.2) to find an expression for charge in terms of current:
q(t) =
∫ (^) t 0
i(x) dx
We are given the expression for current, i, which can be substituted into the above expression. To find the total charge, we let t → ∞ in the integral. Thus we have
qtotal =
∫ (^) ∞ 0
20 e−^5000 x^ dx =
e−^5000 x
∣∣ ∣∣
∞ 0
(e∞^ − e^0 )
= 0. 004 C = 4000 μC
AP 1.4 Recall from Eq. (1.2) that current is the time rate of change of charge, or i = dqdt. In this problem we are given an expression for the charge, and asked to find the maximum current. First we will find an expression for the current using Eq. (1.2):
i = dq dt
d dt
[ 1
α^2
( (^) t α
α^2
) e−αt
]
d dt
α^2
) − d dt
( (^) t α e−αt
) − d dt
α^2 e−αt
)
α e−αt^ − α t α e−αt
) −
( −α
α^2 e−αt
)
( −
α
α
) e−αt
= te−αt
Now that we have an expression for the current, we can find the maximum value of the current by setting the first derivative of the current to zero and solving for t: di dt
d dt (te−αt) = e−αt^ + t(−α)eαt^ = (1 − αt)e−αt^ = 0
Since e−αt^ never equals 0 for a finite value of t, the expression equals 0 only when (1 − αt) = 0. Thus, t = 1/α will cause the current to be maximum. For this value of t, the current is
i =
α e−α/α^ =
α e−^1
Remember in the problem statement, α = 0. 03679. Using this value for α,
i =
e−^1 ∼= 10 A
1–4 CHAPTER 1. Circuit Variables
Substitute the expression for power, p, above. Note that to find the total energy, we let t → ∞ in the integral. Thus we have
w =
∫ (^) ∞ 0
2 × 105 e−^10 ,^000 x^ dx =
2 × 105
e−^10 ,^000 x
∣∣ ∣∣ ∣
∞
0
=
2 × 105
(e−∞^ − e^0 ) =
2 × 105
2 × 105
= 20 J
AP 1.7 At the Oregon end of the line the current is leaving the upper terminal, and thus entering the lower terminal where the polarity marking of the voltage is negative. Thus, using the passive sign convention, p = −vi. Substituting the values of voltage and current given in the figure,
p = −(800 × 103 )(1. 8 × 103 ) = − 1440 × 106 = − 1440 MW
Thus, because the power associated with the Oregon end of the line is negative, power is being generated at the Oregon end of the line and transmitted by the line to be delivered to the California end of the line.
Chapter Problems
P 1.1 To begin, we calculate the number of pixels that make up the display:
npixels = (1280)(1024) = 1, 310 , 720 pixels
Each pixel requires 24 bits of information. Since 8 bits comprise a byte, each pixel requires 3 bytes of information. We can calculate the number of bytes of information required for the display by multiplying the number of pixels in the display by 3 bytes per pixel:
nbytes = 1 , 310 , 720 pixels 1 display
3 bytes 1 pixel = 3, 932 , 160 bytes/display
Finally, we use the fact that there are 106 bytes per MB:
3 , 932 , 160 bytes 1 display
1 MB
106 bytes = 3. 93 MB/display
Problems 1–
P 1.2 c = 3 × 108 m/s so
c = 1. 5 × 108 m/s
- 5 × 108 m 1 s
5 × 106 m x s
so x =
5 × 106
1. 5 × 108
= 33. 3 ms
P 1.3 We can set up a ratio to determine how long it takes the bamboo to grow 10 μm First, recall that 1 mm = 10^3 μm. Let’s also express the rate of growth of bamboo using the units mm/s instead of mm/day. Use a product of ratios to perform this conversion: 250 mm 1 day
1 day 24 hours
1 hour 60 min
1 min 60 sec
mm/s
Use a ratio to determine the time it takes for the bamboo to grow 10 μm: 10 / 3456 × 10 −^3 m 1 s
10 × 10 −^6 m x s so x =
10 × 10 −^6
10 / 3456 × 10 −^3
= 3. 456 s
P 1.4 Volume = area × thickness
106 = (10 × 106 )(thickness)
⇒ thickness =
10 × 106
= 0. 10 mm
P 1.
300 × 109 dollars 1 year
100 pennies 1 dollar
1 year
- 25 days
1 day 24 hr
1 hr 3600 s
- 5 mm 1 penny
1 m 1000 mm = 1426 m/s
P 1.6 Our approach is as follows: To determine the area of a bit on a track, we need to know the height and width of the space needed to store the bit. The height of the space used to store the bit can be determined from the width of each track on the disk. The width of the space used to store the bit can be determined by calculating the number of bits per track, calculating the circumference of the inner track, and dividing the number of bits per track by the circumference of the track. The calculations are shown below.
Width of track = 1 in 77 tracks
25 , 400 μm in = 329. 87 μm/track
Bits on a track =
1.4 MB
2 sides
8 bits byte
1 side 77 tracks = 72, 727. 273 bits/track
Circumference of inner track = 2π(1/ 2 ′′)(25, 400 μm/in) = 79, 796. 453 μm
Width of bit on inner track = 79 , 796. 453 μm 72 , 727. 273 bits
= 1. 0972 μm/bit
Area of bit on inner track = (1.0972)(329.87) = 361. 934 μm^2
Problems 1–
P 1.11 p = (6)(100 × 10 −^3 ) = 0. 6 W; 3 hr · 3600 s 1 hr = 10, 800 s
w(t) =
∫ (^) t 0 p dt w(10,800) =
∫ (^10) , 800 0
- 6 dt = 0.6(10,800) = 6480 J
P 1.12 Assume we are standing at box A looking toward box B. Then, using the passive sign convention p = vi, since the current i is flowing into the + terminal of the voltage v. Now we just substitute the values for v and i into the equation for power. Remember that if the power is positive, B is absorbing power, so the power must be flowing from A to B. If the power is negative, B is generating power so the power must be flowing from B to A. [a] p = (20)(15) = 300 W 300 W from A to B [b] p = (100)(−5) = − 500 W 500 W from B to A [c] p = (−50)(4) = − 200 W 200 W from B to A [d] p = (−25)(−16) = 400 W 400 W from A to B
P 1.13 [a]
p = vi = (−20)(5) = − 100 W Power is being delivered by the box. [b] Leaving [c] Gaining
P 1.14 [a] p = vi = (−20)(−5) = 100 W, so power is being absorbed by the box.
[b] Entering [c] Losing
P 1.15 [a] In Car A, the current i is in the direction of the voltage drop across the 12 V battery(the current i flows into the + terminal of the battery of Car A). Therefore using the passive sign convention, p = vi = (−40)(12) = − 480 W. Since the power is negative, the battery in Car A is generating power, so Car B must have the ”dead” battery.
1–8 CHAPTER 1. Circuit Variables
[b] w(t) =
∫ (^) t 0 p dx; 1. 5 min = 1. 5 · 60 s 1 min = 90 s
w(90) =
∫ (^90) 0 480 dx
w = 480(90 − 0) = 480(90) = 43, 200 J = 43. 2 kJ
P 1.16 p = vi; w =
∫ (^) t 0
p dx Since the energy is the area under the power vs. time plot, let us plot p vs. t.
Note that in constructing the plot above, we used the fact that 60 hr = 216, 000 s = 216 ks
p(0) = (6)(15 × 10 −^3 ) = 90 × 10 −^3 W
p(216 ks) = (4)(15 × 10 −^3 ) = 60 × 10 −^3 W
w = (60 × 10 −^3 )(216 × 103 ) +
(90 × 10 −^3 − 60 × 10 −^3 )(216 × 103 ) = 16. 2 kJ
P 1.17 [a] To find the power at 625 μs, we substitute this value of time into both the equations for v(t) and i(t) and multiply the resulting numbers to get p(625 μs): v(625 μs) = 50e−1600(625×^10 − (^6) ) − 50 e−400(625×^10 − (^6) ) = 18. 394 − 38 .94 = − 20. 546 V
i(625 μs) = 5 × 10 −^3 e−1600(625×^10 − (^6) ) − 5 × 10 −^3 e−400(625×^10 − (^6) )
= 0. 0018394 − 0 .003894 = − 0. 0020546 A
p(625 μs) = (− 20 .546)(− 0 .0020546) = 42. 2 mW
[b] To find the energy at 625 μs, we need to integrate the equation for p(t) from 0 to 625 μs. To start, we need an expression for p(t): p(t) = v(t)i(t) = (50)(5 × 10 −^3 )(e−^1600 t^ − e−^400 t)(e−^1600 t^ − e−^400 t)
1–10 CHAPTER 1. Circuit Variables
P 1.19 [a] 0 s ≤ t < 1 s:
v = 5 V; i = 20t A; p = 100t W 1 s < t ≤ 2 s: v = 0 V; i = 20 A; p = 0 W 2 s ≤ t < 3 s: v = 0 V; i = 20 A; p = 0 W 3 s < t ≤ 4 s: v = − 5 V; i = 80 − 20 t A; p = −400 + 100t W 4 s ≤ t < 5 s: v = − 5 V; i = 80 − 20 t A; p = −400 + 100t W 5 s < t ≤ 6 s: v = 5 V; i = −120 + 20t A; p = −600 + 100t W 6 s ≤ t < 7 s: v = 5 V; i = −120 + 20t A; p = −600 + 100t W t > 7 s: v = 0 V; i = 20 A; p = 0 W
[b] Calculate the area under the curve from zero up to the desired time: w(1) = 12 (1)(100) = 50 J w(6) = 12 (1)(100) − 12 (1)(100) + 12 (1)(100) − 12 (1)(100) = 0 J w(10) = 12 (1)(100) − 12 (1)(100) + 12 (1)(100) − 12 (1)(100) + 12 (1)(100) = 50 J
P 1.20 [a] p = vi = (100e−^500 t)(0. 02 − 0. 02 e−^500 t) = (2e−^500 t^ − 2 e−^1000 t) W
dp dt = − 1000 e−^500 t^ + 2000e−^1000 t^ = 0 so 2 e−^1000 t^ = e−^500 t
2 = e^500 t^ so ln 2 = 500t thus p is maximum at t = 1.4 ms
Problems 1–
pmax = p(1. 4 ms) = 0. 5 W
[b] w =
∫ (^) ∞ 0 [2e−^500 t^ − 2 e−^1000 t] dt =
[ 2
e−^500 t^ −
e−^1000 t
∣∣ ∣∣
∞ 0
]
= 2 mJ
P 1.21 [a] p = vi = 200 cos(500πt)4.5 sin(500πt) = 450 sin(1000πt) W Therefore, pmax = 450 W [b] pmax(extracting) = 450 W [c] pavg =
4 × 10 −^3
∫ (^4) × 10 − 3 0 450 sin(1000πx) dx =
4 × 10 −^3
[ (^) − cos 1000πt 1000 π
] 4 × 10 −^3 0
=
4 π [cos 4π − cos 0] =
4 π
[1 − 1] = 0 W
[d] pavg =
4 π
[cos 15π − cos 0] =
4 π
[− 1 − 1] =
4 π
= 71. 62 W
P 1.22 [a] q = area under i vs. t plot
=
= 15 ,000 + 30,000 + 30,000 + 120,000 + 20,000 = 215, 000 C
[b] w =
∫ pdt =
∫ vi dt v = 0. 2 × 10 −^3 t + 8 0 ≤ t ≤ 20 ks 0 ≤ t ≤ 5000 s i = 20 − 1. 2 × 10 −^3 t p = (8 + 0. 2 × 10 −^3 t)(20 − 1. 2 × 10 −^3 t) = 160 − 5. 6 × 10 −^3 t − 2. 4 × 10 −^7 t^2 w 1 =
∫ (^5000) 0 (160 − 5. 6 × 10 −^3 t − 2. 4 × 10 −^7 t^2 ) dt
=
( 160 t −
5. 6 × 10 −^3
t^2 −
2. 4 × 10 −^7
t^3
)∣∣ ∣∣ ∣
5000
0
= 720 kJ 5000 ≤ t ≤ 15 , 000 s i = 17 − 0. 6 × 10 −^3 t p = (8 + 0. 2 × 10 −^3 t)(17 − 0. 6 × 10 −^3 t) = 136 − 1. 4 × 10 −^3 t − 1. 2 × 10 −^7 t^2 w 2 =
∫ (^15) , 000 5000 (136 − 1. 4 × 10 −^3 t − 1. 2 × 10 −^7 t^2 ) dt
=
( 136 t −
1. 4 × 10 −^3
t^2 −
1. 2 × 10 −^7
t^3
)∣∣ ∣∣ ∣
15 , 000
5000
= 1090 kJ
Problems 1–
Use a calculator to find the two solutions to this quadratic equation: t 1 = 8. 453 s; t 2 = 31. 547 s Now we must find which of these two times gives the minimum power by substituting each of these values for t into the equation for p(t): p(t 1 ) = (8. 453 − 0 .025(8.453)^2 )(4 − 0. 2 · 8 .453) = 15. 396 W p(t 2 ) = (31. 547 − 0 .025(31.547)^2 )(4 − 0. 2 · 31 .547) = − 15. 396 W Therefore, maximum power is being delivered at t = 8. 453 s.
[b] The maximum power was calculated in part (a) to determine the time at which the power is maximum: pmax = 15. 396 W (delivered)
[c] As we saw in part (a), the other “maximum” power is actually a minimum, or the maximum negative power. As we calculated in part (a), maximum power is being extracted at t = 31. 547 s.
[d] This maximum extracted power was calculated in part (a) to determine the time at which power is maximum: pmaxext = 15. 396 W (extracted)
[e] w =
∫ (^) t 0 pdx =
∫ (^) t 0 (4x − 0. 3 x^2 + 0. 005 x^3 )dx = 2t^2 − 0. 1 t^3 + 0. 00125 t^4 w(0) = 0 J w(30) = 112.50 J w(10) = 112.50 J w(40) = 0 J w(20) = 200 J To give you a feel for the quantities of voltage, current, power, and energy and their relationships among one another, they are plotted below:
1–14 CHAPTER 1. Circuit Variables
1–16 CHAPTER 1. Circuit Variables
P 1.27 [a] From the diagram and the table we have
pa = −vaia = −(900)(− 22 .5) = 20, 250 W pb = −vbib = −(105)(− 52 .5) = 5512. 5 W pc = −vcic = −(−600)(−30) = − 18 , 000 W pd = vdid = (585)(− 52 .5) = − 30 , 712. 5 W pe = −veie = −(−120)(30) = 3600 W pf = vf if = (300)(60) = 18, 000 W pg = −vgig = −(585)(82.5) = − 48 , 262. 5 W ph = −vhih = −(−165)(82.5) = 13, 612. 5 W ∑ Pdel = 18 ,000 + 30, 712 .5 + 48, 262 .5 = 96, 975 W ∑ Pabs = 20 ,250 + 5512.5 + 3600 + 18,000 + 13, 612 .5 = 60, 975 W Therefore,
∑ Pdel =
∑ Pabs and the subordinate engineer is correct. [b] The difference between the power delivered to the circuit and the power absorbed by the circuit is 96 , 975 − 60 ,975 = 36, 000 One-half of this difference is 18 , 000 W, so it is likely that pc or pf is in error. Either the voltage or the current probably has the wrong sign. (In Chapter 2, we will discover that using KCL at the top node, the current ic should be 30 A, not − 30 A!) If the sign of pc is changed from negative to positive, we can recalculate the power delivered and the power absorbed as follows: ∑ Pdel = 30 , 712 .5 + 48, 262 .5 = 78, 975 W ∑ Pabs = 20 ,250 + 5512.5 + 18,000 + 3600 + 18,000 + 13, 612 .5 = 78, 975 W Now the power delivered equals the power absorbed and the power balances for the circuit.
P 1.28 pa = vaia = (9)(1.8) = 16. 2 W
pb = −vbib = −(−15)(1.5) = 22. 5 W pc = −vcic = −(45)(− 0 .3) = 13. 5 W pd = −vdid = −(54)(− 2 .7) = 145. 8 W pe = veie = (−30)(−1) = 30 W pf = −vf if = −(−240)(4) = 960 W pg = −vgig = −(294)(4.5) = − 1323 W ph = vhih = (−270)(− 0 .5) = 135 W
Problems 1–
Therefore, ∑ Pabs = 16.2 + 22.5 + 13.5 + 145.8 + 3 − +960 + 135 = 1323 W
∑ Pdel = 1323 W
∑ Pabs =
∑ Pdel
Thus, the interconnection satisfies the power check
P 1.29 pa = vaia = (−160)(−10) = 1600 W
pb = vbib = (−100)(−20) = 2000 W pc = −vcic = −(−60)(6) = 360 W pd = vdid = (800)(−50) = − 40 , 000 W pe = −veie = −(800)(−20) = 16, 000 W pf = −vf if = −(−700)(14) = 9800 W pg = −vgig = −(640)(−16) = 10, 240 W ∑ ∑^ Pdel^ = 40,^000 W Pabs = 1600 + 2000 + 360 + 16,000 + 9800 + 10,000 = 40, 000 W Therefore,
∑ Pdel =
∑ Pabs = 40, 000 W
P 1.30 [a] From an examination of reference polarities, the following elements employ the passive convention: a, c, e, and f. [b] pa = − 56 W pb = − 14 W pc = 150 W pd = − 50 W pe = − 18 W pf = − 12 W ∑ Pabs = 150 W;
∑ Pdel = 56 + 14 + 50 + 18 + 12 = 150 W.
