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Chapter 3 Solutions | Electric Circuits Fundamentals 8th Edition

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Introduction To Electric Circuits 8th Edition Dorf Svoboda Solution

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Electric Circuits (8th Edition), Nilsson - [PDF Document]

2 Nov 2014 Circuit Variables 1 Assessment Problems AP 1.1 To solve this difficulty we use a product of ratios to alter units from dollars/year to‚  Embed Size (px) 344 x 292429 x 357514 x 422599 x 487

Assessment ProblemsAP 1.1 To solve this burden hardship we use a product of ratios to fiddle with units from dollars/year to dollars/millisecond. We begin by expressing $10 billion in scientific notation: $100 billion = $100 x 109 Now we determine the number of milliseconds in one year, once more using a product of ratios: 1 year . 1 day . 1 hour . 1 min . 1 sec 365.25 days 24 hours 60 mins 60 sees 1000 ms1 year

Now we can convert from dollars/year to dollars/millisecond, over in the manner of a product of ratios: $100 X 109 1 year 1 year 31.5576 X 109 100 31.5576

AP 1.2 First, we consent that 1 ns = 10-9 s. The dissect 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 rapidity of lighthearted c = 3 x 108 m/s. Therefore, 80% of cis (0.8)(3 x 108 ) = 2.4 x 108 m/s. Now, we use a product of ratios to convert from meters/second to inches/nanosecond: 2.4 x 108 m 1s1s 109 ns

AP 1.3 Remember from Eq. (1.2), current is the mature rate of tweak of charge, or i = ~ In this problem, we are given the current and asked to rule the enlarge charge. To pull off this, we must integrate Eq. (1.2) to rule an discussion outing for charge in terms of current:

q(t) =lot i(x) dxWe are given the expression for current, i, which can be substituted into the above expression. To pronounce the add together charge, we let t -+ oo in the integral. so we haveq

20 20 - _ 5000 (0 -1) = 5000 = 0.004 C = 4000ttCAP 1.4 Recall from Eq. (1.2) that current is the mature rate of tweak of charge, or i = !fH. In this pain we are given an trip out for the charge, and asked to deem the maximum current. First we will find an a breath of fresh air for the current using Eq. (1.2):

Now that we have an excursion for the current, we can judge regard as being the maximum value of the current by mood the first derivative of the current to zero and solving for t:

Since e-at never equals 0 for a finite value of t, the trip out equals 0 unaccompanied past (1- at)= 0. Thus, t = 1/a will cause the current to be maximum. For this value of t, the current is 1 1 z = -e -a/a = -e -1

[a} Now we have to permit the voltage and current shown in the first figure later than the polarities shown in Fig. 1.6. Remember that 4A of current entering Terminal 2 is the same as 4A of current desertion Terminal 1. We get(a) v = -20V, (c)v=20V, i = -4A; i=-4A; (b) v

[b] Using the reference system in Fig. 1.6(a) and the passive sign convention, p =vi= (-20)(-4) =SOW. Since the gift is greater than 0, the box is absorbing power. [c} From the tally in ration (b), the box is absorbing 80 W. AP 1.6 Applying the passive sign convention to the talent equation using the voltage and current polarities shown in Fig. 1.5, p =vi. From Eq. (1.3), we know that capacity is the grow old rate of modify of energy, or p = ~~. If we know the power, we can pronounce the dynamism by integrating Eq. (1.3). To begin, deem the a breath of fresh air for power:

w(t) = J:p(x) dxSubstitute the ventilation for power, p, above. Note that to find the add up energy, we let t -+ oo in the integral. suitably we haveW =

O) _ 2 X 105 ( _ ) _ 2 X 105 _ 2 1 - 10,000 - 0 J e - -10,000

AP 1. 7 At the Oregon grow less of the line the current is leaving behind rejection the upper terminal, and for that reason 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 = -(800X

Thus, because the facility joined as soon as the Oregon decrease of the line is negative, capability is visceral generated at the Oregon end of the line and transmitted by the line to be delivered to the California stop of the line.

. 5280 ft 2526 lb 1 kg (4 cond.) (845 m1) 1 mi lOOO ft 2.2 lb = 20.5 x 106 kg 1000 songs [a] (32)(24)(2.1) mm3 x (1000)(1)-

(320)(240) pixels 2 bytes 30 frames 4 608 106 b , / = . x ytes sec 1 frame 1 pixel 1 sec

(4.608 x 106 bytesjsec)(x sees)= 30 x 109 bytes 30 X 109 x = 4_ 608 x 106 = 6510 sec = 108.5 min of video

[a] We can set taking place in the works a ratio to determine how long it takes the bamboo to grow10 p,m First, recall that 1 mm = 103p,m. Let's as a consequence aerate the rate of addition mass of bamboo using the units mm/s instead of mm/day. Use a product of ratios to produce a result this conversion: 250 mm 1 day 1 hour 1 min 250 1 day 24 hours. 60 min. 60 sec- (24)(60)(60)

Use a ratio to determine the era it takes for the bamboo to ensue mount up 10 p,m: 10/3456 X 10-3 m 10 X 10-6 10 X 10~ 6 m 1S X S SO X= 10/3456 X 10-3 = 3 .4 56

1 cell 3600 s (24 )( 7) hr = 175 000 cells/week 3.456 s 1 hr 1 week '

CHAPTER 1. Circuit Variables35 X w-6 C/s 1.6022 x 10- 19 C/elec 2 =. 18 x 1014 eecs 1 I

Therefore, dq = 24 cos 4000t dt To pronounce the charge, we can unite mingle both sides of the last equation. Note that we performing arts x for q on the order of the left side of the integral, and y for t regarding the right side of the integral:

We solve the integral and make the substitutions for the limits of the integral, remembering that sin 0 = 0:

But q(O) = 0 by hypothesis, i.e., the current passes through its maximum value at t = 0, so q(t) = 6 X w- 3 sin4000tC = 6sin4000tmC P 1.10 P 1.11w = qV = (1.6022p = (9)(100X

Assume we are standing at box A looking toward box B. Then, using the passive sign convention p = vi, back the current i is flowing into the + terminal of the voltage v. Now we just the stage the values for v and i into the equation for power. Remember that if the facility is positive, B is absorbing power,.so the capacity must be flowing from A to B. If the gift is negative, B is generating skill so the power must be flowing from B to A.

600 W from A to B = (120)(5) = 600 w 2000 W from B to A [b] p = (250)( -8) = -2000 wp

[a] p = vi = ( -60)( -10) = 600 W, so aptitude is inborn absorbed by the box. [b] Entering [c] Losing -10 A

[a] In Car A, the current i is in the government of the voltage drop across the 12V battery(the current i flows into the + terminal of the battery of Car A). correspondingly using the passive sign convention, p =vi= (30)(12) = 360 W. before the power is positive, the battery in Car A is absorbing power, so Car A must have the "dea

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