Circuit Variables. AP 1.1 Use a product of ratios to convert two-thirds the speed of light from meters per second to miles per second: 1 ft 12 in

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1 Circui Variables 1 Assessmen Problems AP 1.1 Use a produc of raios o conver wo-hirds he speed of ligh from meers per second o miles per second: ( ) m 3 1 s 1 cm 1 m 1 in 2.54 cm 1 f 12 in 1 mile 124, miles = 528 fee 1 s Now se up a proporion o deermine how long i akes his signal o ravel 11 miles: 124, miles 1 s Therefore, x = = 11 miles x s , =.885 = s = 8.85 ms AP 1.2 To solve his problem we use a produc of raios o change unis from dollars/year o dollars/millisecond. We begin by expressing $1 billion in scienific noaion: $1 billion = $1 1 9 Now we deermine he number of milliseconds in one year, again using a produc of raios: 1 year days 1 day 24 hours 1 hour 6 mins 1 min 6 secs 1 sec 1 ms = 1 year ms Now we can conver from dollars/year o dollars/millisecond, again wih a produc of raios: $ year 1 year ms = = $3.17/ms 215 Pearson Educaion, Inc., Upper Saddle River, NJ. All righs reserved. This publicaion is proeced by Copyrigh and wrien permission should be obained recording, or likewise. For informaion regarding permission(s), wrie o: Righs and Permissions Deparmen, Pearson Educaion, Inc., Upper Saddle River, NJ

2 1 2 CHAPTER 1. Circui Variables AP 1.3 Remember from Eq. (1.2), curren is he ime rae of change of charge, or i = dq In his problem, we are given he curren and asked o find he oal d charge. To do his, we mus inegrae Eq. (1.2) o find an expression for charge in erms of curren: q() = i(x) dx We are given he expression for curren, i, which can be subsiued ino he above expression. To find he oal charge, we le in he inegral. Thus we have q oal = 2e 5x dx = 2 = 2 2 ( 1) = e 5x =.4 C = 4µC = 2 5 (e e ) AP 1.4 Recall from Eq. (1.2) ha curren is he ime rae of change of charge, or i = dq. In his problem we are given an expression for he charge, and asked o d find he maximum curren. Firs we will find an expression for he curren using Eq. (1.2): i = dq d = d [ ( 1 d α 2 α + 1 ) ] e α α 2 ( ) ( ) ( ) 1 1 α 2 α e α α 2e α = d d d d d d ( 1 = α e α α ) α e α ( α 1 ) α 2e α = ( 1 α ) e α α = e α Now ha we have an expression for he curren, we can find he maximum value of he curren by seing he firs derivaive of he curren o zero and solving for : di d = d d (e α ) = e α + ( α)e α = (1 α)e α = Since e α never equals for a finie value of, he expression equals only when (1 α) =. Thus, = 1/α will cause he curren o be maximum. For his value of, he curren is i = 1 α e α/α = 1 α e Pearson Educaion, Inc., Upper Saddle River, NJ. All righs reserved. This publicaion is proeced by Copyrigh and wrien permission should be obained recording, or likewise. For informaion regarding permission(s), wrie o: Righs and Permissions Deparmen, Pearson Educaion, Inc., Upper Saddle River, NJ 7458.

3 Problems 1 3 Remember in he problem saemen, α = Using his value for α, i = e 1 = 1 A AP 1.5 Sar by drawing a picure of he circui described in he problem saemen: Also skech he four figures from Fig. 1.6: [a] Now we have o mach he volage and curren shown in he firs figure wih he polariies shown in Fig Remember ha 4A of curren enering Terminal 2 is he same as 4A of curren leaving Terminal 1. We ge (a) v = 2V, i = 4A; (b) v = 2V, i = 4A (c) v = 2V, i = 4A; (d) v = 2V, i = 4A [b] Using he reference sysem in Fig. 1.6(a) and he passive sign convenion, p = vi = ( 2)( 4) = 8 W. Since he power is greaer han, he box is absorbing power. [c] From he calculaion in par (b), he box is absorbing 8 W. AP 1.6 [a] Applying he passive sign convenion o he power equaion using he volage and curren polariies shown in Fig. 1.5, p = vi. To find he ime a which he power is maximum, find he firs derivaive of he power wih respec o ime, se he resuling expression equal o zero, and solve for ime: p = (8,e 5 )(15e 5 ) = e 1 dp d = e e 1 = 215 Pearson Educaion, Inc., Upper Saddle River, NJ. All righs reserved. This publicaion is proeced by Copyrigh and wrien permission should be obained recording, or likewise. For informaion regarding permission(s), wrie o: Righs and Permissions Deparmen, Pearson Educaion, Inc., Upper Saddle River, NJ 7458.

4 1 4 CHAPTER 1. Circui Variables Therefore, = Solving, = = = 2 ms [b] The maximum power occurs a 2 ms, so find he value of he power a 2 ms: p(.2) = (.2) 2 e 2 = mw [c] From Eq. (1.3), we know ha power is he ime rae of change of energy, or p = dw/d. If we know he power, we can find he energy by inegraing Eq. (1.3). To find he oal energy, he upper limi of he inegral is infiniy: w oal = = x 2 e 1x dx ( 1) 3 e 1x [( 1) 2 x 2 2( 1)x + 2) = ( 1) 3 e ( + 2) = 2.4 mj AP 1.7 A he Oregon end of he line he curren is leaving he upper erminal, and hus enering he lower erminal where he polariy marking of he volage is negaive. Thus, using he passive sign convenion, p = vi. Subsiuing he values of volage and curren given in he figure, p = (8 1 3 )( ) = = 144 MW Thus, because he power associaed wih he Oregon end of he line is negaive, power is being generaed a he Oregon end of he line and ransmied by he line o be delivered o he California end of he line. 215 Pearson Educaion, Inc., Upper Saddle River, NJ. All righs reserved. This publicaion is proeced by Copyrigh and wrien permission should be obained recording, or likewise. For informaion regarding permission(s), wrie o: Righs and Permissions Deparmen, Pearson Educaion, Inc., Upper Saddle River, NJ 7458.

5 Problems 1 5 Chaper Problems P 1.1 P 1.2 ( )(54) 1 9 = 14.4 gigawa-hours (48)(32) pixels 1 frame 2 byes 1 pixel 3 frames 1 sec = byes/sec ( byes/sec)(x secs) = byes x = = 3728 sec = 62 min 1 hour of video P 1.3 [a] [b] 2, phoos (11)(15)(1) mm = x phoos 3 1 mm 3 x = (2,)(1) (11)(15)(1) = 121 phoos byes (11)(15)(1) mm 3 = x byes (.2) 3 mm 3 x = (16 23 )(.8) (11)(15)(1) = 832,963 byes P 1.4 P 1.5 (4 cond.) (845 mi) 528 f 1 mi Volume = area hickness 2526 lb 1 f 1 kg 2.2 lb = kg Conver values o millimeers, noing ha 1 m 2 = 1 6 mm = (1 1 6 )(hickness) hickness = 16 =.1 mm 1 16 P 1.6 [a] We can se up a raio o deermine how long i akes he bamboo o grow 1µm Firs, recall ha 1 mm = 1 3 µm. Le s also express he rae of growh of bamboo using he unis mm/s insead of mm/day. Use a produc of raios o perform his conversion: 25 mm 1 day 1 day 24 hours 1 hour 6 min 1 min 6 sec = 25 (24)(6)(6) = mm/s Use a raio o deermine he ime i akes for he bamboo o grow 1µm: 1/ m 1 s = m x s so x = = s 1/ Pearson Educaion, Inc., Upper Saddle River, NJ. All righs reserved. This publicaion is proeced by Copyrigh and wrien permission should be obained recording, or likewise. For informaion regarding permission(s), wrie o: Righs and Permissions Deparmen, Pearson Educaion, Inc., Upper Saddle River, NJ 7458.

6 1 6 CHAPTER 1. Circui Variables P 1.7 [b] 1 cell lengh s 36 s 1 hr (24)(7) hr 1 week = 175, cell lenghs/week [a] Firs we use Eq. (1.2) o relae curren and charge: i = dq d =.125e 25 Therefore, dq =.125e 25 d To find he charge, we can inegrae boh sides of he las equaion. Noe ha we subsiue x for q on he lef side of he inegral, and y for on he righ side of he inegral: q() q() dx =.125 e 25y dy We solve he inegral and make he subsiuions for he limis of he inegral: q() q() =.125 e 25y = (1 e 25 ) 25 Bu q() = by hypohesis, so q() = 5(1 e 25 )µc [b] As, q T = 5µC. [c] q( ) = (5 1 6 )(1 e ( 25)(.5) ) = µC. P 1.8 Firs we use Eq. (1.2) o relae curren and charge: i = dq d = 2cos 5 Therefore, dq = 2 cos 5 d To find he charge, we can inegrae boh sides of he las equaion. Noe ha we subsiue x for q on he lef side of he inegral, and y for on he righ side of he inegral: q() q() dx = 2 cos5y dy We solve he inegral and make he subsiuions for he limis of he inegral, remembering ha sin = : sin 5y q() q() = 2 5 = sin 5 sin 5() = sin5 Bu q() = by hypohesis, i.e., he curren passes hrough is maximum value a =, so q() = sin5 C = 4sin 5 mc 215 Pearson Educaion, Inc., Upper Saddle River, NJ. All righs reserved. This publicaion is proeced by Copyrigh and wrien permission should be obained recording, or likewise. For informaion regarding permission(s), wrie o: Righs and Permissions Deparmen, Pearson Educaion, Inc., Upper Saddle River, NJ 7458.

7 Problems 1 7 P 1.9 [a] Firs we use Eq. (1.2) o relae curren and charge: i = dq d = 4e 5 Therefore, dq = 4e 5 d P 1.1 n = To find he charge, we can inegrae boh sides of he las equaion. Noe ha we subsiue x for q on he lef side of he inegral, and y for on he righ side of he inegral: q() q() dx = 4 ye 5y dy We solve he inegral and make he subsiuions for he limis of he inegral: q() q() = 4 e 5y ( 5y 1) = e 5 ( 5 1) ( 5) 2 = (1 5e 5 e 5 ) Bu q() = by hypohesis, so q() = 16(1 5e 5 e 5 )µc [b] q(.1) = (16)[1 5(.1)e 5(.1) e 5(.1) = 14.4µC C/s C/elec = elec/s P 1.11 P 1.12 w = qv = ( )(6) = =.961 aj [a] p = vi = (4)( 1) = 4 W Power is being delivered by he box. [b] Enering [c] Gaining P 1.13 [a] p = vi = ( 6)( 1) = 6 W, so power is being absorbed by he box. [b] Enering 215 Pearson Educaion, Inc., Upper Saddle River, NJ. All righs reserved. This publicaion is proeced by Copyrigh and wrien permission should be obained recording, or likewise. For informaion regarding permission(s), wrie o: Righs and Permissions Deparmen, Pearson Educaion, Inc., Upper Saddle River, NJ 7458.

8 1 8 CHAPTER 1. Circui Variables [c] Losing P 1.14 Assume we are sanding a box A looking oward box B. Use he passive sign convenion o ge p = vi, since he curren i is flowing ino he + erminal of he volage v. Now we jus subsiue he values for v and i ino he equaion for power. Remember ha if he power is posiive, B is absorbing power, so he power mus be flowing from A o B. If he power is negaive, B is generaing power so he power mus be flowing from B o A. [a] p = (3)(6) = 18 W [b] p = ( 2)( 8) = 16 W [c] p = ( 6)(4) = 24 W [d] p = (4)( 9) = 36 W 18 W from A o B 16 W from A o B 24 W from B o A 36 W from B o A P 1.15 [a] In Car A, he curren i is in he direcion of he volage drop across he 12 V baery(he curren i flows ino he + erminal of he baery of Car A). Therefore using he passive sign convenion, p = vi = (3)(12) = 36 W. Since he power is posiive, he baery in Car A is absorbing power, so Car A mus have he dead baery. [b] w() = w(6) = pdx; 6 P 1.16 p = vi; w = 36dx 1 min = 6 s w = 36(6 ) = 36(6) = 21,6 J = 21.6 kj pdx Since he energy is he area under he power vs. ime plo, le us plo p vs Pearson Educaion, Inc., Upper Saddle River, NJ. All righs reserved. This publicaion is proeced by Copyrigh and wrien permission should be obained recording, or likewise. For informaion regarding permission(s), wrie o: Righs and Permissions Deparmen, Pearson Educaion, Inc., Upper Saddle River, NJ 7458.

9 Problems 1 9 Noe ha in consrucing he plo above, we used he fac ha 4 hr = 144, s = 144 ks p() = (1.5)(9 1 3 ) = W p(144 ks) = (1)(9 1 3 ) = W w = (9 1 3 )( ) ( )( ) = 162 J P 1.17 p = (12)(1 1 3 ) = 1.2 W; 4 hr 36 s 1 hr = 14,4 s w() = p d w(14,4) = 14,4 1.2d = 1.2(14,4) = kj P 1.18 [a] p = vi = (15e 25 )(.4e 25 ) =.6e 5 W p(.1) =.6e 5(.1) =.6e 5 =.44 = 4.4 mw [b] w oal = p(x)dx =.6e 5x dx =.6 =.12(e e ) =.12 = 1.2 mj 5 e 5x P 1.19 [a] p = vi = (.5e 1 )(75 75e 1 ) = (3.75e e 2 ) W dp d = 375e e 2 = so 2e 2 = e 1 2 = e 1 so ln2 = 1 hus p is maximum a = µs [b] w = p max = p(693.15µs) = mw [3.75e e 2 ] d = = = mj [ e ] 2 e 2 P 1.2 [a] p = vi =.25e 32.5e e 8 p(625µs) = 42.2 mw [b] w() = (.25e 32.5e e 8 ) = e e e 8 µj w(625 µs) = µj [c] w oal = µJ 215 Pearson Educaion, Inc., Upper Saddle River, NJ. All righs reserved. This publicaion is proeced by Copyrigh and wrien permission should be obained recording, or likewise. For informaion regarding permission(s), wrie o: Righs and Permissions Deparmen, Pearson Educaion, Inc., Upper Saddle River, NJ 7458.

10 1 1 CHAPTER 1. Circui Variables P 1.21 [a] p = vi dp d = [(15 + 1)e 75 ](.4e 75 ) = (6 +.4)e 15 = 6e 15 15e 15 (6 +.4) = 9,e 15 Therefore, dp d = when = so p max occurs a =. [b] p max = [(6)() +.4]e =.4 [c] w = = 4 mw P 1.22 [a] p = vi pdx w = 6xe 15x dx +.4e 15x dx = 6e 15x ( 15x 1) +.4 e 15x ( 15) 2 15 When = all he upper limis evaluae o zero, hence 6 w = = 53.33µJ. dp d = [( )e 1 ][( )e 1 ] = e 2 [512, ] = e 2 [1,24, + 124] 2e 2 [512, ] = e 2 [ ,24,] Therefore, dp d = when = so p max occurs a =. [b] p max = e [ ] [c] w = = 512 mw pdx w = 512,x 2 e 2x dx + 124xe 2x dx + = 512,e 2x [4 1 6 x 2 + 4x + 2] e 2x ( 2x 1) e 2x 2.512e 2x dx 215 Pearson Educaion, Inc., Upper Saddle River, NJ. All righs reserved. This publicaion is proeced by Copyrigh and wrien permission should be obained recording, or likewise. For informaion regarding permission(s), wrie o: Righs and Permissions Deparmen, Pearson Educaion, Inc., Upper Saddle River, NJ 7458.

11 Problems 1 11 When all he upper limis evaluae o zero, hence w = (512,)(2) w = = 64µJ. P 1.23 [a] We can find he ime a which he power is a maximum by wriing an expression for p() = v()i(), aking he firs derivaive of p() and seing i o zero, hen solving for. The calculaions are shown below: p = <, p = > 4 s p = vi = (1.25)(4.2) = W 4 s dp = =.15( ) d dp = when = d 1 = s; 2 = s (using he polynomial solver on your calculaor) p( 1 ) = 4(8.453).3(8.453) 2 +.5(8.453) 3 = W p( 2 ) = 4(31.547).3(31.547) 2 +.5(31.547) 3 = W Therefore, maximum power is being delivered a = s. [b] The maximum power was calculaed in par (a) o deermine he ime a which he power is maximum: p max = W (delivered) [c] As we saw in par (a), he oher maximum power is acually a minimum, or he maximum negaive power. As we calculaed in par (a), maximum power is being exraced a = s. [d] This maximum exraced power was calculaed in par (a) o deermine he ime a which power is maximum: p max = W (exraced) [e] w = pdx = (4x.3x 2 +.5x 3 )dx = w() = J w(3) = J w(1) = J w(4) = J w(2) = 2 J To give you a feel for he quaniies of volage, curren, power, and energy and heir relaionships among one anoher, hey are ploed below: 215 Pearson Educaion, Inc., Upper Saddle River, NJ. All righs reserved. This publicaion is proeced by Copyrigh and wrien permission should be obained recording, or likewise. For informaion regarding permission(s), wrie o: Righs and Permissions Deparmen, Pearson Educaion, Inc., Upper Saddle River, NJ 7458.

12 1 12 CHAPTER 1. Circui Variables P 1.24 [a] v(1 ms) = 4e 1 sin2 = V i(1 ms) = 5e 1 sin2 = 1.67 A p(1 ms) = vi = W 215 Pearson Educaion, Inc., Upper Saddle River, NJ. All righs reserved. This publicaion is proeced by Copyrigh and wrien permission should be obained recording, or likewise. For informaion regarding permission(s), wrie o: Righs and Permissions Deparmen, Pearson Educaion, Inc., Upper Saddle River, NJ 7458.

13 Problems 1 13 [b] p = vi = 2e 2 sin 2 2 = 2e 2 [ cos4 ] = 1e 2 1e 2 cos 4 w = 1e 2 d 1e 2 cos4 d = 1 e 2 2 { e 2 } 1 [ 2cos 4 + 4sin 4] (2) 2 + (4) 2 [ ] 2 = 5 1 = w = 4 J P 1.25 [a] p = vi = 2 cos(8π) sin(8π) = 1 sin(16π) W Therefore, p max = 1 W [b] p max (exracing) = 1 W [c] p avg = [d] = [ cos16π 16π 1 sin(16π) d ] p avg = [ cos 16π = π = 25 [1 cos4π] = π 1 sin(16π) d ] = 4 [1 cos 25π] = W π P 1.26 [a] q = area under i vs. plo = 1 2 (8)(12,) + (16)(12,) (16)(4) = 48, + 192, + 32, = 272, C [b] w = pd = vi d v = ks 12,s: i = p = w 1 = 12, ( )d = ( )1 3 = 2256 kj 215 Pearson Educaion, Inc., Upper Saddle River, NJ. All righs reserved. This publicaion is proeced by Copyrigh and wrien permission should be obained recording, or likewise. For informaion regarding permission(s), wrie o: Righs and Permissions Deparmen, Pearson Educaion, Inc., Upper Saddle River, NJ 7458.

14 1 14 CHAPTER 1. Circui Variables 12, s 16, s: i = p = w 2 = 16, 12, ( )d = ( )1 3 = kj w T = w 1 + w 2 = = kj P 1.27 [a] s < 1 ms: v = 8 V; i = 25 A; p = 2 W 1 ms < 3 ms: v = 8 V; i =.5 25 A; p = 2 4 W 3 ms < 4 ms: v = V; i = 25 ma; p = W 4 ms < 6 ms: v = 8 V; i = A; p = 2 1 W > 6 ms: v = V; i = 25 ma; p = W [b] Calculae he area under he curve from zero up o he desired ime: w(.1) = 1 (2)(.1) = 1 mj 2 w(.3) = w(.1) 1(2)(.1) + 1 (2)(.1) = 1 mj 2 2 w(.8) = w(.3) 1 (2)(.1) + 1 (2)(.1) = 1 mj Pearson Educaion, Inc., Upper Saddle River, NJ. All righs reserved. This publicaion is proeced by Copyrigh and wrien permission should be obained recording, or likewise. For informaion regarding permission(s), wrie o: Righs and Permissions Deparmen, Pearson Educaion, Inc., Upper Saddle River, NJ 7458.

15 Problems 1 15 P 1.28 [a] [b] i() = ma, 1 ks i() = 15 ma, 1 ks 2 ks i() = ma, 2 ks 3 ks i() =, > 3 ks p = vi = 12i so p() = mw, 1 ks p() = 18 mw, 1 ks 2 ks p() = 3.6 mw, 2 ks 3 ks p() =, > 3 ks 215 Pearson Educaion, Inc., Upper Saddle River, NJ. All righs reserved. This publicaion is proeced by Copyrigh and wrien permission should be obained recording, or likewise. For informaion regarding permission(s), wrie o: Righs and Permissions Deparmen, Pearson Educaion, Inc., Upper Saddle River, NJ 7458.

16 1 16 CHAPTER 1. Circui Variables [c] To find he energy, calculae he area under he plo of he power: w(1 ks) = 1 (.6)(1,) + (1.2)(1,) = 15 kj 2 w(2 ks) = w(1 ks) + (1.8)(1,) = 33 kj w(1 ks) = w(2 ks) + 1 (.6)(1,) + (1.2)(1,) = 48 kj 2 P 1.29 We use he passive sign convenion o deermine wheher he power equaion is p = vi or p = vi and subsiue ino he power equaion he values for v and i, as shown below: p a = v a i a = (4)( ) = 16 mw p b = v b i b = ( 24)( ) = 96 mw p c = v c i c = ( 16)(4 1 3 ) = 64 mw p d = v d i d = ( 8)( ) = 12 mw p e = v e i e = (4)( ) = 1 mw p f = v f i f = (12)( ) = 3 mw Remember ha if he power is posiive, he circui elemen is absorbing power, whereas is he power is negaive, he circui elemen is developing power. We can add he posiive powers ogeher and he negaive powers ogeher if he power balances, hese power sums should be equal: Pdev = = 42 mw; Pabs = = 42 mw Thus, he power balances and he oal power absorbed in he circui is 42 mw. P 1.3 p a = v a i a = ( 3)( ) =.75 W p b = v b i b = (4)( ) = 1.6 W p c = v c i c = (1)(4 1 6 ) =.4 W p d = v d i d = (1)( ) =.15 W p e = v e i e = ( 4)(2 1 6 ) =.8 W p f = v f i f = (4)(5 1 6 ) =.2 W Therefore, Pabs = = 1.95 W Pdel = = 1.95 W = P abs Thus, he inerconnecion does saisfy he power check. 215 Pearson Educaion, Inc., Upper Saddle River, NJ. All righs reserved. This publicaion is proeced by Copyrigh and wrien permission should be obained recording, or likewise. For informaion regarding permission(s), wrie o: Righs and Permissions Deparmen, Pearson Educaion, Inc., Upper Saddle River, NJ 7458.

17 Problems 1 17 P 1.31 [a] From he diagram and he able we have p a = v a i a = (46.16)(6) = W p b = v b i b = (14.16)(4.72) = W p c = v c i c = ( 32)( 6.4) = 24.8 W p d = v d i d = (22)(1.28) = W p e = v e i e = ( 33.6)( 1.68) = W p f = v f i f = (66)(.4) = 26.4 W p g = v g i g = (2.56)(1.28) = W p h = v h i h = (.4)(.4) =.16 W Pdel = = W Pabs = = W Therefore, P del P abs and he subordinae engineer is correc. [b] The difference beween he power delivered o he circui and he power absorbed by he circui is = 52.8 W One-half of his difference is 26.4 W, so i is likely ha p f is in error. Eiher he volage or he curren probably has he wrong sign. (In Chaper 2, we will discover ha using KCL a he node connecing componens f and h, he curren i f should be.4 A, no.4 A!) If he sign of p f is changed from negaive o posiive, we can recalculae he power delivered and he power absorbed as follows: Pdel = = W Pabs = = W Now he power delivered equals he power absorbed and he power balances for he circui. P 1.32 [a] Remember ha if he circui elemen is absorbing power, he power is posiive, whereas if he circui elemen is supplying power, he power is negaive. We can add he posiive powers ogeher and he negaive powers ogeher if he power balances, hese power sums should be equal: P sup = = 25 W; Pabs = = 25 W Thus, he power balances. [b] The curren can be calculaed using i = p/v or i = p/v, wih proper applicaion of he passive sign convenion: 215 Pearson Educaion, Inc., Upper Saddle River, NJ. All righs reserved. This publicaion is proeced by Copyrigh and wrien permission should be obained recording, or likewise. For informaion regarding permission(s), wrie o: Righs and Permissions Deparmen, Pearson Educaion, Inc., Upper Saddle River, NJ 7458.

18 1 18 CHAPTER 1. Circui Variables i a = p a /v a = ( 6)/(4) = 1.5 A i b = p b /v b = ( 5)/( 1) =.5 A i c = p c /v c = (4)/(2) = 2. A i d = p d /v d = ( 6)/(3) = 2. A i e = p e /v e = (1)/( 2) =.5 A i f = p f /v f = (2)/(5) = 4. A i g = p g /v g = ( 125)/( 5) = 2.5 A P 1.33 [a] If he power balances, he sum of he power values should be zero: p oal = = Thus, he power balances. [b] When he power is posiive, he elemen is absorbing power. Since elemens a, b, c, e, and f have posiive power, hese elemens are absorbing power. [c] The volage can be calculaed using v = p/i or v = p/i, wih proper applicaion of he passive sign convenion: v a = p a /i a = (.175)/(.25) = 7 V v b = p b /i b = (.375)/(.75) = 5 V v c = p c /i c = (.15)/(.5) = 3 V v d = p d /i d = (.32)/(.4) = 8 V v e = p e /i e = (.16)/(.2) = 8 V v f = p f /i f = (.12)/(.3) = 4 V v g = p g /i g = (.66)/(.55) = 12 V P 1.34 p a = v a i a = (12)( 1) = 12 W p b = v b i b = (12)(9) = 18 W p c = v c i c = (1)(1) = 1 W p d = v d i d = (1)( 1) = 1 W p e = v e i e = ( 1)( 9) = 9 W p f = v f i f = ( 1)(5) = 5 W p g = v g i g = (12)(4) = 48 W p h = v h i h = ( 22)( 5) = 11 W Pdel = = 228 W 215 Pearson Educaion, Inc., Upper Saddle River, NJ. All righs reserved. This publicaion is proeced by Copyrigh and wrien permission should be obained recording, or likewise. For informaion regarding permission(s), wrie o: Righs and Permissions Deparmen, Pearson Educaion, Inc., Upper Saddle River, NJ 7458.

19 Problems 1 19 Pabs = = 228 W Therefore, P del = P abs = 228 W Thus, he inerconnecion now saisfies he power check. P 1.35 [a] The revised circui model is shown below: [b] The expression for he oal power in his circui is v a i a v b i b v f i f + v g i g + v h i h = (12)( 1) (12)(1) ( 12)(3) + 12i g + ( 24)( 7) = Therefore, 12i g = = 36 so i g = = 3 A Thus, if he power in he modified circui is balanced he curren in componen g is 3 A. 215 Pearson Educaion, Inc., Upper Saddle River, NJ. All righs reserved. This publicaion is proeced by Copyrigh and wrien permission should be obained recording, or likewise. For informaion regarding permission(s), wrie o: Righs and Permissions Deparmen, Pearson Educaion, Inc., Upper Saddle River, NJ 7458.