Section 2.2 Charge and Current 2.6 b) The current direction is designated as the direction of the movement of positive charges.

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1 Chaper Soluions Secion. Inroducion. Curren source. Volage source. esisor.4 Capacior.5 Inducor Secion. Charge and Curren.6 b) The curren direcion is designaed as he direcion of he movemen of posiive charges..7 The relaionship of charge and curren is so q q q ( ) i( ) d + q( ) ( ) sin( π ) d + q( ) π ( ) cos( π ) + q( ).8 The coulomb of one elecron is denoed by e and So If and q ( ),.9 n q ( ) i( ) d + q( ) e ( ) q( ) / e d + q( ) 6 n e ( ) 4 Pearson Educaion, Inc., Upper Saddle iver, NJ. All righs reserved. This publicaion

2 ( ) q q id ( ) 5 d q 5 ( ). q 5 ( ) [ 5] 5( 5) 5( ) 5 Coulombs. Using he definiion of curren-charge relaionship, he equaion can be rewrien as: i dq d Δn e Δ Thus, he curren flow wihin and ime inerval is, 9 ( 5.75 ) 9 i (.6 ) A The negaive sign shows he curren flow in he opposie direcion wih respec o he elecric charge.. Assuming he area of he meal surface is S, The mass of he nickel wih deph d.5mm is m ρ d S Meanwhile, using he elecro-chemical equivalen, he mass of he nickel can be expressed as m k I where I σ S. Equaing he wo expressions of he mass, he coaing ime is found: ρ d / σ.4 5 s 4.4 hour Secion. Volage. By he definiion of volage, when a posiive charge moves from high volage o low volage, is poenial energy decreases. So a is +, b is -. In oher words, u ab V..4 The curren i() is defined as: i ( ) < elsewhere Therefore, he charge is 4 Pearson Educaion, Inc., Upper Saddle iver, NJ. All righs reserved. This publicaion

3 g d C The energy in Joules is given by: J V C 5 5J.5 elecron 9.6 Coulombs. Therefore, here are elecrons in a coulomb. 6 Coulombs of5 elecrons Therefore, he volage is V J 5 C 8 875V.6.7 J q ( ) sin π d 4 cos π.444c π V C J J q C V i dq C. A d 4s 5 Secion.4 especive Direcion of Volage and Curren.8 True.9 False. True Secion.5 Kirchoff s Curren Law. According o KCL 4 Pearson Educaion, Inc., Upper Saddle iver, NJ. All righs reserved. This publicaion

4 7 i Therefore,. Noice ha Therefore, I can be seen ha i 4A + i i4 A i5 A 4 i 6 can now be found i 6. i 6 8A 5kΩ i i ma i kω kω + v - Figure S.: Circui for Problem.. According o KCL, and Therefore, ma i + i ki 8ki i 9mA, i ma V I k ma V.4 Using he KCL mehod, we have, 4 Pearson Educaion, Inc., Upper Saddle iver, NJ. All righs reserved. This publicaion

5 Also, a he oher end of he node, we have, i + i i i + 4 i i, combine hese wo equaions we obain, i i 4.5 The algebraic sum of currens flowing ino he node in Figure. is i + i - i - i 4 where i and i flow in, and i and i 4 flow ou As curren is defined as he rae of variaion of he charge, i + i - i - i 4 dq/d By law of he conservaion of charge, charges canno be sored in he node and charges can neiher be desroyed nor creaed. Therefore he variaion rae of he charge is zero and furher he KCL law holds, i + i - i - i 4.6 The resisance across he volage source is // + 5Ω 4 + So he curren flowing hrough he volage source is V 5V I. A 5Ω Therefore,.7 By KCL, Noe ha, Therefore, 4 i. 5mA 6 V V + 5 V I V I x x I x I 5 I 5 x + I x x I x I x 4 Pearson Educaion, Inc., Upper Saddle iver, NJ. All righs reserved. This publicaion

6 I I.5A x x.8 The volage across node C and D (node C is posiive) is U - U, he volage across node C and B is U AB - U, and he volage across node D and B is U AB - U The KCL equaion for node C is U / (U - U ) / + (U AB - U ) / which is 5U - U U AB The KCL equaion for node D is U / + (U - U ) / (U AB - U ) / which is 5U - U U AB Solving U and U, U 9U AB /, U U AB / Thus, by he KCL equaion a node A I AB U / + U / AB U AB /I AB 7 /5 A + I AB + + C - U U - D U AB - B Figure S.8: Circui for Problem.8. Secion.6 Kirchoff s Volage Law.9 By KVL, v A vb vc + v v A v A V. Using KVL, D + v 6+ 4 v 4V 4 Pearson Educaion, Inc., Upper Saddle iver, NJ. All righs reserved. This publicaion

7 . Using KVL, Using KVL,. Using KVL, Using KVL,. Using KVL, Using KVL, Using KVL, v D vd 8V 8 + v + 8 C vc V v C vc 5V 5 + v 5 6 E ve V v + 4 D vd V v C vc 5V 5 v + E ve 8V.4 Using KVL, Using KVL, Using KVL, 7 v + E ve 8V v B vb V 5 v 7 D 4 Pearson Educaion, Inc., Upper Saddle iver, NJ. All righs reserved. This publicaion

8 Using KVL.5 Using KVL, vd V v + A va V VH + VH 8V + VH 5+ VI VI 5V V V V C C VD 4 + VC VD 5 + V.6 Ω V 6V + - i i 4 4Ω i i i 6Ω Figure S.6: Circui for Problem.6. a) 6Ω, Ω By KCL, i + i i i 4 + i i Applying KVL on he loop made by 6V source, and, 6i + i 6 Applying KVL on he loop made by 6V source, Ω resisor and 6Ω resisor, i + 6i 4 6 Applying KVL on he loop made by Ω resisor, 4Ω resisor and, i + 4i 6i Now we have 5 equaions for 5 variables and i can be solved using Cramer s rule or variable eliminaion. we ge i /4A, i 7/A, i 5/6A, i 5/6A, i 4 7/A b) i > i i, i i 4. 4 Pearson Educaion, Inc., Upper Saddle iver, NJ. All righs reserved. This publicaion

9 Moreover, he volage across he 4Ω resisor is zero. This also leads o he volage across equals o he volage across he Ω resisor and he volage across equals o he volage across he 6Ω resisor, i i i 6i 4 Therefore, / /6 /.7 The curren from node A o node C is I AB - I, he curren from node D o node B is I - I, and he curren from node C o node B is I AB - I + I. he KVL equaion for loop ADC is I + I (I AB - I ) which is I + I I AB The KVL equaion for loop DBC is (I - I ) I + (I AB - I + I ) which is I - 4I I AB Solving I and I, I I AB /5, I -I AB /5 Thus, U AB I + (I - I )7I AB /5 AB U AB /I AB 7 /5 A + I AB I C I D U AB - B Figure S.7: Circui for Problem.7. Secion.7 Ohm s Law.8 By KCL, he curren hrough he 4Ω resisor is A. Using Ohm s law, V I 4 V.9 By KVL, he volage across he resisor is 5V. Using Ohm s law: 4 Pearson Educaion, Inc., Upper Saddle iver, NJ. All righs reserved. This publicaion

10 V 5. 5Ω I.4 Using Ohm s law: V 5 I A 5.4 The resisiviy of copper is, ρ.7 8 Ω m The resisance of he copper wire is, ρl 7Ω A.86 Using he definiion of Ohm s Law, we can obain he maximum allowable volage, which is: V I 7 7V max max.4 Using Ohm s law, he resisance of he issue is Because, Therefore, 84. 5kΩ 6.4 ρl A 75Ω m L 84.5kΩ.468 m L 5.94 m.4 The resisance across he volage source is + // + + V Using Ohm s law I, and Vs ( + ) i I + + ( + ) + V s ( + ) + 4 Pearson Educaion, Inc., Upper Saddle iver, NJ. All righs reserved. This publicaion

11 .44 By KCL: Therefore, by Ohm s Law: By KVL: V A V V B C I 5A 5 V V 5 V V S V.45 Using Ohm s law: I 4.mA.46 For shor circui, Using ohm s law For open circui Using ohm s law v() shor v()/i() i() open v()/i() infiniy.47 V ou Vs(s /(s + c)) - Vs(b /(a+b)) By produc rule of derivaion dv ou /ds Vsc/(s+c) When s, he above expression is minimum. However, you canno conrol over sensor resisance, bu you can choose c o achieve he maximum slope. We differeniae dv ou /ds wih respec o c, d(dv ou /ds) /dc Vs(s-c) /(s+c) The maximum slope is achieved when c s. c can be chosen o have he same value as he nominal resisance of he sensor..48 Using daa pairs in he able, we ge hree equaions in A, B, C (/7) A + B ln(6) + C (ln(6)) (/98) A + B ln(5) + C (ln(5)) (/) A + B ln(8) + C (ln(8)) Solving hose equaions, we ge A.84 B.64-4 C Pearson Educaion, Inc., Upper Saddle iver, NJ. All righs reserved. This publicaion

12 Secion.8 Power and Energy.49 a) The volage and he curren are in associaed direcion, P u i - -W > acive elemen b) The volage and he curren are no in associaed direcion, P -u i 4W > passive elemen c) The volage and he curren are no in associaed direcion, P -u i - -6W > acive elemen.5 a) The volage and he curren are in associaed direcion, P u i.5e - 5e - W > acive elemen b) The volage and he curren are no in associaed direcion, P -u i - sin -4sin W > >, acive elemen; <, passive elemen.5 The curren can be found using he power formula: P 5 I.5A V And he resisance can hen be found using Ohm s law: V Ω I.5.5 P V I I 5 5W.5 Applying KCL, he curren hrough each resisor is 4mA. Given P I, ( 4 ) mw P 5 8 ( 4 ) mw P 4 4 ( 4 ) mw P 5.54 According o KVL: By KCL:.5ki ki ma i + i i ma 4 Pearson Educaion, Inc., Upper Saddle iver, NJ. All righs reserved. This publicaion

13 V ma kω V.55 From he problem.54, we have i ma and i ma. ecall P I. Therefore, P ( ).5 6mW ( ) mw ( ) mw.5k P k P k.56 I + V - Figure S.56: esisor wih curren direcion and volage polariy. Consider Figure S.56, he power absorbed by he resisor is PVI By Ohm s law, VI Thus PI V / Therefore P is always a nonnegaive number when is posiive..57 The volage and he curren are in associaed direcion, so P u i and he waveform of he power is shown below P (W) - (s) The consumed energy is Figure S.57: Power curve. P W 4 Pearson Educaion, Inc., Upper Saddle iver, NJ. All righs reserved. This publicaion

14 which is resuled from he negaive symmery of P around..58 a) Le he curren hrough he -5V volage source be in associaed direcion wih he volage, which is from he posiive side o he negaive side. By KCL a he node below he -5V volage source, he curren hrough i equals A+4A6A. Therefore he power of he -5V volage source is P -5V 6A W b) To zero he power above, he curren hrough he -5V volage has o be zero. So he 4A curren source needs o be changed o -A by KCL. Secion.9 Independen and Dependen Sources.59 In Figure S.59, i axis is he curren hrough he source and u axis is he volage across he source. u s denoes he volage provided by he ideal volage source and i s denoes he curren provided by he ideal curren source. I is shown ha for ideal volage source he volage does no change wih he curren hrough i; for ideal curren source he curren does no change wih he volage. u Ideal curren source u s Ideal volage source i s i Figure S.59: Volage VS curren curves..6 Because he wo curren sources are in series, he currens hrough hem are he same, i A.6 Because he Ω resisor is in series wih he curren source, he curren hrough he resisor is A and, by Ohm s law, is volage is u Ω A V By KVL, he volage across he curren source is u + V 5V The power supplied by he curren source is (he curren and he volage are no in associaed direcion) 4 Pearson Educaion, Inc., Upper Saddle iver, NJ. All righs reserved. This publicaion

15 P -5V A -5W.6 Using KVL: Therefore, + V + V x.5 x V x.6 Using KVL and Ohm s law, 6 ( I x.5 ) + I x 6.48 I x I x. 74mA.64 The volage across he Ω resisor is -V, where he volage polariy is associaed wih he curren direcion. So by Ohm s law I -V/Ω -.5A Therefore, he curren of he dependen source is I -A.65 According o KCL, he curren hrough is β ix According o Ohm s law, i x is given by: Therefore, v i v x β i x vs + β v s +.66 According o Ohm s law, he curren flow (in associaed direcion wih he volage) hrough he 5Ω resisor is i 4.9V/5Ω.98A. Because he CCCS is in series wih he 5Ω resisor, i.98i > i A which is he curren hrough he 6Ω. By KCL, he curren hrough he.ω resisor is i-.98i.a. So he volage across he.ω resisor is 4 Pearson Educaion, Inc., Upper Saddle iver, NJ. All righs reserved. This publicaion

16 u.ω.a.v. Applying KVL in he loop made by u s,.ω and 6Ω u s u +6i.+66.V The power supplied by he dependen source is P (.98i) u.98 (.-4.9).98 (-4.898) -4.8W Secion. Analysis of Circuis using PSpice.67 A PSpice circui is shown in Figure S.67. Choose he Bias Poin for he simulaion analysis. I TOTAL TOTAL + VTOTAL TOTAL 4 +.A - 5Ω Using Ohm s law, we obain: Apply he KCL: The i can be calculaed as: i i i i I i + i. i + i 4 i.5a Figure S.67: PSpice Circui for Problem Pearson Educaion, Inc., Upper Saddle iver, NJ. All righs reserved. This publicaion

17 .68 Applying KCL: i i + i A For parallel volage, he curren relaionship is given by: i i i Therefore i 4A i 6A The oal volage across he circui is: VTOTAL iii + i 5V The PSpice seup wih curren source is shown in Figure S.68 a. Figure S.68 b shows he curren source is replaced by volage source ha produces he same curren and volage across each resisance as hose in Figure S.68 a. Figure S.68 a: PSpice Circui wih Curren source for Problem.68. Figure S.68 b: PSpice Circui wih Curren source for Problem Applying KVL equaion, we have: 4 Pearson Educaion, Inc., Upper Saddle iver, NJ. All righs reserved. This publicaion

18 (.) (.) (.) Applying KCL equaion, we have: (.4) (.5) (.6) From equaion (.), he curren relaionship can be wrien as: Using he relaionship obained above and subsiue ino equaion (.), we obain: (.7) Expressing he equaion (.) for i in erm of i : (.8) Subsiue equaion (.8) ino equaion (.4), i is calculaed: (.9) Subsiue equaion (.9) ino equaion (.7), 6 and i 5 can be deermined: (.) (.) A PSpice schemaic is shown in Figure S.69 o verify he answers obained: 4 Pearson Educaion, Inc., Upper Saddle iver, NJ. All righs reserved. This publicaion

19 Figure S.69: PSpice Circui wih Curren source for Problem Applying he KCL euqaion: (.) (.) (.4) (.5) Applying he KVL equaion: (.6) (.7) (.8) (.9) Subsiuing equaion (.) ino equaion (.), we obain: (.) Subsiuing equaion (.) ino equaion (.6), we obain: Using he curren relaionship in equaion (.) and subsiuing i ino equaion (.7), we ge: (.) (.) 4 Pearson Educaion, Inc., Upper Saddle iver, NJ. All righs reserved. This publicaion

20 Subsiuing equaion (.4) and equaion (.8), we obain: (.) Applying he curren relaionship in equaion (.4) and (.5) ino equaion (.9), and i can be rewrien as: We can express he equaions (.) o (.4) by using a marix equaion of, which is given below: (.4) Marix X can obained, A PSpice schemaic is shown in Figure S.7: 4 Pearson Educaion, Inc., Upper Saddle iver, NJ. All righs reserved. This publicaion

21 Figure S.7: PSpice Circui wih Curren source for Problem.7. 4 Pearson Educaion, Inc., Upper Saddle iver, NJ. All righs reserved. This publicaion