Chapter 1, Solution 1 (a) q = 6.482x10 17 x [-1.602x10 -19 C] = -0.10384 C (b) q = 1. 24x10 18 x [-1.602x10 -19 C] = -0.19865 C (c) q = 2.46x10 19 x [-1.602x10 -19 C] = -3.941 C (d) q = 1.628x10 20 x [-1.602x10 -19 C] = -26.08 C Chapter 1, Solution 2 (a) i = dq/dt = 3 mA (b) i = dq/dt = (16t + 4) A (c) i = dq/dt = (-3e -t + 10e -2t ) nA (d) i=dq/dt = 1200 120 π π cos t pA (e) i =dq/dt = −+ − et t4 80 50 1000 50 (cos sin ) A µ t Chapter 1, Solution 3 (a) C 1)(3t +=+= ∫ q(0)i(t)dt q(t) (b) mC 5t)(t 2 +=++= ∫ q(v)dt s)(2tq(t) (c) () q(t) 20 cos 10t / 6 q(0) (2sin(10 / 6) 1) Ct π πµ =++=++ ∫ (d) C 40t) sin 0.12t(0.16cos40e 30t- +−= − + =+= ∫ t)cos 40-t40sin30( 1600900 e10 q(0)t40sin10eq(t) -30t 30t- Chapter 1, Solution 4 () mC 4.698=−= − === ∫∫ π π π 06.0cos1 6 5 tπ6cos 6 5 dt t π 6 5sinidtq 10 0 Chapter 1, Solution 5 µCmC )e1( 2 1 e 2 1 -mC dteidtq 4 2 0 2t-2t- 490=−= === ∫∫ Chapter 1, Solution 6 (a) At t = 1ms, mA 40=== 2 80 dt dq i (b) At t = 6ms, mA 0== dt dq i (c) At t = 10ms, mA 20-=== 4 80 dt dq i Chapter 1, Solution 7 << << << == 8t6 25A, 6t2 25A,- 2t0 A,25 dt dq i which is sketched below: Chapter 1, Solution 8 C 15 µ110 2 110 idtq =×+ × == ∫ Chapter 1, Solution 9 (a) C 10=== ∫∫ 1 0 dt 10idtq (b) C 22.5=−+= ×+ × −+×== ∫ 251015 15 2 15 10110idtq 3 0 (c) C 30=++== ∫ 101010idt 5 0 q Chapter 1, Solution 10 qixt x xx == = − 8 10 15 10 120 36 C µ Chapter 1, Solution 11 q = it = 85 x10 -3 x 12 x 60 x 60 = 3,672 C E = pt = ivt = qv = 3672 x1.2 = 4406.4 J Chapter 1, Solution 12 For 0 < t < 6s, assuming q(0) = 0, qt idt q tdt t tt () ( ) . =+= += ∫∫ 03015 0 2 0 At t=6, q(6) = 1.5(6) 2 = 54 For 6 < t < 10s, qt idt q dt t tt () ( ) =+= +=− ∫∫ 61854185 66 4 66 At t=10, q(10) = 180 – 54 = 126 For 10<t<15s, qt idt q dt t tt () ( ) ( ) =+ =−+=−+ ∫∫ 10 12 126 12 246 10 10 At t=15, q(15) = -12x15 + 246 = 66 For 15<t<20s, qt dt q t () ( ) =+= ∫ 015 15 Thus, qt t t t () . , = − −+ 15 18 54 12 246 66 2 C, 0 < t < 6s C, 6 < t < 10s C, 10 < t < 15s C 15 < t < 20s The plot of the charge is shown below. 0 5 10 15 20 0 20 40 60 80 100 120 140 t q(t) Chapter 1, Solution 13 kJ 2.486-= −= −= == == ∫ ∫∫ 216sin 4 1 12008sin 8 2 1200 1)-2x cos 2xcos (since,1)dt -t8cos2(1200 dtt 4cos1200vidtw 2 0 2 0 2 2 0 2 2 0 tt Chapter 1, Solution 14 (a) ( ) ( ) () C 2.131=−+= +=== ∫∫ 2e2110 2et10dte-110idtq 0.5- 1 0 0.5t- 1 0 0.5t- (b) p(t) = v(t)i(t) p(1) = 5cos2 ⋅ 10(1- e -0.5 ) = (-2.081)(3.935) = -8.188 W Chapter 1, Solution 15 (a) () C 1.297=−−= − === ∫∫ 1e5.1 e 2 3 dt3eidtq 2- 2 0 2t 2 0 2t- (b) We90 )( t4− −== −=−== vip e305e6 dt di5 v 2t-2t (c) J 22.5−= − − === ∫∫ 3 0 4t- 3 0 4t- e 4 90 dt e-90pdtw Chapter 1, Solution 16 mJ 916.7 )( ( ((( , = ++ +−−++= −+ ++= −+−+++== << << << = << << = ∫ ∫∫∫∫∫ 4 3 2 2 4 3 2 3 2 2 2 1 1 0 3 4 3 3 2 2 1 1 0 3 t 4t-t1625028 2 9 122503 2 250 3 250 dtt)4250 2 t -t4250 2 250 t 3 250 25t)mJ-t)(1001040t)dt2510010t)dt2510(25t)dt10v(t)i(t)dtw 4t3 V10t -40 3t1 V 10 1t0 V10t v(t) 4t2 mA25t -100 2t0 mA25t i(t) Chapter 1, Solution 17 Σ p = 0 → -205 + 60 + 45 + 30 + p 3 = 0 p 3 = 205 – 135 = 70 W Thus element 3 receives 70 W. Chapter 1, Solution 18 p 1 = 30(-10) = -300 W p 2 = 10(10) = 100 W p 3 = 20(14) = 280 W p 4 = 8(-4) = -32 W p 5 = 12(-4) = -48 W Chapter 1, Solution 19 pIxxx ss =→−−−+=→= ∑ 0 4 2 6 13 2 5 10 0 3 A I Chapter 1, Solution 20 Since Σ p = 0 -30×6 + 6×12 + 3V 0 + 28 + 28×2 - 3×10 = 0 72 + 84 + 3V 0 = 210 or 3V 0 = 54 V 0 = 18 V Chapter 1, Solution 21 nA8. (. C/s100.8 C/s106110 8 4 electron) / C1061 photon electron 8 1 sec photon 104 8-1911 1911 =×=×××= ×⋅ ⋅ ×= ∆ ∆ = − t q i Chapter 1, Solution 22 It should be noted that these are only typical answers. (a) Light bulb 60 W, 100 W (b) Radio set 4 W (c) TV set 110 W (d) Refrigerator 700 W (e) PC 120 W (f) PC printer 18 W (g) Microwave oven 1000 W (h) Blender 350 W Chapter 1, Solution 23 (a) W12.5=== 120 1500 v p i (b) kWh 1.125. =×=⋅×××== kWh 60 45 1.5J60451051 3 ptw (c) Cost = 1.125 × 10 = 11.25 cents Chapter 1, Solution 24 p = vi = 110 x 8 = 880 W Chapter 1, Solution 25 cents 21.6 cents/kWh 930hr 6 4 kW 1.2 Cost =×××= Chapter 1, Solution 26 (a) mA 80 . = ⋅ = 10h hA80 i (b) p = vi = 6 × 0.08 = 0.48 W (c) w = pt = 0.48 × 10 Wh = 0.0048 kWh Chapter 1, Solution 27 ∫∫ =××==== ×== kC 43.2 36004 3 T33dt idt q 36005 4 4h T Let (a) T 0 [] kJ 475.2 . . . )(( = ××+×= += +=== × ∫∫∫ 3600162503600403 3600 250 103 dt 3600 t50 103vidtpdtW b) 36004 0 2 0 T 0 t t T cents 1.188 ( =×= == cent 9kWh 3600 475.2 Cost Ws)(J kWs, 475.2 W c) Chapter 1, Solution 28 A 0.25=== 120 30 (a) V P i $31.54 ( =×= = ××== 262.8 $0.12Cost kWh 262.8Wh 2436530ptW b) Chapter 1, Solution 29 cents 39.6 . =×= =+= + + ++ == 3.3 cents 12Cost kWh 3.30.92.4 hr 60 30 kW 1.8hr 60 45)1540(20 kW21 ptw Chapter 1, Solution 30 Energy = (52.75 – 5.23)/0.11 = 432 kWh Chapter 1, Solution 31 Total energy consumed = 365(4 +8) W Cost = $0.12 x 365 x 12 = $526.60 Chapter 1, Solution 32 cents 39.6 . =×= =+= + + ++ == 3.3 cents 12Cost kWh 3.30.92.4 hr 60 30 kW 1.8hr 60 45)1540(20 kW21 ptw Chapter 1, Solution 33 C 6=××==→= ∫ 3 1032000idtq dt dq i Chapter 1, Solution 34 (b) Energy = ∑ = 200 x 6 + 800 x 2 + 200 x 10 + 1200 x 4 + 200 x 2 pt = 10,000 kWh (c) Average power = 10,000/24 = 416.67 W Chapter 1, Solution 35 kWh 10.4 )(( =+= ×+×××+×+×== ∫ 28007200 24002120012200210006400W a) dttp W/h 433.3( = h 24 kW 10.4 b) Chapter 1, Solution 36 days 6,667 , ( A 4 === = ⋅ = day / 24h h000160 0.001A 160Ah tb) 40 hA160 i (a) Chapter 1, Solution 37 ( ) J 901.2. −=×−== −=×−×= − 12180W C 180106021105q 1920 qv