Solutions to Chapter 1 Electrical Engineering Problems, Exams of Mathematics

Download Solutions to Chapter 1 Electrical Engineering Problems and more Exams Mathematics in PDF only on Docsity! Chapter 1, Solution 1 (a) q = 6.482x1017 x [-1.602x10-19 C] = –103.84 mC (b) q = 1. 24x1018 x [-1.602x10-19 C] = –198.65 mC (c) q = 2.46x1019 x [-1.602x10-19 C] = –3.941 C (d) q = 1.628x1020 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 cos t pA (e) i =dq/dt =  e t tt4 80 50 1000 50( cos sin ) A Chapter 1, Solution 5 10 2 0 101 25 C 02 4 t q idt tdt     Chapter 1, Solution 6 (a) At t = 1ms,  2 30 dt dq i 15 A (b) At t = 6ms,  dt dq i 0 A (c) At t = 10ms,    4 30 dt dq i –7.5 A Chapter 1, Solution 7          8t6 25A, 6t2 25A,- 2t0 A,25 dt dq i which is sketched below: Chapter 1, Solution 10 q = it = 10x103x15x10-6 = 150 mC Chapter 1, Solution 11 q= it = 90 x10-3 x 12 x 60 x 60 = 3.888 kC E = pt = ivt = qv = 3888 x1.5 = 5.832 kJ Chapter 1, Solution 12 For 0 < t < 6s, assuming q(0) = 0, q t idt q tdt t t t ( ) ( ) .     0 3 0 15 0 2 0 At t=6, q(6) = 1.5(6)2 = 54 For 6 < t < 10s, q t idt q dt t t t ( ) ( )      6 18 54 18 5 6 6 4 66 At t=10, q(10) = 180 – 54 = 126 For 10<t<15s, q t idt q dt t t t ( ) ( ) ( )        10 12 126 12 246 10 10 At t=15, q(15) = -12x15 + 246 = 66 For 15<t<20s, q t dt q t ( ) ( )   0 15 15 Thus, q t 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. Chapter 1, Solution 14 (a)      2e2102.02et02.0dte-10.02idtq 0.5-1 0 0.5t-1 0 0.5t-    = 4.261 mC (b) p(t) = v(t)i(t) p(1) = 10cos(2)x0.02(1–e–0.5) = (–4.161)(0.007869) = –32.74 mW Chapter 1, Solution 15 (a)       1e003.0 e 2 006.0 dt0.006eidtq 4- 2 0 2t2 0 2t- 2.945 mC (b) 2t-2t- e12.0)10(e012.0 dt di10 v  V this leads to p(t) = v(t)i(t) = (-0.12e-2t)(0.006e-2t) = –720e–4t µW (c) 3 0 64t-3 0 4t- 10e 4 720 dt e-0.72pdtw      = –180 µJ Chapter 1, Solution 16 (a) 30 mA, 0 < t <2 ( ) 120-30t mA, 2 < t<4 t i t     5 V, 0 < t <2 ( ) -5 V, 2 < t<4 v t     150 mW, 0 < t <2 ( ) -600+150t mW, 2 < t<4 t p t     which is sketched below. p(mW) 300 4 t (s) -300 1 2 (b) From the graph of p, 4 0 0 JW pdt  Chapter 1, Solution 19 I = 8 –2 = 6 A Calculating the power absorbed by each element means we need to find vi for each element. p8 amp source = –8x9 = –72 W pelement with 9 volts across it = 2x9 = 18 W pelement with 3 bolts across it = 3x6 = 18 W p6 volt source = 6x6 = 36 W One check we can use is that the sum of the power absorbed must equal zero which is what it does. Chapter 1, Solution 20 p30 volt source = 30x(–6) = –180 W p12 volt element = 12x6 = 72 W p28 volt e.ement with 2 amps flowing through it = 28x2 = 56 W p28 volt element with 1 amp flowing through it = 28x1 = 28 W pthe 5Io dependent source = 5x2x(–3) = –30 W Since the total power absorbed by all the elements in the circuit must equal zero, or 0 = –180+72+56+28–30+pinto the element with Vo or pinto the element with Vo = 180–72–56–28+30 = 54 W Since pinto the element with Vo = Vox3 = 54 W or Vo = 18 V. Chapter 1, Solution 21 60 0.5 A 120 p p vi i v      q = it = 0.5x24x60x60 = 43.2 kC 18 236.24 10 2.696 10 electronseN qx x x  Chapter 1, Solution 24 W = pt = 60 x24 Wh = 0.96 kWh = 1.44 kWh C = 8.2 centsx0.96 = 11.808 cents Chapter 1, Solution 25 cents/kWh 2.830hr 60 3.5 kW 1.5 Cost  = 21.52 cents Chapter 1, Solution 26 (a)    10h hA8.0 i 80 mA (b) p = vi = 6  0.08 = 0.48 W (c) w = pt = 0.48  10 Wh = 0.0048 kWh 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 kW21ptw Chapter 1, Solution 30 Monthly charge = $6 First 250 kWh @ $0.02/kWh = $5 Remaining 2,436–250 kWh = 2,186 kWh @ $0.07/kWh= $153.02 Total = $164.02 Chapter 1, Solution 31 Total energy consumed = 365(120x4 + 60x8) W Cost = $0.12x365x960/1000 = $42.05 Chapter 1, Solution 34 (a) Energy = = 200 x 6 + 800 x 2 + 200 x 10 + 1200 x 4 + 200 x 2 pt = 10 kWh (b) Average power = 10,000/24 = 416.7 W Chapter 1, Solution 35 energy = (5x5 + 4x5 + 3x5 + 8x5 + 4x10)/60 = 2.333 MWhr Chapter 1, Solution 36 days 6,667 , ( A 4     day / 24h h000160 0.001A 160Ah tb) 40 hA160 i (a)