ECE109 HW 1 SOLUTIONS

# ECE109 HW 1 SOLUTIONS - ® 1.7 A current of 1200 A exists...

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Unformatted text preview: ® 1.7 A current of 1200 A exists in a copper wire, with a . circular cross- -section (radius—- = 1. 5 mm). The cur- rent is due to free electrons moving through the wire at an average velocity of vmeters/ second If the concentration of free electrons is 101’9 electrons per cubic meter and if they are uniformly dispersed throughout the wire, then what is the average veloc- ity of an electron? 1.6022 x 10-ig C x 102? electrons 1 electron 1 n13 C/m3 = =15022 >< 101°C/m3 ' Cross—sectional area of wire =-7rr2 = «(1.5 x 10’3 m)2 = 7.07 X 10‘6 m2 C/m = (1.6022 x 101°C/m3)(7.07 >< 10-61112) = 113.253 x 103 (3/111 Therefore, i(— C —) = (”113.253 x 103) (9) x avg ve1<2> 58C ' m S 12 1200 loc’ = ———————— = ————‘— = 0.0106131 s Thus’ average ve my 113. 253 x 103 ..113 253 x 103 - / - . ‘L ,. @ 1.9 The current entering the upper terminal of Fig. 1.5 is ’1‘?” ' ' = 24 4000: A. + t cos "U“ Assume the charge at the upper terminal is zero at a: the instant the current is passing through its maxi- mum value. Fmd the expression for q(t). _ dq _. z — E — 24ccs4000t Therefore, dq = 24 cos 4000t dt ‘ To ﬁnd the charge, we can integrate both sides of the last equation. Note that we substitute :1: for q on the left side of the integral, and y for t on the right side of the integral: 00‘) f 0! =24 (0) :1: / cos4000ydy We solve the integral and make the substitutions for the limits of the integral, remembering that sin0= 0: 40:) -q(0) = 2431114000, — 24 24 4000 — m sin4000t -— ——4000 sin4000(0) = TOGO sin4000t But .q(0) = 0 by hypothesis, i.e., the current passes through its maximum - Ivalue at t = 0, so q(t) =3 6 x 10‘3 sin4000tC = 631114000th ». ( <9 [.10 The manufacturer of at 9V dry-cell ﬂashlight battery :3): that the battery will deliver 20 mA for 80 con- umzous hours During that time the voltage will drop from 9 \' to 6 V. Assume the drop in voltage is linear with lime How much energy does the battery deliver in this 80 h interval? r I' .1 vi; w = / pd: - 0 Since the energy E the area under the power vs. time plot, let us plot p vs. t. p ”0110“ . mun“ ' 258“ 1 Note that in constructing the plot above, we used the fact that 80 hr = 288,000 5 = 283 ks p(o) = (9)(20 x 10") = 180 x 10-3 W p’C'ZSS ks) = (6)(20 x 10-3) = 120 x 10-3 w u‘ = (120 x 10‘°)(288 x 103) + \$080 x 10"3 — 120 x 10-3)(283 x 103) = 43.2 M @ 1.17 The voltage and current at the terminals of the cir- cuit element in Fig. 15 are zero fort < O. For: 2 0 they are 1.: = Em — c'm V, i= 30 — 40e‘5‘” + 10c"5°°' 111A. 3) Find the power at! = 1 ms. b) How much energy is delivered to the circuit ele- ment between 0 and 1 ms? c) ﬁnd the total energy delivered to the element. . [a] ”£1“ 7 303:0“ @306-” ‘ 406"” + SOs-W — lee-W m = . In [1’] w“) ‘ = £(30e‘500‘ .. 303—1500: _ 403—1000; + SOs—m — lee-Wm 21.67 — 60e-5°°‘ + 206—150m + 4oe—loom_ 255200“ + 3.33e‘mpJ w(1 ms) = 1.24M 1°] 19m; =12me ll 1.18 Th: voltaz: 212:1 {31:21 a: 1'2: 1:17:21}; cf t: :5:— - m3 suit dunes: 111571;. 1.5 are 2531's: I < E: Ft: : 2 L- thsy are a) Fund ‘32: power absarbc: 3 1‘2: :1:::.::.: a: r = 10 125. 1:) Fund the 2011 energy abscrbec‘. by it: 92:25:; [a] mo ms) = 4002-1 sin2 = 133.5 v {(10 m5) = 55151112: 1.67 A p(10 ms) = vi = 223.79 W [b] p = vi=2000€m \$112200: = 2000e’m[———005400‘LJ E H II if?! H 2%; 1*? Ha; ”\$72 E? 3. g H H g n: I! /——2°0 } 5 1 _ = _ "’1000L1x10L;16><10*l w=4J @ L19 The voltage and current a: the terminals of the Cir- suit element in ﬁg. 1.5 are shown in Fxg.P1.19. a) Sketch the power versus (plot for 0 s r S 50 s. b) Calculate the energy delivered to the circuit ele- ment at! = 4. 12, 36.2md 50 s Figure P1.” ‘L‘(V) H O 2024283236404448525616) Nouuo‘m (a) 43 52 56 1(5) F L19 [n] 03 \$f<d a: unlﬁt V; inn um; pn- 2.5tpW 4s<ts832 u=10V; inOA; Il=0w Ss\$t<lﬁsz . v = —2.5t + 30 V; 1' == —1}LA; p = 2.51—30pW 168<t52052 1l=—10V; i=0A; p=0W 208\$t<3631 u=t~30 V; i=0.d;LA; p=0.4t—12;1W 36\$<ts4652 umGV; i=0A; p=0W 46s_<_t<5082 v=-1.5t+75 V; i= -0.6pA; p=0.9t—45pW t>5031 v=0V; i=0A; p 0W ll 9M [b] Calculate the area. under the curve from zero up to the desired time: w(4) = awe) = 20m w(12) = 111(4) -— %(4)(10) = O J . w(36) = w(12) + %(4)(10) — §(10)(4) + %(6)(2.4) = 7.2 pJ w(50) = w(36) —- %(4)(3.6) = 0 J Q L20 The voltage and current at the terminals of the cir- "L" cuit element in Fig. 15 are zero for t < O. For I 2 0 they are 1) = 75 — 75e'1m v, i ? 50 {lmmﬁ 3) Find the maximum value of the power delivered to the circuit. b) Find the total energy delivered to the element. ‘ [a] p = vi = (0.05e-1mx75 — 755100“) = (375510” — 3.755200“) W 26—2000: = 8—1000: d d—‘t’ = —3750e-1°°°t + 75005200“ = 0 so 2=e‘°°°‘ so m2=1000t thus pismaximumatt=69315ps pm = p(693.15 ps) = 937.5 mW [b] w = [0 0013-7584000: — 3.755200% dt = [ 3‘75 e~1°°°‘ — 3'75 e—mlw] O —1000 -—2000 3.75 3.75 —m—m—l.875 111.] 8 1,21 The voltage and current at the terminals of the ele- rsrmt ment in Fig.15 are v=36sin2007rtV, i=25cosZOOrrtA a) Fmd the maximum value of the power being delivered to the element b) Find the maximum value of the power being extracted from the element. c) Fmd the average value of p in the interval 0 s r s 5 ms. d) Fmd the average value of p in the interval 0 s t s 6.25 ms. [a] p = vi = 9005in(2007rt) cos(2007rt) = 450 sin(4007rt) W Therefore, pm = 450 W lb] pau(extracting) = 450 W 5x10"3 [C] p»; = 200 /o 4508it1(4007rt) dt —oos4007rt “do" 225 = 9X10‘[—-——0 = T[1—cos27r]=0 180 1 [‘1]va = T[1 -0082.51r] = g = 57.3 W 1.22 The voltage and current a: the terminal: of an auto- “: mobile baztcr‘j during a charge, cycle are shown in Fag. P121 a) Calculate the total charge uamfcrxcd to the battery E) Calculat: the total energy transferrcd to the ban: 1 figure 91.22 0 4 8 12 16 20 ((ks) ﬁw— g fI. 12' [a] q = area under 1' vs. 1 plot C = WW 1410)“) + 11810) + (81(6) + 1131(5)} x 103 = [10+ 40+ 10+ 40+9]103 =123.000 C [b] w 2 jpdt=/vidt v = 0.2x10‘3t+9 1351:3155 0 g 1 5 4000.1 1' = 15 —1.25 x 1041 p = 135 — 8.25 x 10-31- 0.25 y 10-51? 1000 w, = f (135 - 3.25 x 10-31— 0.25 x 104:2)41 0 = (540 — 66 — 5.3333)103 = 468.667 11.1 1/ 4000 g t 5 12,000 i = 12 — 0.5 x 10-31 p .-= 108 - 2.1 x 10-31— 0.1 x 10-612 12,000 w, = j (108- 2.1 x 10-11—01 x 10-612)dz 4000 = (864 — 134.4 - 55.467)10’ = 674.133 11.1 / 12,000 5 t 3 15,000 ' i = 30 -— 2 x 10‘3t p = 270 — 12 x 10-5: — 0.4 x 10-“:2 15.000 w: = fume (270 —- 12 x 10“°t — 0.4 x 104g) dt = (810 — 436 — 219.6)10’ = 1044 H w w, = w1+uh+uu=468.667+674.133+10¢4=1247.211.) v = (16,000: + 20km“ v, i = (128: + 0.16)e"°°' A a) At what instant of time: is maximum power delivered to the element? b) Fmd thc maximum power in warts. C) ﬁnd the total 0.11ch delivered to the element in millijoulcs [a] p = m‘ = [16,000t + 20)e”3°°‘][(128t + O.16)e’8°°‘] = 2048 x 103t2e-1m + 5120te“°°°‘ + 3261”“ = 3.2516001[540mm2 + 1600i + 1] d_p = 3.2{e'1m[1280 x 10% + 1600] — 160062—16”[640.0005z + 16001! + 1]} dt . = ~3.2c‘1m[128 x 10‘(800t2 + t)] = —409.6 x 10‘e‘1mt(800t + 1) Therefore, 3:3 = 0 when t = 0 sopmoccursatt=0. [6] pm = 3.2c‘°[0+0+1] = 3.2 W [c1 w = om t 51% = /0‘640.000I’e“°°°'dz+ 0‘1600xe'1mdz-i- 0 e'lmdl‘ ' sumac-“0‘” ' _ muss x 10‘s’+3200m+2] 0+ 1600e"°°°' ' 2‘1““ ' 256x 1m ('lm"l)o+ —1600 0 When t -’ 00 all the upper limits evaluate to zero, hence .2. = (640,000)(2) + 1600- 1 3.2 4096le 256x10‘+1600 w=10-3+2x 10‘3+2x10‘3=5 mJ. GD 1-25 The voltage and current at the terminals of the cir- .""“ cunt element in Fig. 1.5 are zero for t < 0. For t a 0 they are A . v = (10,000: + 5)e‘4°°' v, z 2 0; i: (40: + 0.05)e'4°°' A, t 2 0. a) Find the time (in milliseconds) when the power delivered to the circuit element is maximum. b) Find the maximum value of p in milliwatts. c) Find the total energy delivered to the circuit ele- ment in millijoules P 1.25 [a] p = m' -= 400 x maﬁa-800‘ + mom-300‘ + 0.258‘800‘: = ¢=3‘8°°‘[400,000t2 + 700t + 0.25] d —p {e-W[800 x 10% + 700] — 23003-30“[400,000:2 + 700:: + 0.25)} (it = [—3,200,000t2 + 24001: + 511002-890t d Therefore, d—lf) = 0 when 3,200,0001E2 — 240015 — 5 = 0 so pm“ occurs at t = 1.68 ms. . [b] pm '= [400,000(.00168)2+700(.00168) +0.251e-800coows) 666 mW t [c] w = /0-pd:1: t t t /O 400,00012e-8de+ / 700ze‘300‘d:c+ / 0.253‘300‘dx 0 t O '11) 400,000e‘8°°‘ 7006-300“ ‘ 8‘80”": ‘ —— — — .2 64x10“( 8001: 1)o+0 5—8000 When t = 00 all the upper limits evaluate to zero, hence _ (400,000)(2) 700 0.25 _ w“ 512x105 +64x104+ 800 ‘2'97mJ' , . 1‘4 I. - 6—,, , , la". n... . . ...
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