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Solution Manual for Fundamentals of Electric Circuits, 6th Edition - Document preview page 1

Solution Manual for Fundamentals of Electric Circuits, 6th Edition - Page 1

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Solution Manual for Fundamentals of Electric Circuits, 6th Edition

Master your textbook with Solution Manual for Fundamentals of Electric Circuits, 6th Edition, offering detailed solutions to every question.

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Page 1 of 16
Solution Manual for Fundamentals of Electric Circuits, 6th Edition - Page 1 preview imageSolution 1.1(a)q = 6.482x1017x [-1.602x10-19C] =–103.84 mC(b) q = 1.24x1018x [-1.602x10-19C] =–198.65 mC(c) q=2.46x1019x [-1.602x10-19C] =–3.941 C(d) q = 1.628x1020x [-1.602x10-19C] =–26.08 C
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Solution Manual for Fundamentals of Electric Circuits, 6th Edition - Page 2 preview image
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Solution Manual for Fundamentals of Electric Circuits, 6th Edition - Page 3 preview imageSolution 1.2Determine the current flowing through an element if the charge flow is given by(a)( )( )mC3tq=(b)( )C4)-20t(4ttq2+=(c)( )()2e15etq18t-3t−−=nC(d) q(t) = 5t2(3t3+ 4) pC(e) q(t) = 2e-3tsin(20πt) μC(a)i = dq/dt = 0 mA(b)i = dq/dt = (8t + 20) A(c)i = dq/dt = (–45e-3t+ 36e-18t) nA(d)i=dq/dt = (75t4+ 40t) pA(e)i =dq/dt = {-6e-3tsin(20πt) + 40πe-3tcos(20πt)} μA
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Solution Manual for Fundamentals of Electric Circuits, 6th Edition - Page 4 preview imageSolution 1.3(a)C1)(3t+=+=∫q(0)i(t)dtq(t)(b)mC5t)(t2+=++=∫q(v)dts)(2tq(t)(c)()q(t)20 cos10t/ 6q(0)(2sin(10/ 6)1)Ctππμ=++=++∫(d)C40t)sin0.12t(0.16cos40e30t-+−=−+=+=∫t)cos40-t40sin30(1600900e10q(0)t40sin10eq(t)-30t30t-
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Solution Manual for Fundamentals of Electric Circuits, 6th Edition - Page 5 preview imageSolution 1.4Since i is equal to Δq/Δt then i = 300/30 =10 amps.
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Solution Manual for Fundamentals of Electric Circuits, 6th Edition - Page 6 preview imageSolution 1.5102010125 C024tqidttdt====∫∫
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Solution Manual for Fundamentals of Electric Circuits, 6th Edition - Page 7 preview imageSolution 1.6(a) At t = 1ms,===230dtdqi15 A(b) At t = 6ms,==dtdqi0 A(c) At t = 10ms,=−==430dtdqi–7.5 A
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Solution Manual for Fundamentals of Electric Circuits, 6th Edition - Page 8 preview imageSolution 1.7<<<<<<−<<==4t310A,3t20A,21,20A1t0A,10dtdqitwhich is sketched below:310–201t(s)i(A)420
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Solution Manual for Fundamentals of Electric Circuits, 6th Edition - Page 9 preview imageSolution 1.8C15μ1102110idtq=×+×==∫
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Solution Manual for Fundamentals of Electric Circuits, 6th Edition - Page 10 preview imageSolution 1.9(a)C10===∫∫10dt10idtq(b)C5.2255.7151521510110idtq30=++=×+×−+×==∫(c)C30=++==∫101010idtq50
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Solution Manual for Fundamentals of Electric Circuits, 6th Edition - Page 11 preview imageSolution 1.10q = it = 10x103x15x10-6=150 mC
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Solution Manual for Fundamentals of Electric Circuits, 6th Edition - Page 12 preview imageSolution 1.11q= it = 90 x10-3x 12 x 60 x 60 =3.888 kCE = pt = ivt = qv = 3888 x1.5 =5.832 kJ
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Solution Manual for Fundamentals of Electric Circuits, 6th Edition - Page 13 preview imageSolution 1.12For 0 < t < 6s, assuming q(0) = 0,q tidtqtdtttt( )( ).=+=+=∫∫0301 5020At t=6, q(6) = 1.5(6)2= 54For 6 < t < 10s,q tidtqdtttt( )( )=+=+=−∫∫61854185466At t=10, q(10) = 180 – 54 = 126For 10<t<15s,q tidtqdtttt( )()()=+=−+= −+∫∫1012126122461010At t=15, q(15) = -12x15 + 246 = 66For 15<t<20s,q tdtqt( )()=+=∫0156615Thus,q tttt( ).,=−−+1 5185412246662C,0 < t < 6sC,6 < t < 10sC,10 < t < 15sC15 < t < 20sThe plot of the charge is shown below.
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Solution Manual for Fundamentals of Electric Circuits, 6th Edition - Page 14 preview image05101520020406080100120140tq(t)
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Solution Manual for Fundamentals of Electric Circuits, 6th Edition - Page 15 preview imageSolution 1.13(a)i= [dq/dt] = 20πcos(4πt) mAp=vi= 60πcos2(4πt) mWAt t=0.3s,p=vi= 60πcos2(4π0.3) mW =123.37 mW(b)W=W= 30π[0.6+(1/(8π))[sin(8π0.6)–sin(0)]] =58.76 mJ
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Solution Manual for Fundamentals of Electric Circuits, 6th Edition - Page 16 preview imageSolution 1.14The voltagev(t)across a device and the currenti(t)through it arev(t)= 20sin(4t) volts andi(t)= 10(1 + e-2t) m-amps.Calculate:(a)the total charge in the device att= 1 s, assume q(0) = 0.(b)the power consumed by the device att =1 s.(a)()()()11-2t-2t-200qidt0.01 1edt0.01 t0.5e0.01 10.5e0.5==+=−=−+∫∫= 0.01(1 – 0.135335 + 0.5) =13.647 mC.(b)p(t) =v(t)i(t);v(1)= 20sin(4) = 20sin(229.18°) = –15.135 volts;andi(1)= 10(1+e-2)(10–3) = 10(1.1353)(10–3) = 11.353 m-ampsp(1) = (–15.125)(11.353)(10–3) =–171.71 mW
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Subject
Electrical Engineering & Electronics
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Fundamentals Of Electric Circuits b...

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