电路基础习题答案.doc

Problem 6.110(1 - 3t)e-3t A09/20/1999p = vi = 10(1-3t)e-3t 2t e-3t = 20t(1 - 3t)e-6t WProblem 6.2w2 = 160 kWProblem 6.3i = C 480 mAProblem 6.4 = 1 - 0.75 cos 4tProblem 6.5v = For 0 t 1, i = 4t,dt + 0 = 100t2 kVv(1) = 100 kVFor 1 t 2, i = 8 - 4t, = 100 (4t - t2 - 3) + 100 kVThus v (t) = Problem 6.6 x slope of the waveform.For example, for 0 t 2,i = Thus the current i is sketched below.i(t) (mA)-15046821012150t (msec)Problem 6.7 = 0.04k2 + 10 VProblem 6.8 v(t) = v(2) = 12(2 + e-2) = 25.62 Vp = iv = 12 (t + e-t) 6 (1-e-t) = 72(t-e-2t)p(2) = 72(2-e-4) = 142.68 WProblem 6.9i(t) = i.e. twice the slope of v.The value of i(t) is plotted as shown below:i(t) (A)-1023415710t (s)6Problem 6.10v = For 0 t 1, kVv(1) = 10 kVFor 1 t 2,For 2 t 3,= -10t + 30kVThusv(t) = Problem 6.11t)= - 0.7e p sin 4pt AP = vi = 60(-0.72)p cos 4p t sin 4p t = -21.6p sin 8p t WW = t dt= = -5.4JProblem 6.12Under dc conditions, the circuit becomes that shown below:10 W30 Wi1+v1-60V+-20 W50 W+v2-i2i2 = 0, i1 = 60/(30+10+20) = 1Av1 = 30i2 = 30V, v2 = 60-20i1 = 40VThus,v1 = 30V, v2 = 40VProblem 6.13(a)Ceq = 4C = 120 mF(b) Ceq = 7.5 mFProblem 6.14In series, as in Fig. (a),v1 = v2 = 100(a)C2+v2-100V+-+v1-C1+ -v1(b)C2+v2-100V+-C1w20 = 0.1Jw30 = 0.15J(b) When they are connected in series as in Fig. (b): v2 = 40w20 = 36 mJw30 = 24 mJProblem 6.15(a)4F in series with 12F = 4 x 12/(16) = 3F3F in parallel with 6F and 3F = 3+6+3 = 12F4F in series with 12F = 3Fi.e. Ceq = 3F(b) Ceq = 5 + 6 | (4 + 2) = 5 + (6 | 6) = 5 + 3 = 8F (c) 3F in series with 6F = (3 x 6)/9 = 6FCeq = 1FProblem 6.16For the capacitors in parallel = 15 + 5 + 40 = 60 mFHenceCeq = 10 mFProblem 6.173 in series with 6 = 6x3/(9) = 22 in parallel with 2 = 44 in series with 4 = (4x4)/8 = 2The circuit is reduced to that shown below:6202186 in parallel with 2 = 88 in series with 8 = 44 in parallel with 1 = 55 in series with 20 = (5x20)/25 = 4ThusCeq = 4 mFProblem 6.184mF in series with 12mF = (4x12)/16 = 3mF3mF in parallel with 3mF = 6mF6mF in series with 6mF = 3mF3mF in parallel with 2mF = 5mF5mF in series with 5mF = 2.5mFHence Ceq = 2.5mFProblem 6.19Combining the capacitors in parallel, we obtain the equivalent circuit shown below:30 mF40 mF60 mF20 mFabCombining the capacitors in series gives , where = 10mFThusCeq = 10 + 40 = 50 mFProblem 6.20(a)3mF is in series with 6mF 3x6/(9) = 2mFv4mF = 1/2 x 120 = 60Vv2mF = 60Vv6mF = 20Vv3mF = 60 - 20 = 40V(b) Hence w = 1/2 Cv2w4mF = 1/2 x 4 x 10-6 x 3600 = 7.2mJw2mF = 1/2 x 2 x 10-6 x 3600 = 3.6mJw6mF = 1/2 x 6 x 10-6 x 400 = 1.2mJw3mF = 1/2 x 3 x 10-6 x 1600 = 2.4mJProblem 6.2120mF is series with 80mF = 20x80/(100) = 16mF09/20/199914mF is parallel with 16mF = 30mF(a) v30mF = 90Vv60mF = 30Vv14mF = 60Vv20mF = 48Vv80mF = 60 - 48 = 12V(b) Since w = w30mF = 1/2 x 30 x 10-6 x 8100 = 121.5mJw60mF = 1/2 x 60 x 10-6 x 900 = 27mJw14mF = 1/2 x 14 x 10-6 x 3600 = 25.2mJw20mF = 1/2 x 20 x 10-6 x (48)2 = 23.04mJw80mF = 1/2 x 80 x 10-6 x 144 = 5.76mJProblem 6.22(a) For the capacitors in series,Q1 = Q2 C1v1 = C2v2 vs = v1 + v2 = Similarly, (b) For capacitors in parallelv1 = v2 = Qs = Q1 + Q2 = orQ2 = i = , Problem 6.23(a) Ceq = C1 + C2 + C3 = 35mF(b) Q1 = C1v = 5 x 150mC = 0.75mCQ2 = C2v = 10 x 150mC = 1.5mCQ3 = C3v = 20 x 150 = 3mC(c)w = = 393.8mJProblem 6.24(a)Ceq = 2.857mF(b) Since the capacitors are in series,Q1 = Q2 = Q3 = Q = Ceqv = 0.5714mV(c)w = 57.143mJProblem 6.25We may treat this like a resistive circuit and apply delta-wye transformation, except that R is replaced by 1/C.20 mFCcCbCa50 mF= Ca = 5mFCb = 15mFCc = 3.75mFCb in parallel with 50mF = 50 + 15 = 65mFCc in series with 20mF = 23.75mF65mF in series with 23.75mF = 17.39mF in parallel with Ca = 17.39 + 5 = 22.39mFHenceCeq = 22.39mFProblem 6.26(a)C in series with C = C/(2)C/2 in parallel with C = 3C/2in series with C = in parallel with C = C + 1.6 C(b)Ceq2C2CCeq = CProblem 6.27vo = For 0 t 1,i = 60t mA,vo(1) = 10kVFor 1 t 2, i = 120 - 60t mA,vo = = 40t 10t2= 40t 10t2 - 20Problem 6.28Ceq = 4 + 6 = 10mFFor 0 t 1,dt + 0 = t2 kVFor 1 t 3,For 3 t 5, Problem 6.29(a) v(0) = 10 = v1(0) + v2(0)10 = 2 + v2(0) v2(0) = 8V(b)Ceq = = -6 + 8e-3t Vv2 = v - v1 = 10e-3t + 6 - 8e-3t= 6 + 2e-3t(b) i = - 480e-3t mA= -180e-3t mAi2 = 50(2)(-3)e-3t mA = -300e-3t mAProblem 6.30i = 6e-t/2 = -30e-t/2 mVv(3) = -300e-3/2 mV = -0.9487 mVp = vi = -180e-t mWp(3) = -180e-3 mW = -0.8 mWProblem 6.31 200 mHProblem 6.32 = - 6 sin 2t mVp = vi = -72 sin 2t cos 2t mWBut 2 sin A cos A = sin 2Ap = -36 sin 4t mWProblem 6.33= 4.8 cos 100t Vp = vi = 4.8 x 4 sin 100t cos 100t = 9.6 sin 200tw = 96 mJProblem 6.34 = Problem 6.35 = At t = ls, i = 10 - 4.7 + 5e-2 = 5.977 Ai2 = 35.72JProblem 6.36For 0 t 1, AFor 1 t 2,i = 0 + i(1) = 1AFor 2 t 3,i = = 2t - 3 AFor 3 t 4,i = 0 + i(3) = 3 AFor 4 t 5,i = = 2t - 5 AThus,Problem 6.37w = L = 144 mJProblem 6.38= 0.8t + sin 2t -1Problem 6.39i(t) = For 0 t 1, v = 5tdt + 0 = 0.25t2 kAFor 1 t 2, v = -10 + 5t = 1 - t + 0.25t2 kAProblem 6.40Under dc conditions, the circuit is as shown below:2 W4 WiL3 A+vC-By current division, 2A,vc = 0V 1J 0J09/20/1999Problem 6.41Under dc conditions, the circuit is equivalent to that shown below:R2 WiL5 A+ vC - If wc = wL, 80 x 10-3R2 = 2R = 5W4 W6 WiL1+ vC1 -30V+-+ vC2 -iL2Problem 6.42Under dc conditions, the circuit is as shown below: 3A 18V 0VProblem 6.43(a) 7H(b) 3H(c) 2H Problem 6.44= 10 + 5|(3 + 2) = 10 + 2.5= 12.5 mHProblem 6.45 L = 10 mH= 7.778 mHProblem 6.46Leq = 20 mHProblem 6.47 Leq = 7HProblem 6.48Since three pairs of inductors can be combined as shown below.L/2L/2LL(a)abLeqL/2L/2L/2(b)abProblem 6.49Hence the given circuit is equivalent to that shown below:LLL/3L/34H5H3H+ -+v2-+ -v12(di/dt)+v-i2i1iProblem 6.50Let (1)(2)i = i1 + i2 i2 = i i1(3)(4)and (5)Incorporating (3) and (4) into (5),Substituting this into (2) givesComparing this with (1),8.375HProblem 6.513 x slope of i(t).Thus v is sketched below:v(t) (V)-62341576t (s)6Problem 6.52(a) (b) Problem 6.53(a)is = i1 + i2 i2(0) = 2mA(b) Using current division: 2.4e-2t mA 3.6e-2t mA(c) -120e-2t mV -144e-2t mV= 24.36nJ = 11.693nJ= 17.54 nJProblem 6.54(a) 40 W 40 W(b) w = w5 + w20 = 80 W(c) = -50e-200tA = -200e-200tA = -200e-200t A(d)i = i1 + i2 = -250e-200t AProblem 6.55 dt + 0 mA 50(1 - cos 4t) mA = 4.8 sin 4t mVProblem 6.56dt, RC = 50 x 103 x 0.04 x 10-6 = 2 x 10-3dtvo = 100 cos 50t mVProblem 6.57dt + v(0), RC = 50 x 103 x 100 x 10-6 = 5vo = The op amp will saturate at vo = -12 = -2t t = 6sProblem 6.58RC = 4 x 106 x 1 x 10-6 = 4For 0 t 1, vi = 20, -5t mVFor 1 t 2, vi = 10, = -2.5t - 2.5mVFor 2 t 4, vi = - 20, = 5t - 17.5 mVFor 4 t 5m, vi = -10, = 2.5t - 7.5 mVFor 5 t 6, vi = 20, = - 5t + 30 mVThus vo(t) is as shown below:v(t) (V)-52341575t (s)6-7.5-2.502.5Problem 6.59One possibility is as follows:Let R = 100 kW, Problem 6.60By combining a summer with an integrator, we have the circuit below:-+R1v3R2R3v2v1CvoFor the given problem, C = 2mF,R1C = 1 R1 = 1/(C) = 1006/(2) = 500 kWR2C = 1/(4) R2 = 1/(4C) = 500kW/(4) = 125 kWR3C = 1/(10) R3 = 1/(10C) = 50 kWProblem 6.61The output of the first op amp is09/20/1999dt = = - 50tdt = = 2500t2At t = 1.5ms, 5.625 mVProblem 6.62Consider the op amp as shown below:Let va = vb = v-+RvRRCvi+-b+vo-RvaAt node a, 2v - vo = 0(1)At node b, (2)Combining (1) and (2),ordtshowing that the circuit is a noninverting integrator.Problem 6.63RC = 0.01 x 20 x 10-3 secThus vo(t) is as sketched below:vo(t) (V)-2122t (ms)3Problem 6.64 -30 mVProblem 6.65 RC = 50 x 103 x 10 x 10-6 = 0.5The input is sketched in Fig. (a), while the output is sketched in Fig. (b).15(a)10055vi(t) (V)t (ms)-1015(b)10055vo(t) (V)t (ms)-+RiiCRf0VviCvoiRiProblem 6.66i = iR + iCHence Thus vi is obtained from vo as shown below:32041 vo(t) (V)t (ms)-4432041dvo(t)/dt t (ms)-443281vi(t) (V)t (ms)-84-4Problem 6.67Thus, by combining integra