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Solution 1.1 (a) q = 6.482×1017 x [-1.602×10–19 C] = –103.84 mC (b) q = 1. 24×1018 x [-1.602×10–19 C] = –198.65 mC (c) q = 2.46×1019 x [-1.602×10–19 C] = –3.941 C (d) q = 1.628×1020 x [-1.602×10–19 C] […]

Solution 1.20 p30 volt source = 30x(–6) = –180 W p12 volt element = 12×6 = 72 W p28 volt e.ement with 2 amps flowing through it = 28×2 = 56 W p28 volt element with 1 amp flowing through […]

°∠= °∠ °∠ = − =43.911.0 43.1–01.40 904 j40 4j o V Therefore, =)t(v o 100 cos(4×104 t + 91.43°) mV Copyright © 2017 McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw–Hill […]

I = 2.0000 2.1210 + 2.1210i >> V=inv(Y)*I V = 5.2793 – 5.4190i 9.6145 – 9.1955i Vs = V1 – V2 = –4.335 + j3.776 = 5.749∠138.94˚ V. Copyright © 2017 McGraw–Hill Education. All rights reserved. No reproduction or distribution […]

To find th V , consider the circuit in Fig. (b). )12060( 2j1 4j )12060( 20j10j5 20j th °∠ + =°∠ +− =V = 107.3∠146.56° V = °∠ °∠ == 7.33–633.21 56.1463.107 th th N Z V I 4.961∠-179.7° A […]

Solution 10.32 Determine Vo and Io in the circuit of Fig. 10.80 using mesh analysis. Figure 10.80 For Prob. 10.32. Solution Consider the circuit below. For mesh 1, 03)30–10(2)42( 1 =+°∠−+ o jVI where )30–10(2 1 IV −°∠= o 3 […]

Solution 10.1 We first determine the input impedance. 1 1 10 10H j L jx j ω  → = = 1 1 1 0.1 10 1 Fj jC jx ω  → = = − 1 1 11 1 […]

Solution 10.83 The schematic is shown below. The frequency is 15.159 2 1000 2/f = π =πω= When the circuit is saved and simulated, we obtain from the output file FREQ VM(1) VP(1) 1.592E+02 6.611E+00 -1.592E+02 Thus, vo = 6.611cos(1000t […]

Solution 10.47 Let 321o iiii ++= , where 1 i , 2 i , and 3 i are respectively due to the 24-V dc source, the ac voltage source, and the ac current source. For 1 i , consider the […]

Solution 11.1 )t50cos(160)t(v = i(t) = –33sin(50t–30˚) = 33cos(50t–30˚+180˚–90˚) = 33cos(50t+60˚) p(t) = v(t)i(t) = 160x33cos(50t)cos(50t+60˚) = 5280(1/2)[cos(100t+60˚)+cos(60˚)] = [1.320+2.640cos(100t+60˚)] kW. P = [VmIm/2]cos(0–60˚) = 0.5x160x33×0.5 = 1.320 kW. Copyright © 2017 McGraw–Hill Education. All rights reserved. No reproduction or […]

Solution 11.83 (a) 2 2 W1.68835cos840cos === o SP θ (b) S = 840 VA (c) VAR 8.48135sin840sin === o SQ θ (d) (lagging) 8191.035cos/=== o SPpf ooo VIS35840)258)(60210( 1 1∠=−∠∠== ∗ Copyright © 2017 McGraw–Hill Education. All rights reserved. […]

Solution 11.54 For the network in Fig. 11.73, find the complex power absorbed by each element. Figure 11.73 For Prob. 11.54. 20 Ω –j20 Ω 1∠0° A Solution Step 1. P10 = (1)210, Qj10 = (1)2(10), and we need to […]

Solution 11.68 Let cLR SSSS ++= where 0jRI 2 1 jQP2 oRRR +=+=S LI 2 1 j0jQP 2 oLLL ω+=+=S C 1 I 2 1 j0jQP2 occc ω ⋅−=+=S Hence, =S          […]

Solution 11.34 472.420 2036 3 9 3 1 2 0 3 == =        += rms f t frms = 4.472 6)3( 3 1 )( 1 3 2 2 2 0 2 0 22 […]

At node 2, )25.025.0(75.00 4 5.0 4 21 2 1 21 jVV j V V VV +−+=→=+ − or V1 = (0.33333–j0.33333)V2 = (0.4714∠–45°)V2 (2) Substituting (2) into (1) leads to (1.25+j0.2)( 0.33333–j0.33333)V2 – 0.25V2 = 25 = [0.41666+0.066666 – […]

Solution 12.1 (a) If 400 ab =V , then =°∠= 30– 3 400 an V V30–231 °∠ = bn V V150–231 °∠ = cn V V270–231 °∠ (b) For the acb sequence, °∠−°∠=−= 120V0V ppbnanab VVV °∠=    […]

Solution 12.70 8004001200PPP 21T =−=+= 1600–1200400–PPQ 12T =−=−= °=θ→===θ -63.43-2 800 1600– P Q tan T T =θ= cospf (leading)4472.0 40 6 240 I V Z L L p= == = p Z Ω°∠ 63.43–40 Copyright © 2017 McGraw–Hill Education. […]

Solution 12.81 °=θ→= 87.36–(leading)8.0pf 1 kVA36.87–150 1°∠=S °=θ→= 00.1pf 2 kVA0100 2°∠=S °=θ→= 13.53(lagging)6.0pf 3 kVA53.13200 3°∠=S kVA95j80 4 +=S 4321 SSSSS +++= kVA21.452.451165j420 °∠=+=S LL IV3S = A7.542 4803 102.451 V3 S I 3 L L= × × == […]

Solution 12.19 For the ∆–∆ circuit of Fig. 12.50, calculate the phase and line currents. Figure 12.50 For Prob. 12.19. Solution °∠=+= ∆18.4362.3110j30Z The phase currents are = °∠ °∠ == ∆ 18.4362.31 0440 Z V I ab AB 13.915∠–18.43° […]

Solution 12.56 Using Fig. 12.63, design a problem to help other students to better understand unbalanced three–phase systems. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in […]

Solution 12.39 Find the real power absorbed by the load in Fig. 12.58. c b B C (36+j28) Ω (36+j28) Ω − + 100∠–30° V 100∠–150° + − (4+j2) Ω (4+j2) Ω Figure 12.58 For Prob. 12.39. Solution To find […]

Solution 13.29 In the circuit of Fig. 13.97, find the value of the coupling coefficient k that will make the 10-Ω resistor dissipate 1.28 kW. For this value of k, find the energy stored in the coupled coils at t […]

Solution 13.87 ZTh = ZL/n2 or n = 300/75Z/Z ThL= = 0.5 Copyright © 2017 McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw–Hill Education. Solution 13.88 n = V2/V1 = I1/I2 or […]

Solution 13.77 (a) This is a single phase transformer. V1 = 13.2 kV, V2 = 120 V n = V2/V1 = 120/13,200 = 1/110, therefore n = 1/110 or 110 turns on the primary to every turn on the secondary. […]

Solution 3.17 3 40 40 40 2666.67 15 10 jL j Lx ωω − =  → = = = 33 12 0.6 12 10 30 10 62.35 mHM k LL x x x −− = = = If 15 […]

Solution 13.1 For coil 1, L1 – M12 + M13 = 12 – 8 + 4 = 8 For coil 2, L2 – M21 – M23 = 16 – 8 – 10 = – 2 For coil 3, L3 + […]

Solution 13.60 (a) Transferring the 40-ohm load to the middle circuit, ZL’ = 40/(n’)2 = 10 ohms where n’ = 2 10||(5 + 10) = 6 ohms We transfer this to the primary side. Zin = 4 + 6/n2 = […]

Solution 13.44 We can apply the superposition theorem. Let i1 = i1’ + i1” and i2 = i2’ + i2” where the single prime is due to the DC source and the double prime is due to the AC source. […]

Solution 14.57 (a) Consider the circuit below. Z V I s = )sRC2(RsC2 sC1 s 1 + = + =Z V II 222 s 1o CRssRC31 )sRC2(sC sRC2 R R++ + ⋅ + == V IV 222 s o CRssRC31 […]

Solution 14.88 The schematic is shown below. We insert a voltage marker to measure Vo. In the AC Sweep box, we set Total Points = 101, Start Frequency = 1, and End Frequency = 100. After simulation, we obtain the […]

Solution 4.39 222 LR LjR CjCj LjR 1 Yω+ ω− +ω=ω+ ω+ = At resonance, 0)Im( =Y , i.e. 0 LR L C22 0 2 0 0= ω+ ω −ω C L LR 22 0 2 =ω+ 2 -36–-32 2 […]

Solution 14.96 22 2 L1 CsR1 sC ||R + ==Z       + ++ =+= 2L 2L 2 L 1 1 1 2CsR1 LCRsRsL || sC 1 )sL(|| sC 1ZZ L R 1 2L 2L 2 […]

Solution 14.74 Ω=== 300100x3RK‘R 1m1 Ω=== k 1100x10RK‘R 2m2 H 200)2( 10 10 L K K ‘L 6 2 f m µ=== nF 1 10 10 1 KK C ‘C 8 fm === Copyright © 2017 McGraw–Hill Education. All rights […]

Solution 14.20 Design a more complex problem than given in Prob. 14.10, to help other students to better understand how to determine the Bode magnitude and phase plots of a given transfer function in terms of jω. Include at least […]

Solution 14.1 Find the transfer function Io/Ii of the RL circuit in Fig. 14.68. Express the transfer function using ωo = R/L. Figure 14.68 For Prob. 14.1. Solution H(ω) = Io/Ii = [RjωL/(R+jωL)]/(jωL) = 1/(1+jωL/R) If we let ωo = […]

Solution 15.19 Since [ ] 1)t( =δL and 2T = , =)s(F s2– e1 1 − Copyright © 2017 McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw–Hill Education. Solution 15.20 Using Fig. […]

Therefore, =)t(y       <<−+ << otherwise,0 6t5,12t5)2t(– 5t3,21 2  <<+− 3t2,2t2)2t( 2 Copyright © 2017 McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw–Hill Education. Solution 15.45 […]

Solution 15.35 (a) Let 2s B 1s A )2s)(1s( 3s )s(G + + + = ++ + = 2A = , -1B = 2t–t– ee2)t(g 2s 1 1s 2 )s(G −=→ + − + = )6t(u)6t(g)t(f)s(Ge)s(F -6s −−=→= =)t(f [ […]

Solution 15.54 Design a problem to help other students to better understand solving second order differential equations with a time varying input. Although there are many ways to solve this problem, this is an example based on the same kind […]

Solution 15.1 Find the Laplace transform of 5 sin(at)cos(bt). [Hint: using the exponential representation for both functions may make this problem easier.] Solution. Step 1. Although we could work with trigonometric identities and the Laplace transforms for sin and cos, […]

Solution 16.74 Design a problem to help other students to better understand how to find outputs when given a transfer function and an input. Although there are many ways to solve this problem, this is an example based on the […]

Solution 16.89 First select the inductor current iL (left to right) and the capacitor voltage vC to be the state variables. Letting vo = vC and applying KCL we get: sCL sLCCs C CL vvi iivvori v vi +−= ++−==−++− […]

Solution 16.19 The switch in Fig. 16.42 moves from position A to position B at t=0 (please note that the switch must connect to point B before it breaks the connection at A, a make before break switch). Find v(t) […]

Solution 16.1 The current in an RLC circuit is described by 2 2 10 25 0 d i di i dt dt + += If i(0) = 7 A and di(0)/dt = 0, find i(t) for t > 0. Solution […]

)40s5.0s)(2s( )20s( 5.7V 2s )20s( 15V)40s5.0s(2 2 2 x 2 x 2 −++ + −= + + −=−+ We should note that this represents an unstable system because at least one root of the denominator has to be positive resulting […]

Solution 16.34 Solve for the mesh currents in the circuit of Fig. 16.57. You may leave your results in the s-domain. Figure 16.57 For Prob. 16.34. Solution In the s-domain, the circuit is as shown below. [20/s] = [1+(s/4)]I1 – […]

°∠= °−∠ = °∠°∠°∠ = 68.880166.1 68.88886.6 7 )90853.2)(33.67546.1)(99.1135613.1( 7 )4265.16347.0( 68.880166.1 )4265.16347.0( 68.880166.1 )2306.1( 38.2333.2 jsjsss V o −+ °∠ + ++ °−∠ + + − += or vo(t) = [2.333–2.38e–1.2306t+1.0166e–0.6347t(e–(1.4265t+88.68˚)+e(1.4265t+88.68˚))]u(t) volts or = [2.333–2.38e–1.2306t+2.033e–0.6347tcos(1.4265t+88.68˚)]u(t) V. Answer does check […]

Solution 17.1 (a) This is periodic with ω = π which leads to T = 2π/ω = 2. (b) y(t) is not periodic although sin t and 4 cos 2πt are independently periodic. (c) Since sin A cos B = […]

Solution 17.15 (a) Dcos ωt + Esin ωt = A cos(ωt – θ) where A = 22 ED + , θ = tan-1(E/D) A = 622 n 1 )1n( 16 + + , θ = tan-1((n2+1)/(4n3)) f(t) = ∑ ∞ […]

enable Fourier. After simulation, we compare the output and output waveforms as shown. The output includes the following Fourier components. FOURIER COMPONENTS OF TRANSIENT RESPONSE V(1) Copyright © 2017 McGraw–Hill Education. All rights reserved. No reproduction or distribution without the […]

Solution 17.57 ao = (6/–2) = –3 = co cn = 0.5(an –jbn) = an/2 = 3/(n3 – 2) f(t) = ∑ ∞ ≠−∞=− +− 0n n nt50j 3e 2n 3 3 Copyright © 2017 McGraw–Hill Education. All rights reserved. […]

Vs = [1 + jnπ/4]Vx + Vo/3 (3) But –Vx – 2Vx + Vo = 0 or Vo = 3Vx Substituting this into (3), Vs = [1 + jnπ/4]Vx + Vx = [2 + jnπ/4]Vx = (1/3)[2 + jnπ/4]Vo = […]

Solution 17.28 Obtain the trigonometric Fourier series for the voltage waveform shown in Fig. 17.66. Figure 17.66 For Prob. 17.28. Solution This is half–wave symmetric since f(t − T/2) = −f(t). ao = 0, T = 2, ωo = 2π/2 […]

Solution 18.35 (a) x(t) = f[3(t-1/3)]. Using the scaling and time shifting properties, )j6( e e 3/j2 1 3 1 )(X 3/j 3/j ω+ = ω+ =ω ω− ω− (b) Using the modulation property,       […]

Solution 18.1 Obtain the Fourier transform of the function in Fig. 18.26. Figure 18.26 For Prob. 18.1. Solution f(t) = 7u(t+2) – 7u(t+1) – 7u(t–1) + 7u(t–2) f’(t) = 7δ(t+2) – 7δ(t+1) – 7δ(t–1) + 7δ(t–2) jωF(ω) = 7(ej2ω – […]

Solution 18.15 (a) =−=ω ω−ω 3j3j ee)(F ω3sinj2 (b) Let ω− =ω−δ= j e2)(G),1t(2)t(g =ω)(F F      ∫∞− tdt)t(g )()0(F j )(G ωδπ+ ω ω = )()1(2 j e2 j ωδ−πδ+ ω = ω− = ω […]

Solution 18.51 In the frequency domain, the voltage across the 2-Ω resistor is 2 2 10 20 () , 2 2 1 ( 1)( 2) s V V sj j j j ss ωω ω ωω = = = = […]

Alternatively, from the given circuit, 211 1.04 VIV −= 212 1.020 VII += Comparing these with the equations for the h parameters show that 4 11 =h , -0.1 12 =h , 20 21 =h , 1.0 22 =h Using […]

Substituting these into (1) and (2), s11 s 1111s1s Zz V IIzZIV + =→=− s11 s21 1212 Zz Vz IzV + == == 2Th VV s11 s21 Zz Vz + Copyright © 2017 McGraw–Hill Education. All rights reserved. No reproduction […]

Checking using nodal analysis we get, –10 + 2Ia + 2(Ia–Ib) = 0 and 2(Ib–Ia) + 1Ib + 5Ib = 0. From the second equation we get Ia = 4Ib and then from the first equation we get 8Ib + […]

Solution 19.93 We apply the same formulas derived in the previous problem. E Eie R h )Rh( 1−+ + = = +×+ = ++ = )3800200(101 )RR(h1 A5– ELoe i 5.144 )RR(h1 )hR1)(RhR(h RhZ ELoe oeELreEfe Eiein ++ +− ++= […]

Solution 19.86 (a) By definition, g11 = 0I 1 1 2 V I = , g21 = 0I 2 1 2 V V = . We let V1 = 1 V and open-circuit the output port. The schematic is shown […]

Solution 19.1 Obtain the z parameters for the network in Fig. 19.65. 10 Ω 10 Ω 10 Ω Figure 19.65 For Prob. 19.1. Solution Step 1. Label the circuit to allow us to determine the z–parameters. The z–parameter equations are, […]

222121o IzIzV += (2) But Lo2L2o Z/VIZIV −=→−= (3) From (2) and (3) ,        +=→−= 21L 22 21 o1 L o 22121o zZ z z 1 VI Z V zIzV (4) Substituting (3) […]

indicates that 0 2=I and that =][y S 5j75.01 105.121050 -6-6       + ×× -6-6 y10)250j65(10)5.12.25j5.77( ×+=×−+=∆ =             ∆ = 11 y 11 […]

Solution 19.50 To get a and c, consider the circuit below. I1=0 2 s I2 + + V1 4/s V2 – – 2 1 2 2 2 21 s25.01 V V aV 4s 4 V s/4s s/4 V+==→ + = […]

Solution 2.57 Find Req and I in the circuit of Fig. 2.121. Figure 2.121 For Prob. 2.57. Solution Rab = Ω== ++ 30 10 300 10 101010101010 xxx Rac = 216/(8) = 27Ω, Rbc = 36 Ω Rde = Ω== […]

Solution 2.71 Figure 2.131 represents a model of a solar photovoltaic panel. Given that Vs = 95 V, R1 = 25 Ω, iL = 2 A, find RL. Figure 2.131 For Prob. 2.71. Step 1. Vs = iL(R1+RL) or RL […]

Solution 2.39 Evaluate Req looking into each set of terminals for each of the circuits shown in Fig. 2.103. 3 Ω 6 Ω 6 Ω 3 Ω 6 kΩ 6 kΩ 2 kΩ (a) (b) Figure 2.103 For Prob. 2.39. […]

Solution 2.21 Applying KVL, -15 + (1+5+2)I + 2 Vx = 0 But Vx = 5I, -15 +8I + 10I =0, I = 5/6 Vx = 5I = 25/6 = 4.167 V Copyright © 2017 McGraw–Hill Education. All rights reserved. […]

Solution 2.1 Design a problem, complete with a solution, to help students to better understand Ohm’s Law. Use at least two resistors and one voltage source. Hint, you could use both resistors at once or one at a time, it […]

Solution 3.36 Use mesh analysis to obtain ia, ib, and ic in the circuit shown in Fig. 3.84. 30 V − + 5 Ω 10 Ω 15 Ω 20 Ω 45 V – + ia ib ic Figure 3.84 For […]

Solution 3.1 Using Fig. 3.50, design a problem to help other students to better understand nodal analysis. Figure 3.50 12 V + − Ix R3 9 V + − For Prob. 3.1 and Prob. 3.39. Solution Given R1 = 4 […]

Solution 3.64 40 Ω i1 i2 i3 – + 250V + – 4i 0 10 Ω 5 A i 0 v0 For mesh 2, 20i2 – 10i1 + 4i0 = 0 (1) But at node A, io = i1 – […]

Solution 3.20 For the circuit in Fig. 3.69, find v1, v2, and v3 using nodal analysis. 20 Ω 20i + − 40 Ω 10 Ω 40 Ω v2 v3 v1 i Figure 3.69 For Prob. 3.20. Step 1. This is […]

Solution 3.50 Use mesh analysis to find the current io in the circuit in Fig. 3.95. 52 V Figure 3.95 For Prob. 3.50. Step 1. We note that we have three unknown loop currents but only two mesh equations (one […]

Solution 3.78 The schematic is shown below. When the circuit is saved and simulated the node voltages are displayed on the pseudo components as shown. Thus, ,V15 V,5.4 V,3 321 −==−= VVV . Copyright © 2017 McGraw–Hill Education. All rights […]

The Thevenin equivalent circuit is shown below. 44k Ω I Ri + 72 V – 2k Ω mA 244 72 R I ++ = i assuming Ri is in k-ohm. (a) When Ri =500 Ω , mA 1.548 5.0244 72 […]

Solution 4.34 Using Fig. 4.102, design a problem that will help other students better understand Thevenin equivalent circuits. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in […]

Solution 4.1 Ω=+ 20)1525(40 , i = [30/(5+20)] = 1.2 and io = i20/40 = 600 mA. Since the resistance remains the same we get can use linearity to find the new value of the voltage source = (30/0.6)5 = […]

For io3, consider the circuit below. 3 + 2 + 4||10 = 5 + 20/7 = 55/7 i2 = [5/(5 + 55/7)]18 = 7, io3 = [–10/(10 + 4)]i2 = –5 io = 12 – 6 – 5 = 1 […]

Solution 4.63 Because there are no independent sources, IN = Isc = 0 A RN can be found using the circuit below. Applying KCL at node 1, v1 = 1, and vo = (20/30)v1 = 2/3 io = (v1/30) – […]

Solution 4.47 Obtain the Thevenin and Norton equivalent circuits of the circuit in Fig. 4.114 with respect to terminals a and b. 20 Ω 20Ix + − 1 amp 20 Ω Ix Figure 4.114 For Prob. 4.47. Solution Step 1. […]

V = 15 V [zero intercept] R = (18.2 – 15)/1 = 3.2 ohms Copyright © 2017 McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw–Hill Education. Solution 4.78 The schematic is shown […]

Solution 5.1 (a) Rin = 1.5 MΩ (b) Rout = 60 Ω (c) A = 8×104 Therefore AdB = 20 log 8×104 = 98.06 dB Copyright © 2017 McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior […]

Solution 5.39 For the op amp circuit in Fig. 5.76, determine the value of v2 in order to make vo = –7.5 V. 10 kΩ − + 50 kΩ v2 vo 50 kΩ 20 kΩ –3 V 5 V Figure […]

Solution 5.56 Using Fig. 5.83, design a problem to help other students better understand cascaded op amps. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the […]

v4 + − 2R 2R 11/4A Since the current through the equivalent 21R/11-ohm resistor is (11/4) amps, the voltage across the 2R-ohm resistor on the right is (21/4)R volts. This means the current going through the 2R- ohm resistor is […]

Solution 5.76 The schematic is shown below. IPROBE is inserted to measure io. Upon simulation, the value of io is displayed on IPROBE as io = –562.5 µA 750 mV 0.750V 375mV –19.358uV –11.25V –936.8mV 2 kΩ 11.25V Copyright © […]

Solution 5.19 We convert the current source and back to a voltage source. 3 4 42 = (1.5/3)V + − − + vo 2 kΩ =            + −= 3 […]

Solution 6.1 ( ) =−== −− tt tee dt dv Ci 33 625.7 15(1 – 3t)e–3t A p = vi = 15(1–3t)e–3t ⋅ 2t e–3t = 30t(1 – 3t)e–6t W. 15(1 – 3t)e–3t A, 30t(1 – 3t)e–6t W Copyright © […]

Solution 6.21 4µF in series with 12µF = (4×12)/16 = 3µF 3µF in parallel with 3µF = 6µF 6µF in series with 6µF = 3µF 3µF in parallel with 2µF = 5µF 5µF in series with 5µF = 2.5µF Hence […]

Solution 6.60 8 15 5//3== eq L ( ) tt eqo ee dt d dt di Lv 22 154 8 15 −− −=== ∫∫ −− +=−+=+= tt tt t ooo edteidttv L I i 00 22 0 5.12)15( 5 1 […]

Solution 6.40 10, 2 4 30 5 , 4 6 i t ms t t ms  = <<  − <<  3 3 5, 0 2 25, 0 2 5 10 0, 2 4 0, 2 4 10 […]

Solution 6.74 RC = 0.01 x 20 x 10-3 sec secm dt dv 2.0 dt dv RCv i o−=−=      <<− << <<− = 4t3,V2 3t1,V2 1t0,V2 v o Thus vo(t) is as sketched below: -2 […]

Solution 7.75 In the circuit of Fig. 7.140, find vo and io, given that vs = 10[1 – e–t]u(t) V. vo vs − + 100 kΩ 100 kΩ 10 µF − Figure 7.140 For Prob. 7.75. Solution Let va = […]

Solution 7.58 For t < 0, 0)t(v o = For t > 0, 10)0(i = , 5 31 20 )(i = + =∞ Ω=+= 431R th , 16 1 4 41 R L th ===τ [ ] τ ∞−+∞= t– […]

Solution 7.19 In the circuit of Fig. 7.99, find i(t) for t > 0 if i(0) = 5 A. Figure 7.99 For Prob. 7.19. Solution To find th R we replace the inductor by a 1-V voltage source as shown […]

Solution 7.1 (a) τ=RC = 1/200 For the resistor, V=iR= 200 200 3 56 56 8Re 10 7 k 8 tt e x R − −− =  → = = Ω 3 11 0.7143 200 200 7 10 C […]

Solution 7.39 (a) Before t = 0, = + =)20( 14 )t(v V4 After t = 0, [ ] τ ∞−+∞= t– e)(v)0(v)(v)t(v 8)2)(4(RC ===τ , 4)0(v = , 20)(v =∞ 8t– e)204(20)t(v −+= =)t(v Ve1620 8t– − 1 (b) […]

Solution 8.1 (a) At t = 0–, the circuit has reached steady state so that the equivalent circuit is shown in Figure (a). VS + − 6 Ω (a) + − v 10 H 10 µF + − vL i(0-) […]

Solution 8.60 Obtain i1 and i2 for t > 0 in the circuit of Fig. 8.106. Figure 8.106 For Prob. 8.60. Solution Since the independent source is equal to zero until t = 0, i1(0) = i2(0) = 0. Applying […]

Solution 8.34 Calculate i(t) for t > 0 in the circuit in Fig. 8.82. 35u(–t) V 5 Ω Figure 8.82 For Prob. 8.34. Solution Before t = 0, the capacitor acts like an open circuit while the inductor behaves like […]

Thus, v(t) = Ldi/dt = [1.323(–Asin1.323t + Bcos1.323t)e–0.5t] + [–0.5(Acos1.323t + Bsin1.323t)e–0.5t] = [1.3229(5sin1.3229t – 1.8898cos1.3229t)e–0.5t] + [(2.5cos1.3229t + 0.9449sin1.3229t)e–0.5t] v(t) = [(–0cos1.323t + 4.536sin1.323t)e–0.5t] V = [(7.559sin1.3229t)e–t/2] V. Copyright © 2017 McGraw–Hill Education. All rights reserved. No reproduction or […]

Solution 8.18 When the switch is off, we have a source-free parallel RLC circuit. 5.0 2 1 ,2 125.0 11 ===== RC xLC o αω 936.125.04case dunderdampe 2 2 d=−=−=→< αωωωα oo Io(0) = i(0) = initial inductor current = […]

Solution 8.71 The schematic is shown below. We use VPWL and IPWL to model the 39 u(t) V and 13 u(t) A respectively. We set Print Step to 25 ms and Final Step to 4s in the Transient box. A […]

Solution 9.1 (a) Vm = 50 V. (b) Period 220.2094 30 Ts ππ ω === = 209.4ms (c ) Frequency f = ω/(2π) = 30/(2π) = 4.775 Hz. (d) At t=1ms, v(0.01) = 50cos(30×0.01rad + 10˚) = 50cos(1.72˚ + 10˚) […]

Solution 9.59 For the network in Fig. 9.66, find Zin. Let ω = 100 rad/s. Zin 100 mH 1 mF Figure 9.66 For Prob. 9.59. Solution At ω = 100 rad/s the capacitor becomes –j10 Ω and the inductor becomes […]

Solution 9.21 (a) oooo jF 86.343236.8758.48296.690304155 ∠=+=−−∠−∠= )86.3430cos(324.8)( o ttf += (b) ooo jG 49.62565.59358.4571.2504908 −∠=−=∠+−∠= )49.62cos(565.5)( o ttg −= (c) ( ) 40,9050010 j 1 Hoo =ω−∠+∠ ω = i.e. ooo 69.1682748.125.125.0j18025.19025.0H −∠=−−=−∠+−∠= h(t) = 1.2748cos(40t – 168.69°) Copyright […]

Solution 9.76 (a) 2 8sin 5 8cos(5 90 ) o vtt= = − v1 leads v2 by 70o. (b) 26sin 2 6cos(2 90 ) o vtt= = − v1 leads v2 by 180o. (c ) 1 4cos10 4cos(10 180 ) […]

Solution 9.40 In the circuit of Fig. 9.47, find io(t) when: (a) ω = 1 rad/s (b) ω = 5 rad/s (c) ω = 10 rad/s io(t) 10 Ω 50 mF 2 H 10 Ω 5 sin (ωt) A Figure […]