Archives: Solution Manual

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 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 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 12.70 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 12.56 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 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 12.1 (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 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 11.68 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 11.54 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 11.34 472.420 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 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 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 […]

°∠= °∠ °∠ = − =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 […]

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 10.47 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 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 […]

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 […]

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 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 9.59 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 9.40 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 […]

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 9.1 (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 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 […]

RELATION BETWEEN EARNINGS PROCESS AND REVENUE RECOGNITION METHODS ➢ In most situations, the revenue recognition criteria are satisfied at the point of product delivery. T5-46 Instructor’s Resource Manual, Chapter 5 5-59 Copyright © 2015 McGraw-Hill Education. All rights reserved. REVENUE […]

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 […]

FRANCHISE ARRANGEMENTS • Assume that TrueTech starts selling TechStop franchises. TrueTech charges franchisees an initial fee in exchange for (a) the exclusive right to operate the only TechStop in a particular area for a five-year period, (b) the equipment necessary […]

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 […]

RECOGNIZING REVENUE FOR CONTRACTS WITH MULTIPLE PERFORMANCE OBLIGATIONS ➢ Goal: Separate complex contracts into parts that can be viewed on a stand-alone basis. Steps 2 and 4 are critical to this process. ➢ Step 2: Identify the performance obligation(s) in […]

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 […]

CHAPTER 5 Revenue Recognition and Profitability Analysis Overview In Chapter 4 we discussed net income and its presentation in the income statement. In Chapter 5 we focus on revenue recognition, which determines when and how much revenue appears in the […]

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 8.1 (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 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 7.39 (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 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 […]

4-18 Intermediate Accounting, 8/e CHANGE IN ACCOUNTING ESTIMATE ► A change in accounting estimate is reflected in the financial statements of the current period and future periods. T4–12 Copyright © 2015 McGraw-Hill Education. All rights reserved. EARNINGS PER SHARE ➢ […]

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 […]

CHAPTER 4 THE INCOME STATEMENT, COMPREHENSIVE INCOME, AND THE STATEMENT OF CASH FLOWS Overview The purpose of the income statement is to summarize the profit-generating activities that occurred during a particular reporting period. Comprehensive income includes net income as well […]

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 […]

3–56 Intermediate Accounting, 8/e IFRS Case 3–4 Requirement 1 A major difference is the format of Vodafone’s balance sheets (statements of financial position). Under U.S. GAAP, we present current assets and liabilities before noncurrent assets and liabilities. IAS No. 1 […]

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 […]

Problem 3–4 authorized; 400,000 shares issued and outstanding …………. $400,000 Retained earnings ………………………………………………………….. 202,000 Total shareholders’ equity ………………………………………… 602,000 Total liabilities and shareholders’ equity …………………. $992,000 (1) Includes $30,000 in U.S. treasury bills. WEISMULLER PUBLISHING COMPANY Balance Sheet At December […]

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 […]

2. (A) in the summary of significant policies note. 3. (C) on the face of the balance sheet. 4. (B) in a separate disclosure note. 5. (B) in a separate disclosure note. 6. (A) in the summary of significant policies […]