Archives: Solution Manual

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 […]

Question 3–1 The purpose of the balance sheet, also known as the statement of financial position, is to present the financial position of the company on a particular date. Unlike the income statement, which is a change statement that reports […]

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 © […]

FINANCIAL DISCLOSURES  Summary of Significant Accounting Policies ➢ A summary of the company’s significant accounting policies is a required disclosure. Examples include principles of consolidations, definition of cash equivalents, valuation of inventory, method for recording depreciation, and policy for […]

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 […]

CHAPTER 3 THE BALANCE SHEET AND FINANCIAL DISCLOSURES Overview Chapter 1 stressed the importance of the financial statements in helping investors and creditors predict future cash flows. The balance sheet, along with accompanying disclosures, provides relevant information useful in helping […]

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 © […]

Problem 2–9 (continued) Insurance expense Utility expense ___________________________ ___________________________ Bal. 0 Bal. 30,000 Adjusting 1,500 _______________ ______________ 12/31 Bal. 1,500 12/31 Bal. 30,000 Maintenance expense ___________________________ Bal. 15,000 _______________ 12/31 Bal. 15,000 Solutions Manual, Vol.1, Chapter 2 2–81 Copyright © […]

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 […]

Problem 2–4 (continued) Sales revenue Interest revenue ___________________________ ___________________________ 148,000 Bal. 0 Bal. 7. 2,000 1,333 4. Closing 146,000 Closing 1,333 _______________ ______________ 0 12/31 Bal. 0 12/31 Bal. Cost of goods sold Salaries and wages expense ___________________________ ___________________________ Bal. […]

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 […]

Exercise 2–24 Transaction Journal 1. Purchased merchandise on account. PJ 2. Collected an account receivable. CR 3. Borrowed $20,000 and signed a note. CR 4. Recorded depreciation expense. GJ 5. Purchased equipment for cash. CD 6. Sold merchandise for cash. […]

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 […]

Exercise 2–9 1. Interest receivable ($90,000 x 8% x 3/12) ………………….. 1,800 Interest revenue …………………………………………….. 1,800 2. Rent expense ($6,000 x 2/3) …………………………………… 4,000 Prepaid rent …………………………………………………… 4,000 3. Rent revenue ($12,000 x 7/12) ………………………………… 7,000 Deferred rent revenue ……………………………………. […]

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 […]

Question 2–1 External events involve an exchange transaction between the company and a separate economic entity. For every external transaction, the company is receiving something in exchange for something else. Internal events do not involve an exchange transaction but do […]

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 […]

ACCRUALS  Accruals involve transactions where the cash outflow or inflow occurs in a period subsequent to expense or revenue recognition. ➢ ACCRUED LIABILITIES Accrued liabilities represent liabilities recorded when an expense has been incurred prior to cash payment. To […]

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 […]

CHAPTER 2 REVIEW OF THE ACCOUNTING PROCESS Overview Chapter 1 explained that the primary means of conveying financial information to investors, creditors, and other external users is through financial statements and related notes. The purpose of this chapter is to […]

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) – […]

Exercise 1–14 Statement Concept 1. d. Monetary unit assumption 2. h. Full-disclosure principle 3. g. Expense recognition 4. e. Historical cost principle 5. c. Periodicity assumption 6. a. Economic entity assumption 7. i. Cost effectiveness 8. j. Materiality 9. f. […]

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. […]

Question 1–1 Financial accounting is concerned with providing relevant financial information about various kinds of organizations to different types of external users. The primary focus of financial accounting is on the financial information provided by profit– oriented companies to their […]

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 […]

A AP PP PE EN ND DI IX X A A D De er ri iv va at ti iv ve es s L Le ec ct tu ur re e O Ou ut tl li in ne e date, […]

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 […]

1-18 Intermediate Accounting, 8/e QUALITATIVE CHARACTERISTICS OF ACCOUNTING INFORMATION PRIMARY QUALITATIVE CHARACTERISTICS ➢ Relevance  Predictive value  Confirmatory value  Enhancing aspect: materiality ➢ Faithful representation  Completeness — All information that is necessary for faithful representation.  Neutrality […]

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 = […]

CHAPTER 1 ENVIRONMENT AND THEORETICAL STRUCTURE OF FINANCIAL ACCOUNTING Overview The primary function of financial accounting is to provide useful financial information to users external to the business enterprise. The focus of financial accounting is on the information needs of […]

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 […]

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 3.50 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 3.36 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 3.20 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 3.1 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 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 2.57 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 2.39 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 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 […]

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] […]

When I1=0 V2= 80 V,V 1= 400 V,I 2= 3 A; h12 =V1 V2I1=0 =400 80 = 5; h21 =I2 I1V2=0 =−1 1=−1; h22 =I2 V2I1=0 =3 80 = 37.5 mS. ZTh =Zg+h11 h22Zg+∆h= 10 Ω. I2=h21Vg (1 + h22ZL)(h11 […]

h12 =−sM R+sL;h21 =sM R+sL (checks). h11h22 −h12h21 =(R+sL)2−s2M2 R+sL · 1 R+sL −(sM)(−sM) (R+sL)2 =(R+sL)2−s2M2+s2M2 (R+sL)2= 1 (checks). · .. a 11 =V1 V2I2=0 =3. I1=V1 −j50 +V1−5V2 100 +V1−V2 20 =V1j 50 +1 100 +1 20−V25 100 +1 […]

I2=V2 15 +V2 5;I2 V2 =y22 =✓4 15◆S; Two-Port Circuits Assessment Problems AP 18.1 With port 2 short-circuited, we have I1=V1 20 +V1 5;I1 V1 =y11 =0.25 S; I2=✓20 25 ◆I1=0.8I1. With port 1 short-circuited, we have 18–1 18 © […]

H(s)= Io Vg =100s s2+50s+ 400 =100s (s+ 10)(s+ 40); H(jω)= 100(jω) (jω+ 10)(jω+ 40); Vg(ω)= 6 jω; Io(ω)=H(jω)Vg(ω)= 600 (jω+ 10)(jω+ 40) =20 jω+1020 jω+40; io(t) = (20e10t20e40t)u(t)A. [b] io(0)=0. [c] io(0+)=0. Problems 17–21 Vo3/s 0.5+0.01s+(Vo+3/s)s 4= 0; P […]

=1 j2⇡t{4ej2t4ej3t+ej2tej2t+4ej3t4ej2t} =1 ⇡t“3ej2t3ej2t j2+4ej3t4ej3t j2# =1 ⇡t(4 sin 3t3 sin 2t). =!0 (s+a)2+!2 0s=jω =!0 (a+j!)2+!2 0 . [b] F(!)=L{f(t)}s=jω=“1 (s+a)2#s=jω =1 (aj!)2. The Fourier Transform Assessment Problems AP 17.1 [a] F(!)=Z0 τ/2(Aejωt)dt +Zτ/2 0Aejωtdt AP 17.2 f(t)= 1 […]

=57,600 T2 t3 3 T/8 0 +400t T/4 T/8 +3200 T t2 2 T/4 T/8 +6400 T2 t3 3 T/4 T/8 =57,600 1536 T+ 400T 8+ 16003T 64 + 6400 7T 1536 =575 3T. Vrms =s1 T✓575 3T◆=s575 3= 13.84 […]

an=−320 π(n2−1) for neven; b1= 20 and bn= 0 for all other n. A1= 20 and −θ1= 90; An=−320 π(n2−1) and θn=0 for all even n. Thus, v(t)=160 π+ 20 cos(ωot+90 )−320 π 1 X n=2,4,6,··· cos(nωot) (n2−1) V. Ak/−θk=ak−jbk=0−jbk=bk/−90. […]