fundamentals of electric circuits 3rd edition solutions manual chapter 5

... cross-section of the conductor in 20 seconds 25 C0 t q=∫idt=∫ tdt= = Trang 46t2 25A, -2t0 A, Trang 5idt Trang 615 152 1510110idt 0 =++ = ×+ A lightning bolt with 8 kA strikes an object for 15 μ s How ... lamp, (b) the cost of operating the light for one non-leap year if electricity costs 12 cents per kWh Chapter 1, Solution 28 A0.25 2436530pt W b) Chapter 1, Problem 29 An electric stove with ... W,30W, 45W, 60W, Trang 15Chapter 1, Problem 19 Find I in the network of Fig 1.30 I 1A + + Trang 17hr 60 4 kW 1.2 Trang 18Chapter 1, Problem 26 A flashlight battery has a rating of 0.8 ampere-hours

Ngày tải lên: 13/09/2018, 13:31

... >> Y=[1.125,0,0,-0.125;0,0.75,-0.25,0;0,-0.25,0.75,0;-0.125,0,0,1.125] Y = 1.1250 0 0 -0.1250 0 0.7500 -0.2500 0 0 -0.2500 0.7500 0 -0.1250 0 0 1.1250 >> I=[4,-4,-2,2]' ... Y=[0.35,-3.25,3;-0.25,0.95,-0.5;0,-0.5,0.5] Y = 0.3500 -3.2500 3.0000 -0.2500 0.9500 -0.5000 0 -0.5000 0.5000 >> I=[-2,0,6]' I = -2 0 6 >> V=inv(Y)*I V = -164.2105 -77.8947 ... 84 4 V 25 0 V 75 0 0 4 V V 2 0 V 2 V 75 0 V 25 0 0 2 2 0 V 4 V V 32 32 2 V 125 1 V 125 0 0 1 0 V 8 V V 4 V 125 1 0 0 125 0 0 75 0 25 0 0 0 25 0 75 0 0 125 0 0 0 125 1 Now we can

Ngày tải lên: 13/09/2018, 13:31

... 4||10 = 5 + 20/7 = 55/7 i2 = [5/(5 + 55/7)]2 = 7/9, io3 = [-10/(10 + 4)]i = -5/9 2 io = (12/9) – (6/9) – (5/9) = 1/9 = 111.11 mA Trang 25Use superposition to obtain v in the circuit of Fig 4.85 Check ... 13/2)]2 = 3/8 = 0.375 i = 2.5 + 0.375 - 1 = 1.875 A p = i2R = (1.875)23 = 10.55 watts Trang 23Given the circuit in Fig 4.84, use superposition to get i oFigure 4.84 Trang 24Chapter 4, Solution ... the voltage source to obtain the circuit in Fig (b) 10||10 = 5 ohms, i = [5/(5 + 4)](2 – 1) = 5/9 = 555.5 mA Trang 36Chapter 4, Problem 23 Referring to Fig 4.91, use source transformation to

Ngày tải lên: 13/09/2018, 13:31

... circuit of Fig 5.51 Assume that the op Trang 14v k 10 0.27mA + 0.018mA = 288 μA Trang 15Determine the output voltage vo in the circuit of Fig 5.53 Figure 5.53 for Prob 5.14 Chapter 5, Solution ... determine the value of v in order to make 23 3 2 2 1 1 5 2 9 ) 1 ( 50 50 20 50 ) 2 ( 10 50 v v v R R v R R v R − = Thus, V 3 5 2 9 5 Trang 42+ _ Figure 5.77 For Prob 5.40 Chapter 5, Solution 40 ... For Prob 5.59 Trang 67Determine v in the circuit of Fig 5.88 oTrang 68Chapter 5, Problem 62 Obtain the closed-loop voltage gain v /v of the circuit in Fig 5.89 o i Figure 5.89 Chapter 5, Solution

Ngày tải lên: 13/09/2018, 13:31

... L + L = 2L L L L L Lx L L L L 5 0 2 5 0 2 5 0 // + + = + (b) L//L = 0.5L, L//L + L//L = L Leq = L//L = 500 mL Trang 50= +5 L L 3 5 Lx L 3 2 L L 8 5 Trang 51Determine the Leq that can be ... (6) 2.5( 6) 12.5 2.5 2.5 4 10 t x v t dt v t t x − = ∫ + = − + = − ⎪ ⎧ < < < < < < < < − − + s 8 t 6 s 6 t 4 s 4 t 2 s 2 t 0 , V 5 2 t 5 2 , V 5 12 , V t 5 2 5 22 ... 6, Solution 54 ( 10 0 6 12 ) ) 3 9 ( 4 = 4 + 12 ( 0 + 4 ) = 4 + 3 Leq = 7H Trang 49Find Leq in each of the circuits of Fig 6.77 Figure 6.77 Chapter 6, Solution 55 (a) L//L = 0.5L, L + L

Ngày tải lên: 13/09/2018, 13:31

... the prior Chapter 7, Problem 54 Obtain the inductor current for both t < 0 and t > 0 in each of the circuits in Fig 7.120 Figure 7.120 For Prob 7.54 Trang 52Chapter 7, Solution 54. (a) ... V 15 ) 30 ( 1 1 2 2 ) ( + + = ∞ Ω = + = ( 1 1 ) || 2 1 k Rth 4 1 10 4 1 10 C = τ ( 1 e ) , t 0 15 ) t ( v i 1 5 7 ) t ( ( 1 e ) mA 5 7 ) t ( Thus, mA e 5 7 5 7 30 ) t ( = ) t ( Trang 48Chapter ... 1 e ) 0 1054 6 1 ) 1 ( i = − - 1 = 2 1 6 1 3 1 ) ( i ∞ = + = 1) - -(te ) 5 0 1054 0 ( 5 0 ) t ( 1) - -(te 3946 0 5 0 ) t ( Thus, = ) t ( 1 t 0 A e 1 6 1 -1) (t - t Trang 605 0 RLth =

Ngày tải lên: 13/09/2018, 13:31

... reserved No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior t5505.1t 45.6 45 6 Be Ae ) 0 ( di but e 5505 1 Ae 45 6 dt Trang 32PROPRIETARY ... [Acos0.5t + Bsin0.5t], i(0) = 12 = A v = -Ldi/dt, and -v/L = di/dt = 0.5[-12sin0.5t + Bcos0.5t], which leads to -v(0)/L = 0 = B Hence, i(t) = 12cos0.5t A and v = 0.5 However, v = -Ldi/dt = -4(0.5)[-12sin0.5t] ... Figure (c) i(f) = -5(4)/(3 + 5) = -2.5 A v(f) = 5(4 – 2.5) = 7.5 V  Trang 9PROPRIETARY MATERIAL © 2007 The McGraw-Hill Companies, Inc All rights reserved No part of this Manual may be displayed,

Ngày tải lên: 13/09/2018, 13:31

... = − = + = ) 5 2 j 5 2 5 2 ( j 5 4 30 − + = + ) j 5 )( 30 8 Trang 38Find Io in the circuit of Fig 9.60 4 x 6 j Z , 1 j 1 45 6569 5 90 8 4 j 4 4 x − = 5 1 j 5 1 4 j 4 ) 5 1 j 5 9 ( ) 10 ... is 4 j 3 2 j 5 j 1 + + I Therefore, = ) t ( is 25 cos(2t – 53.13 °) A Trang 37If Vo = 8 ∠ 30oV in the circuit of Fig 9.59, find I .5 j 5 j 5 25 j 5 j || + = + s 2 1 1 4 5 2 j 5 12 10 I I ... µ 50 ) 50 )( 0 1 = V ) 50 )( 2 ( ) 100 j 20 j 50 )( 0 1 ( V 80 j 150 100 80 j 50 in in Trang 45For the circuit in Fig 9.70, find the value of ZT⋅.450 j 200 z , 333 13 j 30 15 j 450 j 200 z 45

Ngày tải lên: 13/09/2018, 13:31

... Trang 15= 39 46 1936 1 22 15 86 4 ) 61 61 4502 1 ( 4 2758 1 j 69 4 ) 2758 1 j 6896 0 ( 4 2 j 5 2 xj 5 3 j 4 2 j 5 2 xj 5 3 j 4 VTh 5 Ω – 40 ) 17 67 4123 0 ( V ) 55 46 5235 0 ... 2 Trang 3515 t 5 t 2 20 )t ( i 15 2 15 5 2 2 20 1 I 5 2 2 5 1 I 3 15 5 3 2 2 3 t 3 t t 10 t 100 5 1 I 332 33 ] 33 83 33 83 [ 5 Trang 36Compute the rms value of the waveform depicted in Fig ... waveform of Fig 11.59 as well as the average power absorbed by a 2- Ω resistor when the voltage is applied across the resistor 0 2 2 5 1 V 533 8 ) 8 ( 15 16 3 t 16 5 1 0 3 2 V P 2 Trang 3515 t 5 t

Ngày tải lên: 13/09/2018, 13:31

... its equivalent T-section 5, 25520, 1055) =+ ++ =+ + 74 )4(627)6//( )4 ( j j j j j j j Th j j Trang 25H,5525Trang 26 Determine currents I1, I2, and I3 in the circuit of Fig 13.89 Find the energy ... mesh 1, j12 = (4 + j10 – j5)I 1 + j5I 2 + j5I 2 = (4 + j5)I 1 + j10I 2 (1) For mesh 2, 0 = 20 + (8 + j10 – j5)I 2 + j5I 1 + j5I 1 ++ 10j5j420 12j w = 0.5L 1 i 1 + 0.5L 2 i 2 – Mi 1 i 2 Since ... 40/7.767∠11.89° = 5.15∠–11.89° S = 0.5vsI1* = (20∠0°)(5.15∠11.89°) = 103∠11.89° VA (b) I2 = –I1/n, n = 2.5 I3 = –I2/n’, n = 3 I3 = I1/(nn’) = 5.15∠–11.89°/(2.5x3) = 0.6867∠–11.89° p = 0.5|I2|2(18) =

Ngày tải lên: 13/09/2018, 13:31

... without the prior written consent of McGraw-Hill Education Full file at https://TestbankDirect.eu/ Solution Manual for Fundamentals of Electric Circuits 6th Edition by Alexander Full file at ... without the prior written consent of McGraw-Hill Education Full file at https://TestbankDirect.eu/ Solution Manual for Fundamentals of Electric Circuits 6th Edition by Alexander Full file at ... without the prior written consent of McGraw-Hill Education Full file at https://TestbankDirect.eu/ Solution Manual for Fundamentals of Electric Circuits 6th Edition by Alexander Full file at

Ngày tải lên: 20/08/2020, 12:02

... fundamentals of advanced accounting 6th edition chapter solutions fundamentals of advanced accounting 6th edition chapter solutions fundamentals of advanced accounting 6e solutions manual fundamentals of ... 6th edition pdf fundamentals of advanced accounting 6th edition chapter solutions fundamentals of advanced accounting 6th edition chapter solutions fundamentals of advanced accounting 6th edition ... link: fundamentals of advanced accounting 6th edition solutions pdf fundamentals of advanced accounting 6th edition test bank fundamentals of advanced accounting 6th edition chapter solutions fundamentals

Ngày tải lên: 01/03/2019, 08:49

... edition chapter solutions fundamentals of advanced accounting 6th edition chapter solutions fundamentals of advanced accounting 6e solutions manual fundamentals of advanced accounting 6th edition ... edition pdf fundamentals of advanced accounting 6th edition chapter solutions fundamentals of advanced accounting 6th edition chapter solutions fundamentals of advanced accounting 6th edition solutions ... 6th edition solutions pdf fundamentals of advanced accounting 6th edition test bank fundamentals of advanced accounting 6th edition chapter solutions fundamentals of advanced accounting 6th edition

Ngày tải lên: 27/08/2020, 09:10

... Dividends = $250,250 – 95,000 = $155,250 EPS = Net income / Shares = $250,250 / 90,000 = $2.78 per share DPS = Dividends / Shares = $95,000 / 90,000 = $1.06 per share Taxes = 15($50,000) + 25($25,000) ... value of shareholders’ equity can be negative CHAPTER - 11 18 a Taxes Growth = 15($50,000) + 25($25,000) + 34($14,500) = $18,680 Taxes Income = 15($50,000) + 25($25,000) + 34($25,000) + 39($235,000) ... $68,000 = 15($50,000) + 25($25,000) + 34($25,000) + X($100,000); = $68,000 – 22,250 = $45,750 / $100,000 = 45.75% 25 Cash Accounts receivable Inventory Current assets Balance sheet as of Dec 31,

Ngày tải lên: 27/08/2020, 09:10

... Pads of IHT 1 2 3 4 5 6 7 8 9 10 Freestream velocity, V (m/s) 25 35 45 55 65 75 The effect is significant, with a surface temperature of Ts≈ 70°C corresponding to V = 1 m/s Forvelocities of 1 ... velocity, V(m/s) 650 700 750 800 850 900 950 For T = 275°C = 548 K, the controller would compensate for velocity reductions from 10 to 5 m/s byreducing the power from approximately 935 to 690 W/m ... emissivity of Nichrome wire Electricalcurrent Temperature of air flow and surroundings Velocity of air flow. (a) Surface and centerline temperatures of the wire, (b) Effect of flow velocity and electriccurrent

Ngày tải lên: 08/08/2014, 17:20

... Cost Cost of of Goods Goods Sold Sold When Job is Shipped • To the customer, the cost of the finished job becomes a cost of goods sold • The cost of a completed job is debited to Cost of Goods ... classroom use CHAPTER OBJECTIVES Differentiate the cost accounting systems of service and manufacturing firms and of unique and standardized products Discuss the interrelationship of cost accumulation, ... Accounting for for Cost Cost of of Goods Goods Sold Sold Overhead Variances •It is usually immaterial and is therefore closed to the cost of goods sold account •Cost of goods sold before adjustment

Ngày tải lên: 18/12/2017, 15:27

... $68K = 0. 15( $50 K) + 0. 25( $25K) + 0.34($25K) + X($100K); X($100K) = $68K – 22.25K = $ 45. 75K X = $ 45. 75K / $100K X = 45. 75% 25. Balance sheet as of Dec. 31, 2006 Cash $2 ,52 8 Accounts ... .55 (TA – E) / TA = . 45 (TA / TA) – (E / TA) = .55 (TA / TA) – (E / TA) = . 45 1 – (E / TA) = .55 1 – (E / TA) = . 45 E / TA = . 45 E / TA = .55 E = . 45( TA) E = .55 (TA) Rearranging ROA, ... Sales Growth 35% Sales Growth Sales $1,014,000 $1,098 ,50 0 $1,140, 750 Costs 788,400 854 ,100 886, 950 Other expenses 21,000 22, 750 23,6 25 EBIT $ 204,600 $ 221, 650 $ 230,1 75 Interest 12 ,50 0 12 ,50 0 12 ,50 0 ...

Ngày tải lên: 01/07/2014, 14:22

... 52 8 13.2 Mutual Inductance 52 8 13.3 Energy in a Coupled Circuit 53 5 13.4 Linear Transformers 53 9 13 .5 Ideal Transformers 54 5 13.6 Ideal Autotransformers 55 2 † 13.7 Three-Phase Transformers 55 6 13.8 PSpice ... use Kirchoff’s voltage law to check the results. i 1 = v 1 − 5 2 = 2 − 5 2 =− 3 2 =−1.5A i 2 = i 8 = 0. 25 A i 3 = v 1 + 3 4 = 2 + 3 4 = 5 4 = 1. 25 A i 1 + i 2 + i 3 =−1 .5 + 0. 25 + 1. 25 = 0 (Checks.) Applying ... Measurement 4.11 Summary 153 Review Questions 153 Problems 154 Comprehensive Problems 162 5. 1 Introduction 166 5. 2 Operational Amplifiers 166 5. 3 Ideal Op Amp 170 5. 4 Inverting Amplifier 171 5. 5 Noninverting...

Ngày tải lên: 08/05/2014, 15:37

... differences of 50 to 100°C. Coefficients (W/m 2 ⋅K) T s (°C) ε h r h r,a h 35 0. 05 0.32 0.32 2.1 0.9 5. 7 5. 7 1 35 0. 05 0 .51 0 .50 4.7 0.9 9.2 9.0 100 300 50 0 700 900 Surroundings ... () 2 A DL 0.1m 25m 7.85m . == ì= Hence, () () 22 824444 q 7.85m 10 W/m K 150 25 K 0.8 5. 67 10 W/m K 423 298 K =+ìì ()() 22 q 7.85m 1, 250 1,0 95 w/m 9813 859 2 W 18,4 05 W=+=+= < ... air, P max = 200 W/m 2 ⋅ K(0.0 05 m) 2 ( 85 - 15) ° C = 0. 35 W. < In the dielectric liquid P max = 3000 W/m 2 ⋅ K(0.0 05 m) 2 ( 85- 15) ° C = 5. 25 W. < COMMENTS: Relative...

Ngày tải lên: 08/08/2014, 17:20

... PROBLEM 3 .50 (Cont.) 0.0 35 0.0 45 0. 055 0.0 65 0.0 75 Outer radius of insulation, r3(m) 0 400 800 1200 1600 2000 Heat loss, qprime(W/m) q1 0.0 35 0.0 45 0. 055 0.0 65 0.0 75 Outer radius of insulation, ... D h 52 3K T T 293K ln 75/ 60 0.6m 50 0 W/m K 0.075m 25 W/m K 2 56 .5 W/m K +0.8 0.075m 5. 67 10 W/m K επ σ πππ ππ π π ∞∞ −− − −− =+ + = ìì + ì ì ì ììì 44 4 4 s,o s,o s,o 84 4 s,o T 293 K 52 3 ... solution, T s,o ≈ 50 2K. Hence the heat loss is () ( ) 44 oo s,o ,o o s,o sur q = D h T T D T T πεπσ ∞ ′ −+ − () ( )() 28444 24 W q = 0.075m 25 W/m K 50 2-293 0.8 0.075m 5. 67 10 50 2 243 K mK ππ − ′ ⋅+...

Ngày tải lên: 08/08/2014, 17:20