Solution manual for introduction to mechatronics and measurement systems 5th edition by alciatore

Solution Manual for Introduction to Mechatronics and Measurement Systems 5th Edition By Alciato Full file Solutions at https://TestbankDirect.eu/ Manual INTRODUCTION TO MECHATRONICS AND MEASUREMENT SYSTEMS 5th edition 2018 SOLUTIONS MANUAL David G Alciatore, PhD, PE Department of Mechanical Engineering Colorado State University Fort Collins, CO 80523 Introduction to Mechatronics and Measurement Systems Full file at https://TestbankDirect.eu/ Solution Manual for Introduction to Mechatronics and Measurement Systems 5th Edition By Alciato Full file Solutions at https://TestbankDirect.eu/ Manual This manual contains solutions to the end-of-chapter problems in the fifth edition of "Introduction to Mechatronics and Measurement Systems." Only a few of the open-ended problems that not have a unique answer are left for your creative solutions More information, including an example course outline, a suggested laboratory syllabus, Mathcad/Matlab files for examples in the book, and other supplemental material are provided on the book website at: mechatronics.colostate.edu We have class-tested the textbook for many years, and it should be relatively free from errors However, if you notice any errors or have suggestions or advice concerning the textbook's content or approach, please feel free to contact me via e-mail at David.Alciatore@colostate.edu I will post corrections for reported errors on the book website Thank you for choosing my book I hope it helps you provide your students with an enjoyable and fruitful learning experience in the exciting cross-disciplinary subject of mechatronics Introduction to Mechatronics and Measurement Systems Full file at https://TestbankDirect.eu/ Solution Manual for Introduction to Mechatronics and Measurement Systems 5th Edition By Alciato Full file Solutions at https://TestbankDirect.eu/ Manual 2.1 D = 0.06408 in = 0.001628 m D –6 A = = 2.082  10  = 1.7 x 10-8 m, L = 1000 m L R = - = 8.2 A 2.2 (a) R = 21  10  20% so 168k  R  252k (b) R = 07  10  20% so 5.6k  R  8.4k (c) R s = R + R = 217k  20% so 174k  R s  260k (d) R1 R2 R p = R1 + R2 R 1MIN R 2MIN - = 5.43k R pMIN = R 1MIN + R MIN R 1MAX R 2MAX - = 8.14k R pMAX = R MAX + R MAX 2.3 R = 10  10 , R = 25  10 R1 R2  10  10   25  10  - = 20  10 R = = R1 + R2 10  10 + 25  10 a = = red, b = = black, c = = brown, d = gold 2.4 In series, the trim pot will add an adjustable value ranging from to its maximum value to the original resistor value depending on the trim setting When in parallel, the trim pot could be 0 perhaps causing a short Furthermore, the trim value will not be additive with the fixed resistor 2.5 When the last connection is made, a spark occurs at the point of connection as the completed circuit is formed This spark could ignite gases produced in the battery The negative terminal of the battery is connected to the frame of the car, which serves as a ground reference throughout the vehicle Introduction to Mechatronics and Measurement Systems Full file at https://TestbankDirect.eu/ Solution Manual for Introduction to Mechatronics and Measurement Systems 5th Edition By Alciato Full file Solutions at https://TestbankDirect.eu/ Manual 2.6 No, as long as you are consistent in your application, you will obtain correct answers If you assume the wrong current direction, the result will be negative 2.7 Place two 100 resistors in parallel and you immediately have a 50 resistance 2.8 Put two 50 resistors in series: 5050 2.9 Put a 100 resistor in series with the parallel combination of two 100 resistors: 100100100100100 2.10 From KCL, I s = I + I + I Vs Vs Vs Vs so from Ohm’s Law = + + -R eq R1 R2 R3 R1 R2 R3 1 1 Therefore, = + + so R eq = -R eq R1 R2 R3 R2 R3 + R1 R3 + R1 R2 2.11 Is Is From Ohm’s Law and Question 2.10, V = = -R eq R2 R3 + R1 R3 + R1 R2 -R1 R2 R3 and for one resistor, V = I R R2 R3 Therefore, I =   I s  R R + R R + R R 2 2.12 R1 R2 R1 R2 lim  - = - = R R1 R   R + R 2 2.13 dV dV dV I = C eq - = C = C -dt dt dt From KVL, V = V1 + V2 so dV dV dV - = -1- + -2dt dt dt Introduction to Mechatronics and Measurement Systems Full file at https://TestbankDirect.eu/ Solution Manual for Introduction to Mechatronics and Measurement Systems 5th Edition By Alciato Full file Solutions at https://TestbankDirect.eu/ Manual Therefore, C1 C2 II I 1 = + so = + or C eq = C1 + C2 C eq C1 C2 C eq C1 C2 2.14 V = V1 = V2 dV dV dV dV I = C = C - and I = C = C dt dt dt dt From KCL, dV dV dV I = I + I = C - + C - = -  C + C  dt dt dt dV Since I = C eq dt C eq = C + C 2.15 I = I1 = I From KVL, dI dI dI V = V + V = L - + L - = -  L + L  dt dt dt dI Since V = L eq dt L eq = L + L 2.16 dI dI dI V = L - = L - = L dt dt dt From KCL, I = I + I so V V V Therefore, = + -L L1 L2 dI dI dI - = -1 + -2 dt dt dt L1 L2 1 so - = + or L = -L L1 L2 L1 + L2 2.17 V o = 1V , regardless of the resistance value 2.18 40 From Voltage Division, V o =  – 15  = – 8V 10 + 40 Introduction to Mechatronics and Measurement Systems Full file at https://TestbankDirect.eu/ Solution Manual for Introduction to Mechatronics and Measurement Systems 5th Edition By Alciato Full file Solutions at https://TestbankDirect.eu/ Manual 2.19 Combining R2 and R3 in parallel, R2 R3 23 R 23 = - = = 1.2k R2 + R3 2+3 and combining this with R1 in series, R 123 = R + R 23 = 2.2k (a) Using Ohm’s Law, V in 5V I = = = 2.27mA R 123 2.2k (b) Using current division, R2 I = - I = - 2.27mA = 0.909mA R2 + R3 (c) Since R2 and R3 are in parallel, and since Vin divides between R1 and R23, R 23 1.2 V = V 23 = - V in = - 5V = 2.73V R + R 23 2.2 2.20 (a) From Ohm’s Law, V out – V 14.2V – 10V I = - = - = 0.7mA R 24 6k (b) V = V = V 56 = V out – V = 14.2V – 20V = – 5.8V (a) R 45 = R + R = 5k 2.21 R R 45 R 345 = - = 1.875k R + R 45 R 2345 = R + R 345 = 3.875k R R 2345 R eq = = 0.795k R + R 2345 (b) R 345 V A = - V s = 4.84V R + R 345 Introduction to Mechatronics and Measurement Systems Full file at https://TestbankDirect.eu/ Solution Manual for Introduction to Mechatronics and Measurement Systems 5th Edition By Alciato Full file Solutions at https://TestbankDirect.eu/ Manual (c) VA I 345 = = 2.59mA R 345 R3 I = - I 345 = 0.97mA R + R 45 2.22 This circuit is identical to the circuit in Question 2.21 Only the resistance values are different: (a) R 45 = R + R = 4k R R 45 R 345 = - = 2.222k R + R 45 R 2345 = R + R 345 = 6.222k R R 2345 R eq = = 1.514k R + R 2345 (b) R 345 V A = - V s = 3.57V R + R 345 (c) VA I 345 = = 1.61mA R 345 R3 I = - I 345 = 0.89mA R + R 45 2.23 Using superposition, R2 V R21 = - V = 0.909V R1 + R2 R1 V R22 = - i = 9.09V R1 + R2 V R2 = V R21 + V R22 = 10.0V 2.24 R4 R5 R 45 = - = 0.5k R4 + R5 V1 – V2 I = - = – 0.5mA R1 + R2 Introduction to Mechatronics and Measurement Systems Full file at https://TestbankDirect.eu/ Solution Manual for Introduction to Mechatronics and Measurement Systems 5th Edition By Alciato Full file Solutions at https://TestbankDirect.eu/ Manual R 45 V A = -  V – V  = – 0.238 V R + R 45 2.25 R 45 = R + R = 9k R R 45 R 345 = - = 2.25k R + R 45 R 2345 = R + R 345 = 4.25k R R 2345 R eq = = 0.81k R + R 2345 2.26 Using loop currents, the KVL equations for each loop are: V – I out R = V2 – I5 R5 – I3 R3 – V1 = – I6 R6 + I5 R5 = I R – I 24 R – I 24 R = and using selected KCL node equations, the unknown currents are related according to: I out = I + I + I V1 I V1 = I out –  I + I  I = I + I – I 24 This is now equations in unknowns, which can be solved for Iout and I6 The output voltage is then given by: V out = V – I R 2.27 Applying Ohm’s Law to resistor combination R24 gives: V out – V 4.2V I = - = = 0.7mA R 24 6k The voltage across R5 is: V = V = V 56 = V + – V - = V out – V = – 5.8V 2.28 It will depend on your instrumentation, but the oscilloscope typically has an input impedance of M Introduction to Mechatronics and Measurement Systems Full file at https://TestbankDirect.eu/ Solution Manual for Introduction to Mechatronics and Measurement Systems 5th Edition By Alciato Full file Solutions at https://TestbankDirect.eu/ Manual 2.29 2.30 Since the input impedance of the oscilloscope is M, the impedance of the source will be in parallel, and the oscilloscope impedance will affect the measured voltage Draw a sketch of the equivalent circuit to convince yourself R2 R3 R 23 = R2 + R3 R 23 V out = - V in R + R 23 (a) R 23 = 9.90k , V out = 0.995V in (b) R 23 = 333k , V out = 1.00V in When the impedance of the load is lower (10k vs 500k), the accuracy is not as good 2.31 R2 V out = - V in R1 + R2 (a) 10 V out = - V in = 0.995V in 10.05 (b) 500 V out = V in = 0.9999V in 500.05 For a larger load impedance, the output impedance of the source less error 2.32 The theoretical value of the voltage is: R V theor = V s = - V s R+R The equivalent resistance of the parallel combination of the resistor and the voltmeter input impedance is: R  5R5 -= - R R + 5R And the measured voltage across this resistance is: V meas R = - V s = V s 11 R + - R Therefore, the percent error in the measurement is: V meas – V theor - = – 9% V theor Introduction to Mechatronics and Measurement Systems Full file at https://TestbankDirect.eu/ Solution Manual for Introduction to Mechatronics and Measurement Systems 5th Edition By Alciato Full file Solutions at https://TestbankDirect.eu/ Manual 2.33 It will depend on the supply; check the specifications before answering 2.34 With the voltage source shorted, all three resistors are in parallel, so, from Question 2.10: R1 R2 R3 R TH = -R2 R3 + R1 R3 + R1 R2 2.35 V in =  45 Combining R2 and L in series and the result in parallel with C gives:  R + Z L Z C Z R2 LC = - = 1860.52  – 60.25 = 923.22 – 1615.30j  R2 + ZL  + ZC Using voltage division, Z R2 LC -V V C = -R + Z R2 LC in where R + Z R2 LC = 1000 + 923.22 – 1615.30j = 2511.57  – 40.02 so 1860.52  – 60.25 V C =  45 = 3.70  24.8 = 3.70  0.433rad 2511.57  – 40.02 Therefore, V C  t  = 3.70 cos  3000t + 0.433 V 2.36 With steady state dc Vs, C is open circuit So V C = V s = 10V so V R1 = 0V and V R2 = V s = 10V 2.37 (a) In steady state dc, C is open circuit and L is short circuit So Vs I = - = 0.025mA R1 + R2 (b)  =  6 –j – 10 10 Z C = = - j = – 90  C   5 Z LR2 = Z L + R = jL + R =  10 + 20j  = 10 0.036 10 Introduction to Mechatronics and Measurement Systems Full file at https://TestbankDirect.eu/ Solution Manual for Introduction to Mechatronics and Measurement Systems 5th Edition By Alciato Full file Solutions at https://TestbankDirect.eu/ Manual Z C Z LR2 Z CLR2 = -=  91040 – 28550j  = 95410 – 17.4  Z C + Z LR2 Z eq = R + Z CLR2 =  191040 – 28550j  = 193200 – 8.50  Vs I s = = 0.0259 8.50 mA Z eq ZC I = I s =  0.954 – 17.44 I s = 0.0247 – 8.94 mA Z C + Z LR2 So I  t  = 24.7 cos  t – 0.156  A 2.38 (a) rad   =  , f = = 0.5Hz sec 2 A pp = 2A = 4.0 , dc offset = (b) rad   = 2 , f = = 1Hz sec 2 A pp = 2A = , dc offset = 10.0 (c) rad   = 2 , f = = 1Hz sec 2 A pp = 2A = 6.0 , dc offset = (d) rad   = , f = = 0Hz sec 2 A pp = 2A = , dc offset = sin    + cos    = – 2.39 V rms P = - = 100W R 2.40 V pp V rms =  -    = 35.36V   Introduction to Mechatronics and Measurement Systems Full file at https://TestbankDirect.eu/ 11 Solution Manual for Introduction to Mechatronics and Measurement Systems 5th Edition By Alciato Full file Solutions at https://TestbankDirect.eu/ Manual V rms P = - = 12.5W R 2.41 Vm = 2V rms = 169.7V 2.42 For V rms = 120V , V m = 2V rms = 169.7V , and f = 60 Hz, V  t  = V m sin  2f +   = 169.7 sin  120t +   2.43 From Ohm’s Law, 5V – 2V 3V I = - = R R Since 10mA  I  100mA , 3V 10mA  -  100mA R giving 3V 3V  R  - or 30  R  300 100mA 10mA V For a resistor, P = , so the smallest allowable resistance would need a power rating of R at least:  3V  P = - = 0.3W 30 so a 1/2 W resistor should be specified The largest allowable resistance would need a power rating of at least:  3V  P = - = 0.03W 300 so a 1/4 W resistor would provide more than enough capacity 2.44 Using KVL and KCL gives: V = I R1 R V =  I – I R1 R +  I – I R1 – I R V – V =  I – I R1 – I R – I R 12 Introduction to Mechatronics and Measurement Systems Full file at https://TestbankDirect.eu/ Solution Manual for Introduction to Mechatronics and Measurement Systems 5th Edition By Alciato Full file Solutions at https://TestbankDirect.eu/ Manual The first loop equation gives: V1 I R1 = = 10mA R1 Using this in the other two loop equations gives: 10 =  I – 10m 2k +  I – 10m – I 3k 10 – =  I – 10m – I 3k – I 4k or  5k I –  3k I = 60  3k I –  7k I = 35 Solving these equations gives: I = 12.12mA and I = 0.1923mA 2.45 (a) V out = I R – V = – 4.23V (b) P = I V = 121mW , P = I V = 0.962mW , P = – I V = – 1.92mW Using KVL and KCL gives: V = I R1 R V =  I – I R1 R +  I – I R1 – I R V – V =  I – I R1 – I R – I R The first loop equation gives: V1 I R1 = = 10mA R1 Using this in the other two loop equations gives: 10 =  I – 10m 2k +  I – 10m – I 2k 10 – =  I – 10m – I 2k – I 1k or  4k I –  2k I = 50  2k I –  3k I = 25 Introduction to Mechatronics and Measurement Systems Full file at https://TestbankDirect.eu/ 13 Solution Manual for Introduction to Mechatronics and Measurement Systems 5th Edition By Alciato Full file Solutions at https://TestbankDirect.eu/ Manual Solving these equations gives: I = 12.5mA and I = 0mA 2.46 (a) V out = I R – V = – 5V (b) P = I V = 125mW , P = I V = 0mW , P = – I V = 0mW P avg Vm Im = -  V  t I  t  dt =  sin  t +  V  sin  t +  I  dt T T T T 0 Using the product formula trigonometric identity, T P avg Vm Im =   cos   V –  I  – cos  2t +  V +  I   dt 2T Therefore, Vm Im Vm Im P avg = cos   V –  I  = cos    2 T 2.47 I rms = 1-  I 2m sin  t +  I  dt T Using the double angle trigonometric identity, 2T I rms = Im  -  - – cos   t +  I   dt  T 2 Therefore, I rms = 2.48 I I m  T - - = mT  2 R2 R3 R 23 = - = 5k R2 + R3 R 23 V o = - V i = - sin  2t  R + R 23 This is a sin wave with half the amplitude of the input with a period of 1s 14 Introduction to Mechatronics and Measurement Systems Full file at https://TestbankDirect.eu/ Solution Manual for Introduction to Mechatronics and Measurement Systems 5th Edition By Alciato Full file Solutions at https://TestbankDirect.eu/ Manual 2.49 No A transformer requires a time varying flux to induce a voltage in the secondary coil 2.50 V Np = p = 120V - = Ns Vs 24V 2.51 RL = Ri = 8 for maximum power 2.52 The BNC cable is far more effective in shielding the input signals from electromagnetic interference since no loops are formed Introduction to Mechatronics and Measurement Systems Full file at https://TestbankDirect.eu/ 15 ... dt Introduction to Mechatronics and Measurement Systems Full file at https://TestbankDirect.eu/ Solution Manual for Introduction to Mechatronics and Measurement Systems 5th Edition By Alciato... 40 Introduction to Mechatronics and Measurement Systems Full file at https://TestbankDirect.eu/ Solution Manual for Introduction to Mechatronics and Measurement Systems 5th Edition By Alciato... 345 Introduction to Mechatronics and Measurement Systems Full file at https://TestbankDirect.eu/ Solution Manual for Introduction to Mechatronics and Measurement Systems 5th Edition By Alciato