TEST BANK FOR Introduction to Mechatronics and Measurement Systems 4th Edition By David G. Alciatore, Michael B. Histand

Exam (elaborations) TEST BANK FOR Introduction to Mechatronics and Measurement Systems 4th Edition By David G. Alciatore, Michael B. Histand INTRODUCTION TO MECHATRONICS AND MEASUREMENT SYSTEMS 4th edition 2012(c) SOLUTIONS MANUAL David G. Alciatore and Michael B. Histand Department of Mechanical Engineering Colorado State University Fort Collins, CO 80523 Solutions Manual Introduction to Mechatronics and Measurement Systems 3 2.1 D = 0.06408 in = 0. m.  = 1.7 x 10-8 m, L = 1000 m 2.2 (a) so (b) so (c) so (d) 2.3 , a = 2 = red, b = 0 = black, c = 1 = brown, d = gold 2.4 In series, the trim pot will add an adjustable value ranging from 0 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. A D2 4 = ---------- = 2.082  10–6 R L A = ------ = 8.2 R1 = 21  104  20% 168k  R1  252k R2 = 07  103  20% 5.6k  R2  8.4k Rs = R1 + R2 = 217k  20% 174k  Rs  260k Rp R1R2 R1 + R2 = ------------------- RpMIN R1MINR2MIN R1MIN R2MIN + = ------------------------------- = 5.43k RpMAX R1MAXR2MAX R1MAX R2MAX + = ---------------------------------- = 8.14k R1 = 10  102 R2 = 25  101 R R1R2 R1 + R2 ------------------- 10  102 25 10  1 10  102 + 25  101 = = --------------------------------------------------- = 20  101 Solutions Manual 4 Introduction to Mechatronics and Measurement Systems 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 From KCL, so from Ohm’s Law Therefore, so 2.9 From Ohm’s Law and Question 2.8, and for one resistor, Therefore, 2.10 2.11 From KVL, so Therefore, so or Is = I1 + I2 + I3 Vs Req -------- Vs R1 ------ Vs R2 ------ Vs R3 = + + ------ 1 Req -------- 1 R1 ------ 1 R2 ------ 1 R3 = + + ------ Req R1R2R3 R2R3 + R1R3 + R1R2 = ---------------------------------------------------- V Is Req -------- Is R2R3 + R1R3 + R1R2 R1R2R3 ---------------------------------------------------- = = ---------------------------------------------------- V = I1R1 I1 R2R3 R2R3 + R1R3 + R1R2 ---------------------------------------------------- =   Is R1R2 R1 + R2 -------------------   R1  lim R1R2 R1 = ------------ = R2 I Ceq dV dt ------- C1 dV1 dt --------- C2 dV2 dt = = = --------- V = V1 + V2 dV dt ------- dV1 dt --------- dV2 dt = + --------- I Ceq -------- I C1 ------ I C2 = + ------ 1 Ceq -------- 1 C1 ------ 1 C2 = + ------ Ceq C1C2 C1 + C2 = ------------------- Solutions Manual Introduction to Mechatronics and Measurement Systems 5 2.12 and From KCL, Since 2.13 From KVL, Since 2.14 From KCL, so Therefore, so or 2.15 , regardless of the resistance value. 2.16 From Voltage Division, V = V1 = V2 I1 C1 dV1 dt --------- C1 dV dt = = ------- I2 C2 dV2 dt --------- C2 dV dt = = ------- I I1 + I2 C1 dV dt ------- C2 dV dt + ------- dV dt = = = -------C1 + C2 I Ceq dV dt = ------- Ceq = C1 + C2 I = I1 = I2 V V1 + V2 L1 dI dt ----- L2 dI dt + ----- dI dt = = = -----L1 + L2 V Leq dI dt = ----- Leq = L1 + L2 V LdI dt ----- L1 dI1 dt ------- L2 dI2 dt = = = ------- I = I1 + I2 dI dt ----- dI1 dt ------- dI2 dt = + ------- VL ---- V L1 ----- V L2 ----- + = 1L--- 1 L1 ----- 1 L2 = + ----- L L1L2 L1 + L2 = ------------------ Vo = 1V Vo 40 10 + 40 = ------------------5 – 15 = –8V Solutions Manual 6 Introduction to Mechatronics and Measurement Systems 2.17 Combining R2 and R3 in parallel, and combining this with R1 in series, (a) Using Ohm’s Law, (b) Using current division, (c) Since R2 and R3 are in parallel, and since Vin divides between R1 and R23, 2.18 (a) From Ohm’s Law, (b) 2.19 (a) R23 R2R3 R2 + R3 ------------------- 23 2 + 3 = = ------------ = 1.2k R123 = R1 + R23 = 2.2k I1 Vin R123 ---------- 5V 2.2k = = ---------- = 2.27mA I3 R2 R2 + R3 -------------------I1 25 = = --2.27mA = 0.909mA V3 V23 R23 R1 + R23 ---------------------Vin 1.2 2.2 = = = -------5V = 2.73V I4 Vout – V1 R24 ----------------------- 14.2V – 10V 6k = = ------------------------------- = 0.7mA V5 = V6 = V56 = Vout – V2 = 14.2V – 20V = –5.8V R45 = R4 + R5 = 5k R345 R3R45 R3 + R45 = --------------------- = 1.875k R2345 = R2 + R345 = 3.875k Req R1R2345 R1 + R2345 = ------------------------- = 0.795k Solutions Manual Introduction to Mechatronics and Measurement Systems 7 (b) (c) 2.20 This circuit is identical to the circuit in Question 2.19. Only the resistance values are different: (a) (b) (c) 2.21 Using superposition, VA R345 R2 + R345 = -----------------------Vs = 4.84V I345 VA R345 = ---------- = 2.59mA I5 R3 R3 + R45 = ---------------------I345 = 0.97mA R45 = R4 + R5 = 4k R345 R3R45 R3 + R45 = --------------------- = 2.222k R2345 = R2 + R345 = 6.222k Req R1R2345 R1 + R2345 = ------------------------- = 1.514k VA R345 R2 + R345 = -----------------------Vs = 3.57V I345 VA R345 = ---------- = 1.61mA I5 R3 R3 + R45 = ---------------------I345 = 0.89mA VR21 R2 R1 + R2 = -------------------V1 = 0.909V VR22 R1 R1 + R2 = -------------------i1 = 9.09V VR2 VR21 VR22 = + = 10.0V Solutions Manual 8 Introduction to Mechatronics and Measurement Systems 2.22 2.23 2.24 Using loop currents, the KVL equations for each loop are: and using selected KCL node equations, the unknown currents are related according to: This is now 7 equations in 7 unknowns, which can be solved for Iout and I6. The output voltage is then given by: 2.25 It will depend on your instrumentation, but the oscilloscope typically has an input impedance of 1 M. R45 R4R5 R4 + R5 = ------------------- = 0.5k I V1 – V2 R1 + R2 = ------------------- = –0.5mA VA R45 R3 + R45 = ---------------------V1 – V2 = –0.238V R45 = R4 + R5 = 9k R345 R3R45 R3 + R45 = --------------------- = 2.25k R2345 = R2 + R345 = 4.25k Req R1R2345 R1 + R2345 = ------------------------- = 0.81k V1 – IoutR1 = 0 V2 – I5R5 – I3R3 – V1 = 0 – I6R6 + I5R5 = 0 I3R3 – I24R4 – I24R2 = 0 Iout I2 I3 IV1 = + + IV1 = Iout – I5 + I6 I3 = I5 + I6 – I24 Vout = V2 – I6R6 Solutions Manual Introduction to Mechatronics and Measurement Systems 9 2.26 Since the input impedance of the oscilloscope is 1 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. 2.27 (a) , (b) , When the impedance of the load is lower (10k vs. 500k), the accuracy is not as good. 2.28 (a) (b) For a larger load impedance, the output impedance of the source less error. 2.29 It will depend on the supply; check the specifications before answering. 2.30 With the voltage source shorted, all three resistors are in parallel, so, from Question 2.8: 2.31 Combining R2 and L in series and the result in parallel with C gives: R23 R2R3 R2 + R3 = ------------------- Vout R23 R1 + R23 = ---------------------Vin R23 = 9.90k Vout = 0.995Vin R23 = 333k Vout = 1.00Vin Vout R2 R1 + R2 = -------------------Vin Vout 10 10.05 = -------------Vin = 0.995Vin Vout 500 500.05 = ----------------Vin = 0.9999Vin RTH R1R2R3 R2R3 + R1R3 + R1R2 = ---------------------------------------------------- Vin = 5 45 ZR2LC R2 + ZLZC R2 + ZL + ZC = ------------------------------------- = 1860.52 –60.25 = 923.22 – 1615.30j Solutions Manual 10 Introduction to Mechatronics and Measurement Systems Using voltage division, where so Therefore, 2.32 With steady state dc Vs, C is open circuit. So so and 2.33 (a) In steady state dc, C is open circuit and L is short circuit. So (b) VC ZR2LC R1 + ZR2LC = ---------------------------Vin R1 + ZR2LC = 1000 + 923.22 – 1615.30j = 2511.57 –40.02 VC 1860.52 –60.25 2511.57 –40.02 = --------------------------------------------5 45 = 3.70 24.8 = 3.70 0.433rad