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We are using the random numbers to determine new values of the resistors and voltage

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We are using the random numbers to determine new values of the resistors and voltage. Using these new values we calculate a new Vout. We do this 10 times, and find the average Vout and the std dev of Vout.

 This assignment involves finding the least cost combination of components to use in an electrical circuit to achieve a desired output voltage specification. The circuit for this assignment will be given in class. The output voltage, Vout, has a quality specification that states it needs to be within % % of the time. Each component in the circuit, including the voltage source and all resistors have variability. Each component can be purchased with different levels of precision (variability) with a different cost associated with a different level of precision. The cost structure is given by: 10% resistors cost $100 each 5% resistors cost $150 each 1% resistors cost $250 each 0.1% resistors cost $1000 each For the voltage source the cost is determined by the tolerance percentage of the voltage source according to the following function: Cost of voltage source = $10/Tolerance^2. As an example assume a tolerance of 5%. The cost of this voltage source would then be 10/.05^2 = $4000. You have been asked to find the lowest cost combination of components that will meet the quality specification. You have been asked to do this for two variability assumptions. 1. Assume all components are normally distributed with a mean equal to the value given, and a std dev dependent on the precision level specified. 2. Assume all components are uniformly distributed with a mean equal to the value given and a range dependent on the precision level specified. Present your results in a report. The report must include an executive summary. The report must be well written with all tables and graphs discussed in supporting narrative. Your report should discuss the differences you observe between the normal and uniform assumptions. "Rego frau IRS) + R2 + Rs} 11 R2] + R + Rz Efter att Rot Rc)X2} +R+R Substituting the values +15+49) 1203 (311|20) +10 226 since no unit given} Roqu 150 + 15+ 4 575 25 1139. Ka 51 - TON 602 CHAPTER - ASSIGNING RANDOM NUMBERS. Random NUMBERS COLLECTED From "TABLE OF RANDOM NUMBERS" MATHB ITS NOTEBOOK.com | ALGEBRA 2 | STATISTICS) STRANDOM TABLE - HTML RI RE Ry R2 R3 R4 R5 136518 83775 215 36777 0 5542 29705 89116 2 81639 27973 62413 85652 62817 46132 $3881 3 81390 75635 19428 88048 08727 20092 12615 4 35046 67753 3 69 630 10883 13683 31841 77367 5 40791 97402 27569 90184 02338 39318 54936 34641 95525 86316 , 87384 84180 93793 64953 ? 51472 65358 23701 75230 47200 78176 % 85248 8|90589 74567 22633 78435 37586 07015 987729 9 16703 16224 97661 79907 06611 26501 43389 92725 68158 41859 94198 37182 61354 88857 USING THIS TABLE. NEW RESISTOR VALUES ARE CALCULATED BY: RiNEW = (LOWEST IN- TOLEZANCE VALUE) + (RANGE OF VALUESO. (RANDOM # +42L RESISTORS HAVE A 10% TOLERANCE, R = 5r BUT DUE TO THE 10% TOLERANCE, COULD BE ANY VA BETWEEN 4.5 AND 5.5. ((5)(0.1)) = ²0.52) IN TRIAL #1, R. 4.5+ (180.36518) = 41.5+ 0.36518 R = 4.865182 R: = 2 (22)(0.1) - +20, SO VALUES BETWEEN 18 AND 22. I 3.37% 95%6 Search for table of unfum monte cano Simulation IEE 3100 #s 5 0 20 10 R4 2006 R3 15 S7 10% R5 15 4.5+(1)(.147936) 4.64 M R7 5 4.5 thru RI curent going SS unifom dismbution 5 5.5486 & R IS 20 R2 320 R4 lo RS IS RF Vout = 743 1/10 + 115=6 M lo 5 20-10=10 /R VEL + 132 331 1120+ 1/3) = 12 162 V=IR 20 = 22 16I Voltage drop = 10.97 (R2) Guer I agap 903 Amps which rasister do we 1 houerance on 2 lowest possible