ECE-320 CH02.pdf - Unix.eng.ua.edu

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2Solving the system we obtain:iiiabc16= A364= A9= 12 AVVVVRRRR0123= R i= R1= R0 a= R i= 5.33 V( i − i )2 b3b= 14.22 V( i − i )cab= 5.33 V= 19.55 V( ⊕ up)( ⊕ down)( ⊕ up)( ⊕ up)________________________________________________________________________Solution:Problem 2.37Known quantities:Schematic of voltage divider network shown of Figure P2.37.Find:a) The worst-case output voltages for ± 10 percent toleranceb) The worst-case output voltages for ± 5 percent toleranceAnalysis:a) 10% worst case: low voltageR2 = 4500 Ω, R1 = 5500 Ωv45004500 + 5500OUT , MIN=5 = 2. 2510% worst case: high voltageR2 = 5500 Ω, R1 = 4500 Ωv55004500 + 5500OUT , MAX=5 = 2. 75b) 5% worst case: low voltageR 2 = 4750 Ω, R 1 = 5250 Ωv47504750 + 5250OUT , MIN=5 = 2. 3755% worst case: high voltageR 2 = 5250 Ω, R 1 = 4750 Ωv52505250 + 4750OUT , MAX=5 = 2. 625________________________________________________________________________VVVV2.28