Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Engineering Circuit Analysis, Eighfh Edition Practice Problem Solutions = (60 w)(24 UdaD(60 minlh)(60 s/min) ; (60 WXI kW1000 V't)(24Uday)(7 daylwk) (12.5 centslkWh): 2.4 (pl3) 2.s (pls) VI:-V2 2.6 : (pl7) Pub, 2.7 SO V2:-Vl - (0.220)(4) - (p 17) We have a current - 1.75 A flowing out ofour positive voltage reference terminal, or + A flowing into the positive reference terminal. Thus, applyng the passive sign convention we find Pobs: 1.75 x 2.8 - 3.8 : - 6.65 W or a generated power of - Po*: - (pl7) We see thata current of + 3.2 A flows out of our positive voltage reference terminal, so a current - 3.2 A flows into that positive reference terminal. Applying the passive sign convention, Pobr:8e-lm x P*" (t = 5 ms) = = 3.2: -25.6 e-loot w (15 .6)e-r0o(sxr0-3 ) - 15.53 W Engineering Circuit Analpis, 8ft Edition Copynght @2012 McGraw-Hill, Inc. All Rights Resen/ed 1.75 2.e Engineering Circuit Analysis, Eighth Edition Practice Problem Solutions Chapters One through Six (p20) Moving from left to right and applying the passive sign convention, Pobr: Pob, : Pob, = Pobr: Pob, : (Check 2.r0 -7x8: 2x8: -5x12: 8x20= 0.25v*x20 :[0.25 x - - 56 16 - 60 + 160 - 60:0 ) @24) v - Ri, so R 2.tt + l2fx20: - vli - -441-2xl 0-6: @24) Pub, 2.t2 ll2x1o3 : @24) Pub, 2.13 -'l/R - - i2R - (3xlo-e)'(4.7xlou) _ @27) 28 Thus, wire is 25.7 C) per 1000 ft, so 500 ft is 32.65 C). v- Ri - (32.65X0.1): 3.265V Engineering Circuit Analysis, 8ft Edition Copyright @2012 McGraw-Hill, Inc. All Rights Ressrved