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GATech - ECE 3710 - Class Notes - Week 1

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• What are the reference directions for flow of I? s positive wrrent flows high(+) to low (-) o negative corrent flows low (-) to high (+) Study Sou > _ [§ 1.2 Voltage What is voltage? G when moving through Ē, charges either . gain ou lose energy - moving in same direction " "opposite direction o voltage is the energy gained or lost per coulomWe also discuss several other topics like the oldest application of judicial notice is for

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b of charge how to calculate it? svo o december (&; v) y, "OPE energy charge Quiz A VE energy charge 20 20J 4c -15v (+ bc same direc.) I Quize v = ? 1 h = ov=EJ (- bk opposite direc.) voltage is always measured from a reference point (voltage measurements are relative) Problem 1-2-2 Iakq,92 _no6e-6) (10e-6) 12, 7 z 11 La (20e-zje v2 = .6 z ml.org ( 20e $1.3 Power & Energy power is the rate at which the charges energy (w) changes over time (4) G can also be expressed as vxI - dw - dw da .. De at = da dec = VI => P: VI G unit: Watts (w) O Quiz P-IV = (SA)(2V) - low (consumed)] Study Sour Quiz B PEIV= (-4A) (3x) = Elzw (generated * + power consumed 1-power generated the sum of power generated & consumed in a system = zero GP is the rate of change of energy & - energy is always conserved Quiz A 1-3À Px + 4+8-6=0 vx ? . Px=-ow . P: IV => v= : SV - bu -bo ЗА ow [v=- zul B Quiz ix Px+2 -3 +4=0 Px = - 3 - Bu 1. 0 ZW 4W T: 3 = -F.SA . -3W | Resistance & Ohm's Law when charged particles flow through a co material, they encounter electrical resistance GR is determined by the materiaal's cross sectional area (A), length (L), & resistivity (p) Rze. #circuit flow follows path of least resistance the resistivity is an intrinsic property that quantifies the material's opposition to charge flow units for R: Ă = 2= ohms electrical conductance measures how a material allows charge flow Greciprocal of resistance IG=Ř - OA the conductivity (o) is an intrinsic Study Soup. property that quantifies the material's receptiveness to charge flow lo= 2 Gunit: f = siemens (0) = mhos Quiz R=s= (23610-8)(1)-((2)(105)_2] A 10-3 Quiz B Rz= 2 = Ž Ri=110-s a G= 8z = 1 to lose Ohm's law the voltage produced by the current flow is d to the resistance of the material V: IR o [Ve=by] Problem 1-8-0 Vc = 10 (2+2+3) T 3v + 3 +Vc -VE-VB-0 lo VAT Vc-VE+ VG= Vab... 2+ Vc + 3 + 2 =10 Vc=3v Vac = VA + VOR VH = 2 + 2 + 1 - SV [§ 1.5 Intro to circuit Diagrams open ro Rua wires : lines GR=0 = V 6 G . = node = junction Is Vs see below Vs: independent Is: independent current source voltage source son R: resistors =: ground (reference divec.) • node - any point on a cirwit where 2 or more circit elements meet G different if they have different voltage • junction - the point where wires meet a wire that has the same start & end node as another circit element shouts that element G current will travel through the short instead of the circwit element • an open circuit has terminals that are not connected to other circit elements (no wrrent will flow) (KUL) Module 2 [{2.1 Overview & Kirchoff's Voltage Law • Kirchoffis voltage Low : E Vloop = 0 G trick... when moving around a loop, if you come to a "." first (.e. 10.), then you subtract that term (if 10-), you add KA -VA - V8+ VOR Vc=0 EX. W @ -votvezo Vo = VE Cparallel EX. + VB + IV +(-4v) = 0 Ve = 3V -IV+ 2V + SV - VASO VA = 60 ū ir voltages across parallel elements are equal SM -10 + 100 iz + 200 iz- 0 - 3002 OSAD 1000 - 300 33+ Vi :o solova 2001 ) with an open loop, simply determine polarity ( [$2.2] Kirchoff's current Law Kirchoff's current Law: Eileaving - Erientering 4 Firsiz istizzie <=> riigis UP (KCL) G alternative: Eileaving = 0 & Eientering = •KCL for series elements ... current is the same X 2A -4A ZA Eileaving = 0 → - SA ? ... 2-5-4+1=0 -SA (color-coded nodes) 1 SILA SA entering leaving KCL:22, + izti3 A KUL: Solis) +50(is) -100 (iz) = 0 KVL: 100 12 = 5oi, i.:1, ia: 0., is:D.S SON Problem 2-2-1 (6) RP R204 mm cingos) - 50 202 mm (rs) re 2015) (6) conductance = e = = o.zmno] (c) 2003 - 250 VetR lov: 1258 I-0.4A

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