ECE 201, Section 4Lecture 16Prof. Peter BermelFebruary 17, 2012Exam #1: Question 92/17/2012 ECE 201-4, Prof. BermelExam #1: Question 102/17/2012 ECE 201-4, Prof. BermelIn the following circuit, R1= R2= R3= 1Ω, gm=1S.Use loop (mesh) analysis to express I1in terms of IS.1) I1 = 0.1IS2) *I1 = 0.5IS3) I1 = 0.75IS4) I1 = 1.25IS5) I1 = 1.5IS6) I1 = 2IS7) I1 = 2.5IS8) I1 = 3ISExam #1: Question 112/17/2012 ECE 201-4, Prof. Bermel6 Wix40 V5A+_3 W4 WUsing superposition, find the value for ixin the circuit below (in A):(1) 1 (2) 2 (3) 3 (4) 4(5) 5 (6) *6 (7) 7 (8) 8Exam #1: Question 122/17/2012 ECE 201-4, Prof. BermelWhat are the coefficients α and β in: = +1) α= -1/5, β= 12) α = -1/5, β = 33) α = -1/15, β = 24) α = 1/5, β = 15) α =1/5, β = -16) ∗α =1/15, β = 37) α = 1/30, β = -2Vs+-Is30 Ω10 Ω15 Ω5 Ω60 Ω+-voExam #1: Question 132/17/2012 ECE 201-4, Prof. BermelTwo non-ideal voltage sources are connected in parallel. The first source has voltage V1and resistance R1. The second source has voltage V2and resistance R2. The parallel connection is a non-ideal voltage source with the following voltage and resistance:1) V1+V2, R1+R2.2) V1+V2, (R1R2)/(R1+R2).3) *(R2V1+R1V2)/(R1+R2), (R1R2)/(R1+R2).4) (R2V1+R1V2)/(R1+R2), R1+R2.5) (V1V2)/(V1+V2), R1+R2.6) (V1V2)/(V1+V2), (R1R2)/(R1+R2).7) (R1V1+R2V2)/(R1+R2), (R1R2)/(R1+R2).8) (R1V1+R2V2)/(R1+R2), R1+R2.+-R1+-R2V1V2Power in Inductors• Since the instantaneous power in an inductor: = = • The energy stored between toand t1is: ===12 −• For ac wave with period T=1/f:=0, if =+, for integer m2/17/2012 ECE 201-4, Prof. BermelIntroduction to Capacitors• Capacitors are circuit elements capable of storing charge• Voltage drop proportional to stored charge: V=Q/C, or Q=CV• Capacitance C in units of C/V or F• Current flow given by: == 2/17/2012 ECE 201-4, Prof. BermelCalculating Capacitance• Capacitance determined by geometry and materials present• Calculated using Gauss’ Law: != "Φ$= "%&∙()• For parallel plate capacitor, E=Q/εA=V/d, thus: ="( = • In vacuum, " = "= 8.854∙10.F/m2/17/2012 ECE 201-4, Prof. BermelCalculating Capacitance• For two capacitors in parallel: ===/0/1212 = /0/= += +12= +• For N capacitors in parallel:12=345462/17/2012 ECE 201-4, Prof. BermelC2C1Calculating Capacitance• For two capacitors in series: = = = 12/0/12=+=+112=1+1• For N capacitors in parallel:112=3145462/17/2012 ECE 201-4, Prof. BermelC1C2Example 1• What is the current produced by plugging a 100 µF capacitor into the wall? How much power is dissipated by it?2/17/2012 ECE 201-4, Prof. Bermel+-160 sin(120πt)100 µFSolution = = (10089)∙160∙120<cos 120< = 1.92cos 120< = = 1.92cos 120< ∙160sin 120< = 153.6sin 240<2/17/2012 ECE 201-4, Prof. Bermel+-160 sin(120πt)100 µFExample 2What is the current flow in this circuit as a function after the switch is closed at time t=0?2/17/2012 ECE 201-4, Prof. Bermel+-100 µF200 µF12 V3 kΩSolutionApplying KVL:12 = 3DΩ +/(100+20089) ==12−30089 3DΩ′12−′/(30089)HI=3DΩI2/17/2012 ECE 201-4, Prof. Bermel+-100 µF200 µF12 V3 kΩSolution(−30089) 3DΩ′J−0.0036HI= − 0. 9K ln−0.0036−0.0036= M− M= N./O = M1−N./O = N./O(t>0)2/17/2012 ECE 201-4, Prof. Bermel+-100 µF200 µF12 V3 kΩHere:Qf=3.6 mCT=0.9 sIo=4 mASolution2/17/2012 ECE 201-4, Prof. Bermel00.511.522.533.54-2 -1 0 1 2 3 4Current (mA)Power & Energy in Capacitors• Instantaneous power given by: = = ∙ = • Energy stored over a time interval [to, t1] is: = =PP =12 −2/17/2012 ECE 201-4, Prof. BermelHomework• HW #14 solution posted today• HW #15 due today by 4 pm in EE 325B• HW #16 due Mon.: DeCarlo & Lin, Chapter 7:– Problem 12– Problem 16– Problem 172/17/2012 ECE 201-4, Prof.
View Full Document