ECE 201, Section 4Lecture 14Prof. Peter BermelFebruary 13, 2012Returning Homeworks• HW #10: N=95, µ=82%, σ=22%2/13/2012 ECE 201-4, Prof. BermelExam #1: Thursday, Feb. 9• Will post copy of the exam on Blackboard• Will return scores ASAP2/13/2012 ECE 201-4, Prof. BermelInductors• Inductors resist changes in the flow of current• Physically derived from:– Ampere’s law: a flowing current creates a constant magnetic field– Faraday’s law: a changing magnetic field induces an emf• Mathematically, = /2/13/2012 ECE 201-4, Prof. Bermel = /Example 1• With a wall AC voltage source of = 160sin 120 , turned on at t=0, with an inductor L=40 H, what is the current as a function of time for this circuit?2/13/2012 ECE 201-4, Prof. Bermel160 sin(120πt)+-40 HExample 1: SolutionWith the inductor:0 = 160sin 120 −40() =4sin 120 =1301−cos 1202/13/2012 ECE 201-4, Prof. Bermel160 sin(120πt)+-40 HCalculating Inductance• General formula for inductance: =4 ∙′ −′"#"• For solenoid: = $%&/'where:• µ=magnetic constant (usu. µo=4πx10-7H/m)• N=number of loops• A=cross-sectional area of solenoid• l=length of solenoid2/13/2012 ECE 201-4, Prof. BermelDeriving Inductance• Faraday’s law states:ℰ = $Φ≡ From the definitions for L, Φ, and A: =$Φ=$+ ∙, =$ ∙&"From Ampère’s law: =$ ∙′4 −′"#"2/13/2012 ECE 201-4, Prof. BermelCombining Inductors• Define =-$&• In series: ./= -($0+$%)& ./= 0+ %Alternative proof: 00+ %%= ./By KCL, 0= %= , and 0/ = %/ = /, so: ./= 0+ %2/13/2012 ECE 201-4, Prof. BermelCombining Inductors• In parallel: 00= %%= ./Conservation of current (0+%= ) implies: 0−%= ./ 0− ./ %= ./2/13/2012 ECE 201-4, Prof. BermelCombining Inductors• Assuming / is non-zero: 01− ./ %= ./ 0= 1+ 0 % ./ ./= 01+ 0/ % ./= 0 % 0+ %2/13/2012 ECE 201-4, Prof. BermelCombining Inductors• We can use induction to prove the following:• Combining inductors in series: ./=2 34350• Combining inductors in parallel:1 ./=21 343502/13/2012 ECE 201-4, Prof. BermelExample 2• What is the equivalent inductance, current division, and power dissipated between the 2 branches of this circuit?2/13/2012 ECE 201-4, Prof. Bermel20Θ 7830 H45 H15 HExample 2: Solution ./=09+00:;<:80= 20H300/ = 20/0= 2/3 ; %= 1/3 2/13/2012 ECE 201-4, Prof. Bermel20Θ 7830 H45 H15 HExample 2: Solution= 20 = 78+Θ()781− A/s> = 0∙300/ +%∙60%/> = 20∙/>=400Θ78%(1−)2/13/2012 ECE 201-4, Prof. Bermel20Θ 7830 H45 H15 HHomework• HW #12 solution posted today• HW #13 due today by 4 pm in EE 326B• HW #14 due Mon.: DeCarlo & Lin, Chapter 6:– Problem 48– Problem 53 (check left-hand side of circuit has an independent current source is1pointing upwards)2/13/2012 ECE 201-4, Prof.
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