Chapter 16: Summary. Instructor: Jean-François MILLITHALER.

Transcription

1 Chaper 16: Summary Insrucor: Jean-François MILLITHALER hp://faculy.uml.edu/jeanfrancois_millihaler/funelec/spring2017 Slide 1

2 Curren & Charge Elecric curren is he ime rae of change of charge, measured in amperes (A) i dq d 1 ampere = 1 coulomb/second The charge ransferred beween ime 0 and is Q න 0 id Slide 2

3 Curren & Charge Example The oal charge enering a erminal is given by q = 5 sin4π mc. Calculae he curren a = 0.5 s. Soluion: i = dq d = d d 5 sin4π = (5 sin4π + 20π cos4π)ma A =0.5s i = 5 sin4π π 1 cos4π = π = ma = 5 sin2π + 10π cos2π Slide 3

4 Volage The volage V AB beween wo poins A and B is he energy (or work) needed o move a uni charge from A o B v AB dw dq (V) W is energy in joules (J) and q is charge in coulombs (C). 1 vol = 1 joule/coulomb = 1 newon-meer/coulomb v AB = v BA Slide 4

5 Volage Example An energy source forces a consan curren of 2 A for 10 s o flow hrough a lighbulb. If 2.3 kj is given off in he form of ligh and hea energy, calculae he volage drop across he bulb. Soluion: The oal charge is q = i = 2 10 = 20 C The volage drop is v = w = q 20 = 115 V Slide 5

6 Power Power is he ime rae of expending or absorbing energy, measured in was (W) p dw d p = dw d = dw dq. dq d P = VI = v. i Slide 6

7 Power Example How much energy does a 100 W elecric bulb consume in wo hours? In J and in Wh Soluion: w = p = 100 W 2 h 60 min/h = 720,000 J = 720 s/min w = p = 100 W 2 h = 200 Wh Slide 7

8 Capacior A capacior consiss of wo conducing plaes separaed by an insulaor (or dielecric) q = CV C is he capaciance, q is he charge, V is he volage source Capaciance is he raio of he charge on one plae of a capacior o he volage difference beween he wo plaes, measured in farads (F) C = εa d A is he surface area of each plae, d is he disance beween he plaes, and e is he permiiviy of he dielecric maerial Slide 8

9 Capacior To obain he curren-volage relaionship of he capacior, we ake he derivaive of boh sides of q = Cv To obain dq d = i = Cdv d v = 1 C න id Slide 9

10 The insananeous power delivered o he capacior is p = vi = Cv dv d The energy sored in he capacior is herefore w = න pd = C න Capacior power v dv d d = C න vdv = 1 2 Cv2 w = 1 2 Cv2 = q2 2C Slide 10

11 Capacior Example Deermine he volage across a 2 µf capacior if he curren hrough i is i = 6e 3000 ma Assume ha he iniial capacior volage is zero. Soluion: v = 1 C 0 i d + v(0) and v 0 = 0 v = e 3000 d = ቚ 3000 e 3000 = (1 e 3000 )V 0 Slide 11

12 Inducor An inducor consiss of a coil of conducing wire. v = L di d L is he inducance Inducance is he propery whereby an inducor exhibis opposiion o he change of curren flowing hrough i, measured in henrys (H) L = N2 μa l N is he number of urns, l is he lengh, A is he crosssecional area, and µ is he permeabiliy of he core. Slide 12

13 Inducor The curren-volage relaionship is obained from: v = L di d Inegraing gives di = 1 L vd i = 1 L න 0 v d + i( 0 ) Where i( 0 ) is he oal curren for < < 0 Slide 13

14 Inducor The inducor is designed o sore energy in is magneic field. The power delivered o he inducor is The energy sored is w = න = L න Since i = 0, p = vi = pd = න L di d i L di d id id = 1 2 Li2 1 2 Li2 w = 1 2 Li2 Slide 14

15 Inducor Example The curren hrough a 0.1 mh inducor is i = 10e 5 A. Find he volage across he inducor and he energy sored in i. Soluion: v = 0.1 d d (10e 5 )= e e 5 = e 5 (1 5) V The energy sored is w = 1 2 Li2 = (10e 5 ) 2 = e 10 = 5 2 e 10 J Slide 15

16 Imporan characerisics Relaion Resisor (R) Capacior (C) Inducor (L) v i: i v: v = ir i = v R v = 1 C න 0 id + v( 0 ) i = C dv d v = L di d i = 1 L න 0 vd + i( 0 ) p or w: p = i 2 R = v2 w = 1 R 2 Cv2 w = 1 2 Li2 Series: R eq = R 1 + R 2 1 = L eq = L 1 + L 2 C eq C 1 C 2 Parallel: 1 R eq = 1 R R 2 C eq = C 1 + C 2 1 L eq = 1 L L 2 A dc: Same Open circui Shor circui Slide 16