EDEXCEL NATIONAL CERTIFICATE/DIPLOMA UNIT 67 - FURTHER ELECTRICAL PRINCIPLES NQF LEVEL 3 OUTCOME 2 TUTORIAL 1 - TRANSIENTS

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1 EDEXEL NAIONAL ERIFIAE/DIPLOMA UNI 67 - FURHER ELERIAL PRINIPLE NQF LEEL 3 OUOME 2 UORIAL 1 - RANIEN Uni conen 2 Undersand he ransien behaviour of resisor-capacior (R) and resisor-inducor (RL) D circuis ransien behaviour of R circui: variaion of curren and volage wih ime when charging/scharging; ime consan; graphical deerminaion of growh and decay of volage and curren when charging/scharging; pracical R circui o demonsrae ransien behaviour; demonsrae he effec of he circui ime consan on a recangular waveform e.g. inegraor and ffereniaor circuis; calculaions e.g. ime consan, growh of capacior volage, iniial and seady sae values of curren, decay of resisor volage ransien behaviour of RL circui: variaion of curren and volage wih ime when conneced/sconneced o a D volage source; ime consan; graphical deerminaion of growh and decay of curren and volage when conneced/sconneced o a D volage source; pracical RL circui o demonsrae ransien behaviour; calculaions e.g. ime consan, growh of curren, decay of induced volage, curren decay I is assumed ha sudens have already sued elecrical principles. You will find more useful informaion on capacior charging and scharging a his web sie: hp://www.elecronics-uorials.ws/rc/rc_1.hml D.J.Dunn 1

2 1. HARGING AND DIHARGING OF A APAIOR he circui shows a capacior in series wih a resisor and a D source. When he swich is in posiion 1, he baery will send curren hrough he resisor and charge he capacior. he volage is iniially zero and he volage R is he same as so he capacior charges a a fas rae. As he capacior charges, he volage increases and R decreases and he curren reduces unil he = and he curren is zero. he capacior is hen charged o. If he swich is moved o posiion 2, he charge rushes ou of he capacior hrough he resisor, ssipaing all he energy as hea unil no curren flows and = R =. If we recorded he volage c and R and I agains ime for charging and scharging we ge graphs like his. If he resisance and capaciance are boh large he charging will ake longer like filling a large conainer hrough a ap ha is nearly closed. If hey are small, he charging is fas like filling a small conainer wih he ap fully open. he agram shows his affec on charging. IME ONAN We call he produc R he ime consan and denoe i or someime 'τ'. If you refer o he absolue basic unis of Farads and Ohms you will see ha he unis are indeed seconds. When he volage and curren are changing, we say hey are in a RANIEN AGE. When hey reach a consan value we say hey are in a EADY AE. I is widely acceped ha i akes a ime of 5 or 5R o reach a seady sae. We need mahemaics o derive he equaions of hese curves bu you don' have o follow he derivaion in order o complee his uorial (alhough i helps). he following derivaion uses sraigh forward algebra and basic calculus. HARGING he curren hea flows from he source a rae i amps hrough he resisor and ino he capacior. ΔQ urren is he rae of charge wih ime so i Δ R he volage across he resisor is R = i R i. Equaing R ΔQ R R Δ dq Over an infiniesimally small change in ime his becomes he calculus form R R d dq Hence R R.(A) d D.J.Dunn 2

3 For he capacior, he we know Q = For a small change Q = For an infiniesimally small change dq = d.(b) ubsiue (B) ino (A) d R R d = R hence d d R d d his equaion ells us how varies wih ime bu in order o use i, we mus solve i. d In order o find we mus solve he equaion d Le = x Differeniae and since is a consan we find -d = dx he equaion becomes dx d dx x x Rearranging d x Inegraing 1 dx d ln x x he limis become clear when we subsiue x = 1 dx d ln ln x ln ln ln 1 ake anilogs and e 1 Rearrange 1- e his relaes he volage over he capacior o ime from he momen when he swich is hrown o posiion 1. DIHARGING When he swich in he circui is hrown o posiion 2 he volage over he capacior is a = and if he derivaion is repeaed wih he correc limis of inegraion we find he relaionship is: e he graph below shows he resul for = 2 seconds and = 1. Noe he volage across he resisor is HE MEANING OF HE IME ONAN onsider he charging curve. 1- e he rae of change of volage a any ime is he graen of he curve and simply obained by ffereniaing. d e d onsider ha a = (he sar of he process) d (Remember ha e = 1) d If he change coninued a his rae, he volage would become afer seconds as illusraed on he graph. D.J.Dunn 3

4 We could define as he ime aken o reach he maximum value if he iniial rae of change is mainained. In oher words i is where he iniial graen inerceps he final value as shown. Anoher meaning is obained by examining he value afer seconds. Puing = we have 1 1- e 1- e.633 o he ime consan is he ime aken o change by 63.3% of he final value. his is useful when finng from a graph. If calculae he volage a = 4 we find = 98.2% and his is aken as anoher definiion of. I is widely acceped ha when = 5 he volage has reached is maximum. GRAPHIAL ONRUION MEHOD If we know he ime consan we can consruc he charge or scharge curves wih a graphical mehod as follows. ep 1 - Draw he final value line and mark off he ime consan. onnec he origin o his poin. ep 2 - hoose a poin on he las line drawn and mark off anoher ime consan from his poin. Projec o he final value line and connec he wo poins. Repea his process as shown and you will have a series of angens o he charging curve. his may be drawn in as shown in red. he closer you make your poins he more accurae he graph will be. he consrucion of a scharge curve is he same bu invered. D.J.Dunn 4

5 WORKED EXAMPLE No. 1 A capacior of value 5 μf is charged from zero o 1 hrough a 5 MΩ resisor. alculae he ime consan and he ime aken for he volage o rise o 5. OLUION = R = 5 x 1-6 x 5 x 1 6 = e e e 25 e ln(.5).6931 = 25 x.6931 = seconds WORKED EXAMPLE No. 2 A apaciance of 2 F is conneced in series wih a resisor of 2 k. he volage across he nework is suddenly changed from o 1. alculae he ime consan and deduce he ime aken for he volage on he capacior o rise o 5. OLUION = R = 2 x 1 3 x 2 x 1-6 = 4 seconds =5 = 1(1 e -/4 ).5 =(1 e -/4 ).5 = e -/4 ln(.5) = -/ = -/4 = 2.77 s when = 5 WORKED EXAMPLE No. 3 A apaciance of 1 F is charged up o 5. he capacior is scharged by connecing a 2 M resisor across he erminal. alculae he ime consan and deduce he ime aken for he volage on he capacior o fall o 1. OLUION = R = 2 x 1 6 x 1 x 1-6 = 2 seconds =1 = 5e -/2 1/5 = e -/2 ln(1/5) = -/ = -/2 = s when = 1 D.J.Dunn 5

6 ELF AEMEN EXERIE No he graph shows a charging curve for a capacior and resisance. Work ou he ime consan and deermine he capaciance if he resisor value is 6 kω. (Answers.12 s and 2 μf) 2. alculae he ime consan for an R circui wih a resisance of 22 and capaciance of 47 nf in series. (13 s) How long es i ake for he volage o become a seady value? 3. A capacior of 2 μf is charged o 12 and hen a resisor of 5 Ω is conneced across i. alculae he charge sored. (.24 oulomb) alculae he energy sored. (.144 J) alculae ime aken o scharge o 1 (.25 s) 4. A capaciance of 15 μf is charged o 1 hrough a resisance of 2 MΩ. alculae he ime consan. onsruc he charging curve and deermine he ime aken o charge o 5. How long does i ake o reach a seady value? (21 s and 15 s approx) D.J.Dunn 6

7 2 HARGE DIHARGE OF AN INDUOR When a resisor is conneced in series wih an inducor we ge a similar charging and scharging effec. When he swich is hrown o posiion 1 here is a rush of curren. A his momen he rae of change of curren /d is a maximum so we ge maximum back emf on he inducor. Afer a ime he curren seles down o a maximum value of I = s /R so s = IR Le he curren a any ime be i. R = ir L = L /d ir L d IR ir L d R(I i) L d L (I i) R d If we examined he unis of L/R we would find ha his is seconds and L/R is he ime consan for he circui. = L/R (I i) d d I i If we subsiue x = I i hen dx = - dx dx i d d lnx lni i x x ake anilogs I -i ln(i i) ln(i) ln ln 1 I i i e 1 1 e i I 1 e 1 e I I R his relaes he curren flowing in he circui o ime and he resul is a curren ha rises fas and hen levels off a he consan value. i I he graph shows he resul for = 1, R =.5 Ω and L = 2H giving are no pracical and only used o illusrae he resul) = 4 seconds (he values he ime consan may be defined as he ime aken o reach 63.3% of he maximum curren. Pure inducance is impossible and inducors always have some resisance in he conducor. ircuis usually show a pracical inducor as a pure inducance in series wih a resisance. In iems like ransformers, he resisance of he coil can be quie high. D.J.Dunn 7

8 WORKED EXAMPLE No. 3 An Inducance of 4 mh also has a resisance of.3 Ω. alculae he ime consan. Wha is he seady sae curren when 2 is applied across i? Wha is he curren.2 s afer he volage is suddenly changed from zero o 2? OLUION = L/R =.4/.3 =.13 s I = /R = 2/.3 = 66.7 A.2.13 i I1 e e 51.8 A ELF AEMEN EXERIE No alculae he ime consan for a series R L circui wih an inducance of 6 μh and resisance.2 Ω. (3 ms) 2. An inducor wih inducance 6 mh and resisance.7 Ω suddenly has 2 conneced across i. alculae he seady sae curren. (2.857A) alculae he energy sored and power ssipaed. (.245 J and W) alculae he ime aken for he curren o rise o.5 A. (16 ms) 3. he volage across he inducor shown is iniially zero. how ha when he swich is closed ha he volage across he inducor will be v = E e -/τ where is he ime elapsed afer he swich is closed. Deermine an expression for he ime consan τ in erms of R and L. D.J.Dunn 8

9 3. APAIOR FOR INEGRAING AND DIFFERENIAING d he basic relaionship beween volage and curren flowing ino a capacior is i d capacior can be used inegrae and ffereniae a volage. INEGRAOR IRUI If a volage s is applied o he circui shown, he volage across he capacior is c 1 i d uppose ha a square wave form is applied o he inpu wih a frequency jus sufficien o allow he capacior o charge and scharge on each cycle. he oupu c will be a series a charging and scharging curves. so a If he frequency is slower he capacior will fully charge and dwell before scharging. If he frequency is faser, he capacior will no fully charge or scharge and he waveforms will look like his. If he consans are so arranged, a riangular wave form can be produced. DIFFERENIAOR IRUI If a square wave is applied o his circui, he volage R will be he inverse of he previous oupus ELF AEMEN EXERIE No kech he oupu of an inegraor and ffereniaor circui when a riangular wave form is applied o he inpu. Explain your reasoning. D.J.Dunn 9

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