10. The Series Resistor and Inductor Circuit

Transcription

1 Elecronicsab.nb 1. he Series esisor and Inducor Circui Inroducion he las laboraory involved a resisor, and capacior, C in series wih a baery swich on or off. I was simpler, as a pracical maer, o replace he baery and swich wih a signal generaor producing a square wave. he curren and volage across he resisor and capacior in he circui were calculaed and measured. his lab involves a resisor, and inducor, in series wih a baery and swich and he curren is calculaed as a funcion of ime. Also, he volage drops across he resisor and inducor are calculaed. Again i is easier o sudy an experimenal circui wih he baery and swich replaced by a signal generaor producing a square wave. Kirchoff's oop ule for a Circui he volage, V across an inducor, is given by V = d d i@d (1) where i[] is he curren which depends upon ime,. (Someimes here is a minus sign placed in from of he righ hand side and his is called "enz's aw" bu we shall no need worry abou his as long as we rea equaion like a volage drop across a resisor.) Equaion (1) resuls from Faraday's aw of elecriciy and magneism. Consider he resisor and inducor circui below:

2 Elecronicsab.nb Soluion of he oop Equaion NOE: eading his for he firs ime, you migh wan o skip his secion and go righ o he soluion equaion (11). You can verify ha equaion (11) is he soluion by subsiuing equaion (11) ino equaion (). he Kirchoff oop ule applied o he circui above afer he swich is closed is i@d + d d i@d = V () Afer a long ime he curren sops changing and so di[]/d=0 and equaion () reduces o i[ ]=V which can be solved for he curren afer a long ime i[ ]. he curren iniially righ afer he swich is closed is zero i[0]=0. Equaion () is easily solved by firs dropping he consan V erm on he righ hand side so ha you ge i@d + d d i@d = 0 (3) Equaion (3) can be rearranged di i =- d () which is easily inegraed à âi i o obain =- à â (5)

3 Elecronicsab.nb 3 og@id = - +C () where C is an inegraion consan. Using a propery of ogarihms equaion () is equivalen o i@d = C ExpB- (7) F where C is a consan relaed o C so we have he soluion o equaion (3). Bu wha we wan is he soluion o equaion () which is a lile differen. We make a guess ha he soluion o equaion () is somewha like equaion (7) i@d = f@d ExpB- () F where he coefficien f[] is no consan. Noe ha since i[0]=0 i follows from equaion () wih =0 ha f[0]=0. f[] is deermined by subsiuion of equaion () ino equaion (). Afer doing his one ges d d f@d = V ExpB (9) F Equaion (9) is easily inegraed o yield f@d = f@0d + V ExpB F - 1 () and remember f[0]=0. Finally using equaion () in equaion () yields i@d = V 1 - ExpB- F (11) Noe ha from equaion (11) i[0]=0 and i[ ]=V/ as we expec. An inducive ime consan =/ is someimes inroduces and equaion (11) becomes i@d = V 1 - ExpB- F (1) Noe: =/ is he ime consan for he circui. his ime consan is differen from he ime consan for he C circui where =C. he same symbol is used in boh he and C circuis bu obviously he meaning is differen. he volage across he resisor is jus i[]* ha is V@D = V 1 - ExpB- F (13) Graph of he Soluion Equaion (11) Suppose =kw=000w and = mh=0.00 H and he baery volage is V= vols. We use equaion (11) o graph he curren i[] versus ime.

4 Elecronicsab.nb = 000.; = 0.00; = ; Prin@" =", D V =.; V i@_d := * 1 - ExpB- F =7.5-7 so he ime consan =0.75 msec for his circui. Plo@i@D,, 0, 3 * <, Axesabel "", "i@d"<d i@d he maximum value of he curren is V and afer one ime consan =7.7 - sec he curren is abou % of ha value. he volage across he resisor is jus i[]* so he volage across he resisor has he same shape as he above i[] versus ime graph bu he verical scale is differen. You can measure he volage across he resisor wih your oscilloscope. Aside: You should compare he above graph o he curren (and volage) across a resisor in an C circui

5 Elecronicsab.nb 5 i@d ime his graph is quie a bi differen han for he curren in he circui. For he C circui, he curren sars large a =0 and he curren goes o zero for long imes while he opposie behavior is rue for he curren in he circui since her he curren sars ou small and rises o is maximum value. eurn o he circui. he volage across he inducor is obained by aking he ime derivaive of equaion (1) and using equaion ( o ge = V ExpB- F = V ExpB- F A graph of he volage across he inducor equaion (13) appears := V * ExpB- F (1)

6 0, 3 * <, Axesabel "", eplace he Baery and Swich by a Signal Generaor having a Square Wave he series circui diagram now appears Suppose he and are such ha he ime consan =0.75 msec as before. As a sar, ake he period of he signal generaor o he he same as so he frequency f of he signal generaor is

7 Elecronicsab.nb 7 = ; = ; f = so he signal generaor is se o 1.3 megahz. Suppose he volage ampliude of he signal generaor is vols (he same as he baery volage) he square wave of he signal generaor is graphed using V0 =.; SqWave@_D := IfB <, V0, 0F Plo@SqWave@D,, 0, <, Axesabel "", "V@D"<D V@D Effecively having he signal generaor in he circui is he same as having he baery in he circui for ime 0<<0. msec. he Volage Across he esisor We may graph he volage across he resisor equaion (1) ogeher wih he signal generaor volage and obain

8 Elecronicsab.nb = V0 1 - ExpB- F ; PloBV@D, SqWave@D<, :, 0, >, Axesabel "", If he period of he signal generaor is longer, for example =*, hen he curren in he circui has more ime o increase

9 Elecronicsab.nb 9 = ; PloBV@D, SqWave@D<, :, 0, >, Axesabel "", Furher if he period of he signal generaor is longer sill, for example =3*, hen he curren in he circui has even more ime o increase = 3 ; PloBV@D, SqWave@D<, :, 0, >, Axesabel "", So afer a ime equal o hree ime consans, he curren is abou Vols and is approaching he meximum value of Vols. Afer a sufficienly long ime, he curren in he circui is almos consan so he ime rae of change of he curren in he circui becomes zero afer a long ime. As a resul he volage across he inducor approaches zero afer a long ime and all he signal generaor volage is he same as he volage across he resisor. his increasing behavior of he curren wih ime in he circui should be conrased wih he decreasing curren in he C circui wih ime. For he C circui, iniially here is no charge on he

10 Elecronicsab.nb So afer a ime equal o hree ime consans, he curren is abou Vols and is approaching he meximum value of Vols. Afer a sufficienly long ime, he curren in he circui is almos consan so he ime rae of change of he curren in he circui becomes zero afer a long ime. As a resul he volage across he inducor approaches zero afer a long ime and all he signal generaor volage is he same as he volage across he resisor. his increasing behavior of he curren wih ime in he circui should be conrased wih he decreasing curren in he C circui wih ime. For he C circui, iniially here is no charge on he capacior and since he volage across he capacior is Q/C, he volage iniially across he capacior is zero. Afer a long ime, he capacior charges up o he maximum charge and he volage across he capacior increases o i maximum value. he Volage Across he Inducor We may graph he volage across he inducor equaion (1) ogeher wih he signal generaor volage and obain := V0 * ExpB- F SqWave@D<, :, 0, >, Ploange 0, <, Axesabel "", So he volage across he inducor is decreasing since di/d is decreasing. If he period of he signal generaor is longer, for example =*, hen he curren ges smaller sill and he volage across he inducor is reduced furher

11 Elecronicsab.nb 11 = ; SqWave@D<, :, 0, >, Ploange 0, <, Axesabel "", If he period of he signal generaor is leven onger, for example =3*, hen he curren ges smaller sill and he volage across he inducor is reduced furher

12 Elecronicsab.nb 1 = 3 ; SqWave@D<, :, 0, >, Ploange 0, <, Axesabel "", aboraory Exercises PA A: Place a signal generaor in series wih a resisor and inducor. Prey much any oupu level (he oupu volage) of he signal generaor will do OK bu afer you ge he oscilloscope working properly make a noe of he maximum volage in your lab noebook. Choose a square wave and make he 1 frequency f of he signal generaor such ha f= wih ==/ a firs. Wih channel 1 of he oscilloscope, measure he volage across he signal generaor and wih channel measure he volage across he inducor. Compare wih he graphs of he firs example above. Make he frequency f of he signal generaor smaller ( larger) so he curren in he circui has more ime o increase oward is maximum (he volage across he inducor decreases). he volage across he inducor decreases wih ime. Keep decreasing f. Skech he oscilloscope figures you ge and indicae he values of he volage on he verical scale and he ime on he horizonal scales. Example: Suppose = mh and =.0 kw hen he ime consan =/=0.7 msec. as indicaed below:

13 Elecronicsab.nb 13 = 000.; = 0.00; = which is 0.75 msec and your oscilloscope should be se for somehing like 0. msec per division (bu his migh no be accessible o your oscilloscope). Your signal generaor f is = ; f = and he signal generaor frequency should be 1.3 MHz. NOE: he value of and you use need no be he values given above. I probably would be a good idea o ake he larges inducance you have say 0 mh since hen larger = 000.; = 0.0; = 5. - and your oscilloscope should be se for somehing like msec per division. Your signal generaor f is = ; f = or 00 khz which is easily accessible by your signal generaor. I is no a good idea o ake he resisor oo small since hen he curren is oo large and his migh overload your signal generaor and cause he oupu signal o be disored. Use your digial ohmmeer o measure he value of he resisor and make sure i is he same as given by he color code. Use your digial inducance meer o measured he value of he inducance and i should agree wih he value for prined on he inducor. PA B: Wih channel 1 of he oscilloscope, measure he volage across he signal generaor and wih channel measure he volage across he resisor. Compare wih he graphs of he second example above. Make he frequency f of he signal generaor smaller ( larger) so he capacior has more ime o charge. Keep decreasing f he frequency of he signal generaor. Skech he oscilloscope figures you ge.

14 PA Elecronicsab.nb B: Wih channel 1 of he oscilloscope, measure he volage across he signal generaor and wih channel measure he volage across he resisor. Compare wih he graphs of he second example above. Make he frequency f of he signal generaor smaller ( larger) so he capacior has more ime o charge. Keep decreasing f he frequency of he signal generaor. Skech he oscilloscope figures you ge. PA C: Call he inducor used above. ake a second inducor and call i. Combine he wo inducors in SEIES wihou he signal generaor and oscilloscope aached. he effecive capaciance is given by eff = 1 + Compue he numerical value of he effecive inducance and check i wih he digial inducance meer. Use he SEIES combinaion of 1 and ogeher wih he signal generaor and oscilloscope and repea he measuremens of PA A above. PA D: Combine he wo capaciors in PAAE wihou he signal generaor and oscilloscope aached. he effecive capaciance is given by 1 eff = Compue he numerical value of he effecive inducance and check i wih he digial inducance meer. Use he PAAE combinaion of 1 and ogeher wih he signal generaor and oscilloscope and repea he measuremens of PA A above. 1