9. Capacitor and Resistor Circuits


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1 ElecronicsLab9.nb 1 9. Capacior and Resisor Circuis Inroducion hus far we have consider resisors in various combinaions wih a power supply or baery which provide a consan volage source or direc curren (volage) DC. Now we sar o consider various combinaions of componens and much of he ineresing behavior depends upon ime so we will also consider AC or alernaing curren (volage) sources which are signal generaors. he firs combinaion we consider is a resisor in series wih a capacior and a baery. he RC Circui Consider he resisorcapacior circui indicaed below: When he swich is closed, Kirchoff's loop equaion for his circui is V= Q C + ir (1) for >0 where boh Q[] and i[] are funcions of ime. here are wo unknown quaniies Q[] and i[] in equaion (1) and we need an addiional equaion namely
2 ElecronicsLab9.nb d = d () You can eliminae one of he unknowns beween equaions (1) and () by aking he derivaive of equaion (1) wih respec o ime obaining 0= 1 d d + R C d d (3) and using equaion () o eliminae he derivaive of he charge 0= 1 C d + R d () I is easy enough o solve equaion () since by rearrangemen d d = 1 RC (5) Furher à 1 i âi = 1 RC à â () Inegraion yields LogB i0 F (7) RC where i0 is a consan of inegraion which we will deermine shorly. Using a propery of he exponenial funcion, we obain from equaion (7) = i0 ExpB RC F () Iniially a =0, when he swich is closed, he capacior has zero charge and herefore here is zero poenial across i. he curren in he circui is deermined enirely by he baery poenial V and he resisance R hrough Ohm's law R = V or = V (9) R as iniially he capacior C play no role. Seing =0 in equaion () and using equaion (9) yields = i0 = V R so we have deermined he consan of inegraion. finally he soluion as () Uilizaion of equaion () in equaion () yields
3 ElecronicsLab9.nb 3 à = V R ExpB RC F (11) he produc RC has unis of ime and usually is called he ime consan =RC (1) Graph of he Soluion for he curren Suppose he numerical values V= vols, R=,000 ohms, and C=.5 microfarads as indicaed hen V = ; R = 000.; Cap =.5 *  ; = R * Cap; Consan =",, " sec"d ime Consan =0.0 sec IMPORAN: C is a proec variable assigned o somehing specific in Mahemaica so insead we use Cap as he symbol for capaciance. := V R ExpB F 0, * <, AxesLabel "ime", ime So iniially, righ afer he swich is closed, he curren i[] is a maximum and hereafer is decreases exponenially. he iniial curren is
4 ElecronicsLab9.nb So iniially, righ afer he swich is closed, he curren i[] is a maximum and hereafer is decreases exponenially. he iniial curren is which agrees wih he graph above. he ime consan (=0.0 seconds in his case) deermines he rae of decay. Afer a ime he curren has decreased o and his is equal o equaion (11) wih = namely * ã Noe ha ã so afer one ime = he curren has dropped 37% in is value. Afer =, he curren is * D which is 1% of he original value of he curren since ã and so on. he volage across he resisor is i*r so we may graph his volage as
5 ElecronicsLab9.nb 5 Vr = * R,, 0, * <, AxesLabel "ime", ime Graph of he Charge Q[] Combining equaions () and (11) yields d d = V R ExpB RC F (13) and inegraion yields = + V R à ExpB 0 ' F â' (1) where he iniial charge on he capacior is zero Q[0]=0. he inegral is easily obained ' F â' à ExpB 0  ã Combining his wih equaion (1) and recalling equaion (1) we obain = VC 1  ExpBwhich we can also graph. F (15)
6 ElecronicsLab9.nb V = ; R = 000.; Cap =.5 *  ; = R * Cap; := V * Cap 1  ExpB F 0, * <, AxesLabel "ime", ime Iniially he charge on he capacior is zero 0. and his agrees wih he above graph. Afer a ime = he charge on he capacior is and his can also be obained approximaely from he graph. Afer a very long ime he charge has is maximum value * D which is almos he same as
7 ElecronicsLab9.nb 7 V * Cap he volage across he capacior is VC = PloB Cap Q C and we may graph his using,, 0, * <, AxesLabel "ime", ime Replace he Baery and Swich by a Signal Generaor having a Square Wave he circui diagram now appears
8 ElecronicsLab9.nb Suppose he square wave generaor has a frequency f given by he square of he signal generaor can be graph using = ; 1 f= ; =", f, "Hz"D frequency =50.Hz Noice ha he period of he signal generaor is chosen o be he same as he ime consan in he RC circui. We will discuss his more laer. 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
9 ElecronicsLab9.nb 9 V0 =.; := IfB <, V0, 0F 0, <D Effecively having he signal generaor in he circui is he same as having he baery in he circui for ime 0<< seconds so he equaion (11) for he curren and equaion (15) for he charge hold for hese imes. he Volage Across he Capacior We may graph he volage across he capacior ogeher wih he signal generaor volage and obain
10 ElecronicsLab9.nb = ; PloB: Cap, :, 0, >F If he period of he signal generaor is longer, for example =*, hen he capacior has more ime o charge = ; PloB:, :, 0, >F Cap Furher if he period of he signal generaor is longer sill, for example =3*, hen he capacior has more ime o charge
11 ElecronicsLab9.nb 11 = 3 ; PloB:, :, 0, >F Cap and he capacior almos has ime o fully charge and have all he vols appear across i. Abou vols now appears across he capacior. he Volage Across he Resisor We may graph he volage across he resisor ogeher wih he signal generaor volage and obain = ; R, :, 0, >, PloRange 0, <F If he period of he signal generaor is longer, for example =*, hen he curren ges smaller sill and he volage across he resisor is reduced furher
12 ElecronicsLab9.nb 1 If he period of he signal generaor is longer, for example =*, hen he curren ges smaller sill and he volage across he resisor is reduced furher = ; R, :, 0, >, PloRange 0, <F If he period of he signal generaor is longer, for example =3*, hen he curren ges smaller sill and he volage across he resisor is reduced furher = 3 ; R, :, 0, >, PloRange 0, <F Abou vols now appears across he resisor. he Second Par of he Square Wave of he Signal Generaor.
13 ElecronicsLab9.nb 13 he Second Par of he Square Wave of he Signal Generaor. During he second par of he period of he signal generaor for imes < <, he volage is zero in he original circui. I helps make he analysis simpler o change he wave form a lile and have he signal generaor volage zero during he firs par of he cycle and a consan V0 during he second par of he cycle = ; V0 =.;, 0, V0 F SigGen = 0, <D := IfB < his corresponds o he imes in he range sec < < 0.0 sec in he previous diagram. Effecively for he firs par of he cycle he baery is removed from he circui and replaced by a shoring wire and he circui looks like
14 ElecronicsLab9.nb 1 Kirchoff circui law afer he swich is closed is 0= Q C + ir (1) which is he same as equaion (1) wihou he baery. aking he ime derivaive of equaion (1) and using equaion () yields d d i= i (17) RC Equaion (17) can be solved using he same echniques as before and we obain again equaion (11) d d = = i0 ExpB F (1) However, he iniial condiion i0 is differen his ime as we shall see. Equaion (1) can be inegraed for he charge Q[] obaining = + i0 1  ExpB F (19) he capacior is assumed fully charged iniially (which can happen if he ime consan is shor compared wih he period of he square wave) so iniially = C V (0) and when =0 he par of equaion (19) involving he exponenial funcion vanishes. For long imes here is no charge on he capacior Q[ ]=0. Since ExpA E=0 and equaion (19) reduces o D = C V + i0 = 0 (1) and i follows ha i0 =  CV = V () R Combining equaions (0) and () wih equaion (19) yields = C V + VC ExpB F  1 = V C ExpB F (3) Equaion (3) should make inuiive sence, since during he second half of he square wave cycle, he Q capacior is discharging. he volage across he capacior is V= C iniially so
15 ElecronicsLab9.nb 15 = ; V =.; := IfB < PloB:V * ExpB, 0, VF F, :, 0, >F Furher if he signal generaor is longer say hree imes he ime consan, =3 hen he capacior has even more ime o discharge
16 ElecronicsLab9.nb 1 = 3 * ; V =.; := IfB < PloB:V * ExpB, V0, 0F F, :, 0, >F he Volage Across he Resisor he curren in he circui is obained by aking he derivaive of he charge equaion (3) obaining =  VC ExpB F= V R ExpB F () and he volage across he resisor is jus R*i[]. Graphing he volage across he capacior and he volage across he resisor for he second half he cycle yields
17 ElecronicsLab9.nb 17 = ; V =.; PloB:V * ExpB F,  V * ExpB F>, :, 0, >F Noice he sum of he volage of he capacior and he volage of he resisor is jus zero as required by Kirchoff's law. If he signal generaor period is wice he ime consan hen we obain = * ; V =.; PloB:V * ExpB F,  V * ExpB F>, :, 0, >F
18 ElecronicsLab9.nb 1 If he signal generaor period is hree he ime consan hen we obain = 3 * ; V =.; PloB:V * ExpB F,  V * ExpB F>, :, 0, >F Laboraory Exercises PAR A: Place a signal generaor in series wih a resisor and capacior. 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 ==RC a firs. Wih channel 1 of he oscilloscope, measure he volage across he signal generaor and wih channel measure he volage across he capacior. Compare wih he graphs of he firs example above. Make he frequency f of he signal generaor smaller ( larger) so he capacior has more ime o charge. 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 C=0.1 mf and R=. kw hen he ime consan =RC=0.000 sec. as indicaed below:
19 ElecronicsLab9.nb 19 R =. 3 ; c = ; = R c NOE: he value of R and C you use need no be he values given above. 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 capacior meer o measured he value of he capacior and i should agree wih he capacior code (which is no ha sandardized so check wih he maker of he capacior and use your meer For example, a capacior labeled 50 B means 1 is he firs digi and 5 is he second digi for he capaciance. is he muliplier in powers of so his capacior is C=5 0 mf = 5 mf where mf= F. Capaciors can be much smaller and pf = 1 F is ofen used is he ime i ake he capacior o charge o 7% of he maximum volage (which is he maximum volage of he signal generaor). he signal generaor frequency should be se o have a period = a firs bu wha you acually conrol is he frequency f of he signal generaor where f=1/. For he example above, = ; f = So he frequency f=1,70 Hz= 1.5 khz corresponds o one ime consan. he horizonal ime scale of he oscilloscope had beer be somehing like his frequency f. If he oscilloscope is se a oo high a frequency, he ime will be oo shor o see he volage rise. On he oher hand, if he oscilloscope is se a oo low a frequency, here will no be enough ime o see he volage rise across he capacior. You also mus make sure o se he volage scale a roughly he oupu volage of he oscilloscope which you should have measured firs before connecing he capacior and resisor in he circui. PAR 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. PAR C: Call he capacior used above C1. ake a second capacior and call i C. Combine he wo capaciors in SERIES wihou he signal generaor and oscilloscope aached. he effecive capaciance is given by 1 Ceff = 1 C1 + 1 C Compue he numerical value of he effecive capaciance and check i wih he digial capaciance meer. Noe he effecive capaciance of wo capaciors in SERIES is less han boh C1 and C. Use he SERIES combinaion of C1 and C ogeher wih he signal generaor and oscilloscope and repea he measuremens of PAR A above.
20 ElecronicsLab9.nb 0 Compue he numerical value of he effecive capaciance and check i wih he digial capaciance meer. Noe he effecive capaciance of wo capaciors in SERIES is less han boh C1 and C. Use he SERIES combinaion of C1 and C ogeher wih he signal generaor and oscilloscope and repea he measuremens of PAR A above. PAR D: Call he capacior used above C1. ake a second capacior and call i C. Combine he wo capaciors in PARALLEL wihou he signal generaor and oscilloscope aached. he effecive capaciance is given by Ceff = C1 + C Compue he numerical value of he effecive capaciance and check i wih he digial capaciance meer. Noe he effecive capaciance of wo capaciors in PARALLEL is less han boh C1 and C. Use he PARALLEL combinaion of C1 and C ogeher wih he signal generaor and oscilloscope and repea he measuremens of PAR A above.
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