RC, RL and RLC circuits


 Rolf Barrett
 2 years ago
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1 Name Dae Time o Complee h m Parner Course/ Secion / Grade RC, RL and RLC circuis Inroducion In his experimen we will invesigae he behavior of circuis conaining combinaions of resisors, capaciors, and inducors. We will sudy he way volages and currens change in hese circuis when volages are suddenly applied or removed. To change he volage suddenly, a funcion generaor will be used. In order o observe hese rapid changes we will use an oscilloscope. 1. The square wave generaor Inroducion We can quickly charge and discharge a capacior by using a funcion generaor se o generae a square wave. The oupu of his volage source is shown in Figure T Figure 1: Oupu of squarewave generaor One conrol on he generaor les you vary he ampliude, 0. You can change he ime period over which he cycle repeas iself, T, by adjusing he repeiion frequency f = 1/T. The generaor is no an ideal volage source because i has an inernal resisance 50Ω. Thus, for purpose of analysis, he squarewave generaor may be replaced by he wo circuis shown in Figure 2. When he volage is on, he circui is a baery wih an EMF of 0 vols in series wih a 50Ω resisor. When he volage is off, he circui is simply a 50Ω resisor. R R 0 + On Of f Figure 2: Squarewave generaor equivalen circui 1
2 Procedure To learn how o operae he oscilloscope and funcion generaor, se he funcion generaor for square wave oupu and connec he generaor o he verical inpu of he oscilloscope. Adjus he oscilloscope o obain each of he paerns shown in Figure 3. Try changing he ampliude and repeiion frequency of he generaor and observe wha corresponding changes are needed in he oscilloscope conrols o keep he race on he screen he same. Now se he funcion generaor o a frequency of abou 100 Hz. Observe he paern and adjus he frequency unil he period T = 10.0 ms. Funcion Generaor Oscilloscope Figure 3: Observing he oupu of he squarewave generaor 2. Resisancecapaciance circuis Inroducion We have previously sudied he behavior of capaciors and looked a he way a capacior discharges hrough a resisor. Theory (see exbook) shows ha for a capacior, C, charging hough a resisor, R, he volage across he capacior,, varies wih ime according o ( ) (  RC = 0 1 e ) (1) where 0 is he final seadysae volage. When he same capacior discharges hrough he same resisor, ( )  RC = 0 e (2) The produc of he resisance and capaciance, RC, governs he ime scale wih which he changes ake place. For his reason i is called he ime consan, which we call τ (au). I can be found indirecly by measuring he ime required for he volage o fall o 0 /2 (see Figure 4 below). This ime inerval is called he halflife, T 1/2, and is given by he equaion T 1/2 = (ln2)τ, so T 1 2 T1 2 = = (3) ln
3 / T 1/2 Figure 4: Discharge of a capacior Procedure Assemble he circui shown in Figure 5. R = 10 k C = 0.1 F Funcion Generaor Figure 5: Invesigaing an RC circui Oscilloscope Wih iniial values R = 10 kω, C = 0.1 µf, and f = 100 Hz, observe one period of he charge and discharge of he capacior. Make sure he repeiion frequency is low enough so ha he volage across he capacior has ime o reach is final values, 0 and 0. Figure 6 shows one complee cycle of he inpu squarewave ha is being applied across he resisor and capacior. Superimpose on he squarewave a skech of he waveform you observed, which illusraes he volage across he capacior as a funcion of ime. Figure 6: Capacior volage vs. ime 3
4 Wha is he larges volage, 0, across he capacior? Wha is he larges charge, q 0, on he capacior? 0 = q 0 = Use he ohmmeer o measure R. (Recall ha a resisor should be removed from he circui before you measure is resisance wih an ohmmeer.) R = To measure T 1/2 change oscilloscope gain (vols/cm) and sweep rae (ms/cm) unil you have a large paern on he screen, like he paern shown in Figure 7a. Make sure he sweep speed is in he calibraed posiion so he ime can be read off he x axis. Cener he paern on he screen so ha he horizonal axis is in he cener of he paern. Tha is, so ha he waveform exends equal disances above and below he axis. Move he waveform o he righ unil he sar of he discharge of he capacior is on he verical axis as shown in Figure 7b. The halflife is jus he horizonal disance shown on Figure 7b. Figure 7a and b: Measuring he halflife Measure he halflife, T 1/2, and from his compue he ime consan τ using Equaion 3. Make sure o include unis wih your resuls T 1/2 = τ = 4
5 You have jus deermined his circui s ime consan from he capacior discharging curve. Theoreically, he ime consan is given by he produc of he resisance and capaciance in he circui, RC. Compue RC from componen values. Show your calculaion in he space below. Noe ha, as described above, he squarewave generaor has an inernal resisance of 50Ω. Thus, he oal resisance hrough which he RC circui charges and discharges is R + 50Ω. τ = When his calculaion is carried ou using ohms for resisance and Farads for capaciance, he produc has unis of seconds. Use dimensional analysis o show ha his is indeed he case. Wihin he uncerainies of he olerances (10%) of he resisor and capacior, do your measuremens suppor he equaion τ = RC? (If here is more han 20% disagreemen, consul your insrucor.) Alhough you have been old ha he inernal resisance of he funcion generaor is 50Ω, le s say we had kep his piece of informaion from you. Wihou using an ohmmeer, ouline a procedure for measuring he inernal resisance of your funcion generaor. 5
6 Adjus he funcion generaor o ry differen values of f and hence, T, while keeping τ fixed by no changing eiher R or C. On he lef graph below skech wha you saw when he period T of he square wave was much less han he ime consan, τ. On he righ graph below skech wha you saw when he period T of he square wave was much greaer han he ime consan. >> T = T = << T = T = 3. Resisanceinducance circuis Inroducion In his secion we conduc a similar sudy of a circui conaining a resisor and an inducor, L. Consider he circui shown in Figure 8 below. The ex shows ha if we sar wih he baery conneced o he LR circui, afer a long ime he curren reaches a seadysae value, i 0 = 0 /R. R 0 L Figure 8: A model circui wih an inducor and resisor If we call = 0 he ime when we suddenly hrow he swich o remove he baery, allowing curren o flow o ground, hen curren changes wih ime according o he equaion ( ) ( R/L) i = i0 e (4) If, a a new = 0, we hrow he swich so he baery is conneced, he curren increases according o he equaion ( ) ( ( R/L) i = i0 1 e ) (5) The ime consan for boh equaions is L/R and 6
7 T 1 L 2 = = (6) R We can find he curren as a funcion of ime by measuring he volage across he resisor wih he oscilloscope and using he relaionship i() = ()/R. Noe ha wha we would see firs is he growh of curren given by Equaion 5, where he final curren depends on he squarewave ampliude 0. Then, when he square wave drops o zero, he curren decays according o Equaion 4. The ime consan should be he same in boh cases. Procedure Se up he circui shown in Figure 9 below. L = 25 mh R = 1 k Funcion Generaor Oscilloscope Figure 9: Invesigaing he LR circui Wih iniial values R = 1kΩ and L = 25mH, se he oscilloscope o view one period of exponenial growh and decay. Again, make sure ha f is low enough for he curren o reach is final values, i 0 and 0. Sar wih f = 5 khz. Superimpose a skech of he waveform you observe on he single cycle of he inpu squarewave shown below. Wha is he larges curren hrough he inducor? i 0 = Measure he halflife. From his value, compue he ime consan. T 1/2 = τ = 7
8 Measure he value of R and he dc resisance of he inducor wih an ohmmeer. Finally add he inernal resisance of he squarewave generaor o obain he oal resisance. Compue he value of L/R from he componens values. R (of resisor) = R (of inducor) = R (of funcion generaor) = R (oal) = τ = L/R = Wihin he uncerainies of he manufacuring olerances (10%) of he resisor and inducance, do your measuremens suppor he equaion = L/R? When his calculaion is carried ou using ohms for resisance and Henries for inducance, he raio has unis of seconds. Use dimensional analysis o show ha his is indeed he case. Adjus he funcion generaor o ry differen values of f and hence, T, while keeping τ fixed by no changing eiher R or L. On he lef graph below skech wha you saw when he period T of he square wave was much less han he ime consan, τ. On he righ graph below skech wha you saw when he period T of he square wave was much greaer han he ime consan. 8
9 >> T = T = << T = T = 4. Resisanceinducancecapaciance circuis Inroducion As discussed in he exbook, a circui conaining an inducor and a capacior, an LC circui, is an elecrical analog o a simple harmonic oscillaor, consising of a block on a spring fasened o a rigid wall. L C k M Figure 10: LC Circui and is analog, a mechanical SHM Sysem In he same way ha, in he mechanical sysem, energy can be in he form of kineic energy of he block of mass M, or poenial energy of he spring wih spring consan k; in 1 2 he LC circui energy can reside in he magneic field of he inducor U = 2 Li, or he 1 2 elecric field of he capacior, U = 2 q C. Boh he curren and he charge hen change in a sinusoidal manner. The frequency of he oscillaion is given by 1 0 = (7) LC All circuis have some resisance, and in he same way fricional forces damp mechanical SHM, resisance causes energy loss (i 2 R) which makes he charge decay in ime. ( )  q q e cos( ) (8) = 0 1 ( ) = 0 1 (9) 9
10 where τ = 2L/R or T = ln2(2l R) = 0.693(2L ) (10) 1 2 R For large τ he sysem is underdamped and he charge oscillaes, aking a long ime o reurn o zero. 2 2 Noe from Equaion 9 ha when 0 = 1, 1, which appears in he argumen of he cosine funcion of Equaion 8, is zero a all imes. This condiion is called criical damping. Criical damping occurs when R = 2 L C. When he resisor is larger han he criical value he sysem is overdamped. The charge acually akes longer o reurn o zero han in he criically damped case. The decaying oscillaions in he LRC circui can be observed using he same echnique as used o observe exponenial decay. Again, a squarewave generaor produces he same effec as a baery swiched on and off periodically. The oscilloscope measures he volage across C as a funcion of ime. a. Observing oscillaions in a RLC circui Procedure Assemble he circui of Figure 11. Use a small value of R, say, 47Ω. Be sure o reduce he signal generaor frequency o 100 Hz or below so you can see he enire damped oscillaion. R L = 25 mh C = 0.1 F Funcion Generaor Oscilloscope Figure 11: Invesigaing he LRC circui Measure he period and calculae he frequency of he oscillaions. (The period is NOT 0.01 s = 1/100 Hz, he repeiion frequency of he square wave.) Measured period = Calculaed f 1 = = f = Calculae 0 from componen values. 0 =1 LC = 10
11 Compare he 1 you measured wih 0 ha you calculaed from componen values. In heory, 1, he damped frequency, is only slighly less han 0, he undamped frequency, making his a valid comparison of heory wih experimen. b. Criical damping and overdamping in a RLC circui Procedure Noe ha in he equaions for his circui, R represens he sum of he resisance of he inducor, he inernal resisance of he squarewave generaor, 50Ω, and he resisance of he resisor. To sudy criical damping and overdamping, remove your fixed resisor and pu in is place a 5kΩ variable resisor. Sar wih he variable resisor se o a small value of R. For small R you should see he oscillaions ha are characerisic of underdamping. On he firs graph below skech he waveform ha appears on he oscilloscope. Increase R unil criical damping is reached; ha is, unil he oscillaions disappear. Skech his curve on he middle graph above. Use he ohmmeer o measure he value of he variable resisor a criical damping. (Don forge o disconnec he variable resisor from he circui when measuring is resisance.) R (variable resisor a criical damping) = Calculae he oal resisance in he circui a criical damping by adding he dc resisance of he inducor and 50Ω for he funcion generaor o he resul above. R (oal a criical damping) = Compare his value of he circui resisance a criical damping o he prediced value for criical damping, R = 2 L C. 11
12 Wha happens o he waveform when he resisance is larger han he criicaldamping value? Skech your resuls on he righmos graph above. c. Underdamping in a RLC circui Inroducion When he circui is underdamped, Equaion (8) applies. This means ha he ampliude of he oscillaion will decay exponenially, wih he ime consan for he decay being: 2L (11) R Recall ha when an exponenial decay is ploed on a semilog scale he resuling graph is a sraigh line wih a slope equal o 1/. You can find he slope of a line on a semilog graph by idenifying he wo end poins of he line. Noe he ime and volage a each poin 1 and 1, 2 and 2. Calculae he naural log of he wo volages. Then, ln( 2 ) ln( 1 ) 1 slope 2 1 The following seps describe how o measure he ime consan of he decay of he oscillaions. (12) Procedure Adjus he variable resisor so ha he circui is underdamped and oscillaes abou seven or eigh imes before he oscillaions become oo small o be easily seen on he oscilloscope. Cener he oscillaion paern verically on he screen so ha when he oscillaions have decayed he line on he oscilloscope coincides wih he ime axis. In he able below record he volage of each oscillaion peak, and he corresponding ime for each peak. When your able is complee you should have six or seven ses of daa recorded. Peak Heigh () Time ( ) Table 1 12
13 Creae a semilog graph of your daa. (You can use Excel o creae a semilog graph.) Following he procedure described above, deermine he ime consan for your circui. This is your experimenal value for he ime consan. Show your calculaion and resul here. τ experimen = Remove he variable resisor from he circui and measure is resisance. Add his value o he resisance of he squarewave generaor (50 ohms) and he resisance of he inducor o ge he oal resisance of your circui. Show your calculaion and resul here. R oal = Using Equaion (11) calculae he heoreical decay ime consan for your circui. Show your work. τ heory = Compare his heoreical value o he experimenal value you found above. They should agree wihin en or weny percen. If hey do no, consul your insrucor. 13
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