Practical 1 RC Circuits

Transcription

1 Objecives Pracical 1 Circuis 1) Observe and qualiaively describe he charging and discharging (decay) of he volage on a capacior. 2) Graphically deermine he ime consan for he decay, τ =. Apparaus DC Power Supply Deskop imer Resisor and Capacior Digial Mulimeer Circui Breadboard and Cable Inroducion The Capacior A capacior is a device ha sores elecrical charge. The simples kind is a "parallel plae" capacior ha consiss of wo meal plaes separaed by an insulaing maerial such as dry air, plasic or ceramic. Such a device is shown schemaically below. Meal Plaes Insulaor - + Figure 1. A simple capacior circui. I is sraighforward o see how i could sore elecrical energy. If we connec he wo plaes o each oher wih a baery in he circui, as shown in he figure above, he baery will drive charge around he circui as an elecric curren. Bu when he charges reach he plaes hey can' go any furher because of he insulaing gap; hey collec on he plaes, one plae becoming posiively charged and he oher negaively charged. The volage across he plaes due o he elecric charges is opposie in sign o he volage of he 1

2 baery. As he charge on he plaes builds up, his back-volage increases, opposing he acion of he baery. As a consequence, he curren flowing in he circui decays, falling o zero when he back-volage is exacly equal and opposie o he baery volage. If we quickly remove he wires wihou ouching he plaes, he charge remains on he plaes. Because he wo plaes have differen charge, here is a ne elecric field beween he wo plaes. Hence, here is a volage difference beween he plaes. If, someime laer, we connec he plaes again, his ime wih a ligh bulb in place of he baery, he plaes will discharge: he elecrons on he negaively charged plae will move around he circui o he posiive plae unil all he charges are equalized. During his shor discharge period a curren flows and he bulb will ligh. The capacior sored elecrical energy from is original charge up by he baery unil i could discharge hrough he ligh bulb. The speed wih which he discharge (and conversely he charging process) can ake place is limied by he resisance of he circui connecing he plaes and by he capaciance of he capacior (a measure of is abiliy o hold charge). In his lab you will es he heory ha describes his behavior by measuring some discharge raes wih an oscilloscope and comparing hem o predicions of he heory. circui An circui is simply a circui wih a resisor and a capacior in series conneced o a volage source (baery). R C R C Vo Vo Figure 2a Charging Figure 2b Discharging As wih circuis made up only of resisors, elecrical curren can flow in his circui, wih one modificaion. A baery conneced in series wih a resisor will produce a consan curren. The same baery in series wih a capacior will produce a ime varying curren, which decays gradually o zero. If he baery is removed and he circui reconneced wihou he baery, a curren will flow (for a shor ime) in he opposie direcion as he capacior "discharges". A measure of how long hese ransien currens las in a given circui is given by he ime consan τ. The ime i akes for hese ransien currens o decay depends on he resisance and capaciance. The resisor resiss he flow of curren; i hus slows down he decay. The capaciance measures capaciy o hold charge: like a bucke of waer, a larger capaciy conainer akes longer o empy han a smaller capaciy conainer. Thus, he ime consan of he circui ges larger for larger R and C. In deail: τ (seconds) = R(Ohms) C(Farads)

3 The curren does no fall o zero a ime τ; τ is he ime i akes for he volage of he discharging capacior o drop o 37% is original value. I akes 5 o 6 τ s for he curren o decay o 0 amps. Jus as i akes ime for he charged capacior o discharge, i akes ime o charge he capacior. Due o he unavoidable presence of resisance in he circui, he charge on he capacior and is sored energy only approaches a final (seady-sae) value afer a period of several imes he ime consan of he circui elemens employed. Wha's going on in his Circui 1) Iniially, he swich is open, and no curren is flowing. 2) The swich is closed as in Figure 2. (a). The capacior will charge up, is volage will increase. During his ime, a curren will flow, producing a volage across he resisor according o Ohm s Law, V R = IR. As he capacior is being charged up, he curren will be decreasing, wih a cerain ime consan τ due o he sored charge producing a volage across he capacior. In erms of τ =, he volage across he resisor and he volage across he capacior when he capacior is charging look like: V o V o V c V R 0 τ 2τ 3τ 4τ 5τ 6τ 0 τ 2τ 3τ 4τ 5τ 6τ Figure 3a: Volage across he capacior V c as a funcion of ime. Time consan τ = Figure 3b: Volage across he resisor V R as a funcion of ime. Time consan τ = Vc and V R while he capacior is charging can be expressed as: C () = Vo e V 1 and () V R = V where, e is an irraional number and is he base of he naural logarihm. The value of e is approximaely When = τ =, 1 1 V C = V (1 e ) and = V ( e ) 0 V C = 0.63V V R =. 37V0 0 So, afer = seconds, he capacior has been charged o 63% of is final value and he volage across he resisor has dropped o 37% of is peak (iniial) value. Afer a very long ime, he volage across he capacior will be V 0 and he volage across he resisor will be zero. V R 0 o e

4 3) If we flip he swich as shown in Figure 2(b), we will discharge he capacior. V o V o V c 0 τ 2τ 3τ 4τ 5τ 6τ V R 0 τ 2τ 3τ 4τ 5τ 6τ Figure 4a Volage across he capacior V c for he discharging capacior. τ = Figure 4b Volage across he resisor V R for he discharging capacior. τ = The volage of he resisor is exponenially increasing from -Vo o zero. I is criical o remember ha he oal volage beween he capacior and he resisor mus add up o he applied volage. If he circui is disconneced from power supply, hen he sum of he volage mus be zero. The volage across a capaior and a resisor in a discharging circui is given by: - V o V C = V 0 e and V R = 0 V e Calculaing he naural logarihm of he volage across he capacior (V C ) yields: ln( VC ) = ln V0e = ln( V0) + ln e or ln( VC ) = + ln( V0) Noice, he resul has he same form as he equaion of a sraigh line y=mx+b. In our 1 case, he slope is and he inercep is ln(v 0 ). Procedure In his experimen you will make measuremens using a digial mulimeer and an elecronic imer. You will charge he capacior and measure he volage across he resisor as he capacior discharges. Se he mulimeer o read DC volage. 1. Build he circui shown in Figure 7a. Use he DC power supply, a resisor, and a 6 1,000 ± 200 µf capacior ( 1μF = 10 F). The color code of he resisor you will

5 use will be given o your on your pracical lab answer shee (you ll ge he answer shee when you do he experimen); ener he color code in your Excel spreadshee. Use he color code o deermine he nominal value of he resisance and is uncerainy. Use your Excel spreadshee o do hese calculaions. 2. Use he power supply o charge he capacior o abou 12 V o 13 V. Then disconnec he power supply from he circui and noice ha he volage across he capacior slowly decreases wha could possibly cause his effec (see quesion 5)? 3. Build he circui shown in figure 7b. Afer building he circui shown in 7b, he capacior should sill be charged and you should sill noice ha he volage is slowly decreasing. For he branch of he circui conaining he swich, use a single wire wih a banana plug on each end. For now, only connec one end of his wire. Connecing he second end serves o close he swich. When his is done, he capacior will sar o discharge hrough he resisor. 4. Close he swich and simulaneously sar he deskop imer. Record he volage across he capacior when you closed he swich (a =0)1 in your spreadshee. Coninue recording he volage across he capacior once every 20 seconds unil your imer has reached 100 seconds. 5. Have Excel calculae ln(v C ) and impor your daa ino Kaleidagraph. 6. Make a plo of of ln(v C ) versus ime. Have Kaleidagraph fi your graph wih a bes-fi line. Use he curve fi parameers o deermine he ime consan of he circui and is uncerainy.

6 Uncerainies: The uncerainy in he ime consan τ obained from he graph of lnv c versus ime is given by: δ slope δτ = τ slope From he muliplicaion rule for uncerainies, δr δc δ ( ) = + R C Quesions: 1. Wha is he nominal resisance of your resisor ha you obained from he color code and is uncerainy? 2. From he nominal values of R and C, wha is he value of and is uncerainy? 3. From he equaion for he bes fi line of your graph, wha is he ime consan and is uncerainy? 4. Discuss he consisency of he ime consan obained from your graph and he value of from quesion Afer charging he capacior and disconnecing he power supply, you observed ha he volage measured by he volmeer across he capacior slowly decreased. Wha is a possible explanaion for his observaion? Comparing Daa I is ofen necessary o compare wo differen pieces of daa or resuls of wo differen calculaions and deermine if hey are compaible (or consisen). In jus abou every experimen in his course you will be asked if wo quaniies are compaible or consisen. The following describes how o deermine if wo pieces of daa are consisen (or compaible). Use his procedure o answer he quesion a he end and use i as a reference whenever you are asked if wo pieces of daa are compaible or consisen. Le s denoe he pieces of daa by d 1 and d 2. If d 1 = d 2 or d 1 - d 2 = 0, clearly hey are compaible. We ofen use Δ (pronounced Dela ) o denoe he difference beween wo quaniies: Δ = d 1 d 2 (8) This comparison mus ake ino accoun he uncerainies in he observaion of boh measuremens. The daa values are d 1 ± δd 1 and d 2 ± δd 2. To perform he comparison, we need o find δδ. The addiion/subracion rule for uncerainies is: δδ = δd 1 + δ d 2 (9) Our comparison becomes, is zero wihin he uncerainy of he difference Δ? This is he same hing as asking if: Δ δδ (10)

7 Equaion (9) and (10) express in algebra he saemen d 1 and d 2 are compaible if heir error bars ouch or overlap. The combined lengh of he error bars is given by (9). Δ is he separaion of d 1 and d 2. The error bars will overlap if d 1 and d 2 are separaed by less han he combined lengh of heir error bars, which is wha (10) says. Someimes raher han a second measured value you are comparing your daa o an expeced value. If his is he case, replace d 2 ± δd 2 wih e ± δe, where e ± δe is he expeced value including is uncerainy. RESISTOR COLOR CODES Mos resisors are coded wih color bands around one end of he resisor body. Using he resisor color code sysem is similar o using scienific noaion. Scienific noaion uses a number beween 0 and 9.9 muliplied by some power of en. The resisor color code sysem uses a number beween 01 and 99 muliplied by some power of en. These color bands ell he value of he resisance. Saring from he end, he firs band represens he firs digi of he resisance value and he second band he second digi. The hird band represens he power of en muliplying he firs wo digis. The fourh band represens he olerance. If he forh band is absen, i means he olerance is 20%. The following is a color code char, from which one can ell he resisance of a resisor: Table 2 Resisor Color Codes Color 1 s Digi 2 nd Digi Power of 10 Muliplier Tolerance Black Brown Red Orange ,000 - Yellow ,000 - Green ,000 - Blue ,000,000 - Viole ,000,000 - Gray ,000,000 - Whie ,000,000,000 - Gold - - 5% Silver % None %

8 EXAMPLE: Figure 11 Resisor Suppose he color code on a resisor is yellow, viole, orange and gold like he resisor depiced above in Figure 11. Wha is is resisance and wha is he uncerainy of his resisance? The value of he resisance can be found from he firs hree colors. Form he able above, he firs digi is 4 (corresponding o he yellow band), he second digi is a 7 (corresponding o he viole band) and he power of 10 muliplier is 3 (corresponding o he orange band). So, he resisance is: The fourh color is used o calculae he uncerainy in he resisance. The olerance of his resisor is 5% (corresponding o he gold band). So, he uncerainy of he resisor is: δ R 5 = R olerance = 47,000Ω = 2, 350Ω 100 The resisance of his paricular resisor is 47,000 ± 2,000 Ω or 47 ± 2 kω. Because he olerance is only given o one significan figure, he uncerainy can only be known o one significan figure.

9 Spreadshee for Pracical Lab: Pracical Lab R color code: Time V ln(v) (sec) (vols) Resisance (R) from color code = ± (unis) 0 C = 1000 ± 200 μf 20 = ± (unis) 40 Slope of graph = ± (unis) 60 Time consan (τ) from graph = ± (unis)