How To Solve The \"Kirchoff S Ule\" Problem

Transcription

1 ircui and Time onsan 8M Objec: Apparaus: To invesigae he volages across he resisor and capacior in a resisor-capacior circui ( circui) as he capacior charges and discharges. We also wish o deermine he capaciaive ime consan for he circui. esisor (100Ω), capacior (330 µf), wo volage sensors, power amplifier, pach-cords, compuer, PASO Inerface, and Daa Sudio Sofware. onsider he circui shown in Figure 1: FOWOD a S + _ b V V + _ + _ Figure 1. A resisor-capacior series circui. When he swich S is hrown o posiion a, charges sar o move and we have a curren flow. These charges move hrough he resisor and begin o charge he capacior. A any ime, he sum of he volages around he circui loop mus be zero (Kirchoff s ule), hence we have: (1) V V = 0 Where: = he applied volage (2) V = i = he volage drop across he resisor (3) (4) V = q he volage drop across he capacior When (2) is insered ino equaion (1) we obain: q = i + If we include he relaionship beween he curren and he charge, namely i = dq d 8M 1

2 (5) We obain: = q dq d = ( )I I 0 0( q) ( ) dq ( d ) quaion (5) conains wo variables, he charge on he capacior (q) and he ime (). Afer separaing hese variables we have: + q (6) 1 dq d = ( q) Seing equaion (6) up for inegraion and using he fac ha a he ime = 0 he charge q = 0 as he lower limi and he fac ha a some oher ime he charge is q as he upper limi we have: (7) 1 Le s now make a change of variable by leing u = q and hence du = dq. We will also need o change he limis of he q inegraion by noing ha when q = 0 he lower limi is u = and for he upper limi we replace q wih u = q. This allows us o rewrie (7) as follows: (8) 1 q du ( d = )I I ( u) 0 (9) (10) quaion (8) inegraes o: Insering he limis we have: Taking he anilog of equaion (10) we obain: 1 = lnu 1 q = ln( q) ln() = ln ( ) ( ) (11) e / = ( q) / Which may be solved for q o obain: ( ) (12) q = (1 e / ) Looking a equaion (12), we see ha afer ime >>, he capacior is fully charged and he charge on i is (13) q = = Q max If he charge on he capacior a any ime is given by equaion (12), hen he poenial difference across he capacior a any ime is (14) V = q / = (1 e / ) / = (1 e / ) A plo of V as a funcion of ime is shown in Figure 2. 0 q 8M 2

3 V V vs Figure 2. Plo of he poenial difference V across he capacior in a -circui as a funcion of ime while he capacior is charging The curren in he circui a any ime may be obained from equaion (12) as follows: (15) i = ( ) dq = d[(1 e / )] = e / / = I Max e d d where: (16) I max = / From equaion (15), noe ha a = 0 he curren has he maximum value of / and afer a long ime compared o he curren is zero. If he curren in he circui a any ime is given by equaion (15), he poenial difference across he resisor a any ime is (17) V = i = I max e / / = e A plo of V as a funcion of ime is shown in Figure 3. V V vs Figure 3. Plo of he poenial difference V across he resisor in a -circui as a funcion of ime while he capacior is charging. 8M 3

4 Figure 4 gives a summary of wha happens when he swich S is hrown o posiion a and he poenial difference is applied o he circui charging he capacior. vs (a) V = 0 V vs (b) = 0 V V vs (c) = 0 Figure 4. (a) Volage applied o he -circui as a funcion of ime. (b) Poenial difference V across he capacior as a funcion of ime. (c) Poenial difference V across he resisor as a funcion of ime. Noe ha a all imes V + V =. Now ha we have a good record of wha happens when he swich is hrown o posiion a and he capacior charges, le s hrow he swich o posiion b and le he capacior discharge. Once again, he sum of he volages around he loop mus be zero, hence (18) V + V = 0 where: (19) = O, here is no applied volage V = i = he volage drop across he resisor V = q / c = he volage drop across he capacior When (19) is insered ino equaion (18) we obain (20) q / c = i 8M 4

5 (21) Use he relaionship beween i and q [equaion (4)] o obain Separae he variables q and o obain q dq = d d (22) = dq q Seing equaion (22) up for inegraion and making use of he fac ha a = 0, q = Q max = as he lower limi and a some ime laer he charge on he capacior is q as he upper limi, obain (23) (1/) I d = I dq / q Inegrae equaion (23) o obain (24) (1/) = lnq Inser he upper and lower limis ino equaion (24) o obain (25) (1/) = lnq lnq max = ln (q / Q max ) Take he anilog of equaion (25) o obain (26) e / = (q / Q max ) 0 Solving equaion (26) for q, obain he following expression for he charge on he discharging capacior as a funcion of ime. / (27) q = Q max e If he charge of he capacior a any ime as i discharges is given by equaion (27) he poenial difference across he capacior a any ime as i discharges is (28) V = q / = (Q max / )e / / = e Looking a equaions (27) and (28) we see ha a = 0 he charge on he discharging capacior has is maximum value Q max and he poenial difference across he capacior has a maximum value of. Afer some ime >>, here is no charge on he capacior, he poenial difference across he capacior is zero, and he capacior is oally discharged. A plo of V as a funcion of ime is shown in Figure 5. 0 q Q max q Q max V V vs Figure 5. Plo of he poenial difference V across he capacior in a -circui as a funcion of ime while he capacior is discharging. As he capacior discharges, he curren a any ime obained from equaion (27) is as follows: (29) i = dq / d = d(q max e / ) / d = (Q max / )e / = ( / )e / / = I max e 8M 5

6 If he curren in he circui of he discharging capacior a any ime is given by (29). he poenial difference across he resisor a any ime is (30) V = i = I max e / / = e where: (31) = I max A plo of V as a funcion of ime is shown in Figure 6. V V vs Figure 6. Plo of he poenial difference V across he resisor in a -circui as a funcion of ime while he capacior is discharging. Figure 7 gives a summary of wha happens when he swich S is hrown o posiion b, he applied volage is disconneced, and he capacior is discharged. vs (a) V = 0 V vs (b) (c) = 0 = 0 V vs 8M 6 V Figure 7. (a) Volage applied o he -circui as a funcion of ime. (b) Poenial difference V across he capacior as a funcion of ime. (c) Poenial differencev across he resisor as a funcion of ime. Noe ha a all imes V +V =.

7 Now if he swich is flipped back and forh from a o b o a o b ec., he volage is alernaely applied and disconneced, and he capacior alernaely charges and discharges. All of he preceding informaion allows us o conclude ha a record of, V and V would look like Figure 8. vs (a) V = 0 V vs (b) = 0 V vs V (c) = 0 = 0 is applied Swich o a apacior charging is disconneced Swich o b apacior discharging is applied Swich o a apacior harging Figure 8. For a -circui, he quaniy is a characerisic ime called he capaciaive ime consan. The ime consan is he ime i akes he capacior o charge o 63% of is maximum charge or alernaely he ime for a discharging capacior o lose 63% of is charge. This is shown below by using he ime = in he expression for he charge on a charging capacior [equaion (12)] and in he expression for he charge on a discharging capacior [equaion (27)]. For a charging capacior For a discharging capacior (12) q = Q max (1 e / / ) (27) q = Q max e Insering he ime =, obain: q = Q max (1 e / ) Insering he ime =, obain: / Q max e = Q max (1 e 1 ) = Q max e 1 = (0.63)Q max = (0.37)Q max 8M 7

8 From he above we see ha afer a ime = he charge on a capacior is 63% of is maximum value and he discharging capacior has a charge of 37% of is maximum value (ha is i has los 63% of is maximum charge). As a maer of convenience, he ime consan is represened by he symbol τ, hence (31) τ = Since he poenial difference across he capacior a any ime is V = q /, we may also say ha he ime consan is he ime for he poenial difference across a charging capacior o reach 63% of he applied volage () and/or i is he ime for he poenial difference across a discharging capacior o fall o 37% of he applied volage. Anoher unique ime for a circui is he ime for he charge on he capacior or he poenial difference across a charging capacior o rise o one half is maximum value and/or o fall o one half is maximum value for a discharging capacior. We will refer o his ime as T 1/2. For a charging capacior For a discharging capacior (14) V = (1 e / / ) (28) V = e 1 2 If = T 1/2 hen V =. Also, τ =. If = T 1/2 hen V =. Also, τ =. 1 2 Insering hese values, we obain: 1 2 = (1 e T 1/2 / τ ) Insering hese values, we obain: 1 2 = e T 1/2 / τ 1 2 e T 1/2 / τ = e T 1/2 / τ = 1 2 T 1/2 = 1n2 τ τ T 1/2 = 1n2 (32) τ = T 1/2 (32) τ = T 1/2 1n2 1n2 This gives us an alernae mehod for finding he ime consan. We simply go o he V = / 2 posiion on he V vs plo of a charging or discharging capacior and he corresponding ime is T 1/2. We can hen deermine he ime consan using equaion (32). PODU In his aciviy, a power amplifier is used o produce a low frequency posiive only square wave. When his wave form is applied o a series circui, i has he same effec as connecing and hen disconnecing a D volage source. When he volage source is conneced and hen disconneced, he capacior charges and hen discharges. We will observe he oupu of he power amplifier and use volage sensors o measure he volage across he resisor and capacior as he capacior charges and discharges. We will use a graphical display of hese volages (i.e., V and V ) o invesigae he behavior of he circui while he charge is increasing, seady a is maximum value, and decreasing. We will also deermine he ime consan for he -circui direcly and indirecly. 8M 8

9 Par I. Iniial quipmen and Sofware Se-up 1. Sar Daa Sudio, and selec reae xperimen. 2. onnec he Power Amplifier o he compuer via an analog por on he PASO inerface. Plug he Power Amplifier ino an A oule, and urn he power amplifier on. onnec he power amplifier o an circui consising of a 100Ω esisor and a 330µF apacior. 3. Inform he sofware which analog por you plugged he Power Amplifier ino by selecing he Power Amplifier icon and dragging i o he appropriae analog por. 4. A signal generaor box should appear. hange he wave paern from he defaul Sine Wave o a Posiive Square Wave. Noe ha a Posiive Square Wave and a Square Wave are wo differen ypes of wave forms. Since we are using he Signal generaor as our swich, he wave form needs o be a Posiive Square Wave so ha he volage will alernae beween 0.00 and 4.00 vols. If a normal Square Wave is used, he volage will alernae beween and 4.00 vols. hange he frequency o 0.40Hz. hange he ampliude o 4.00V. The Auo buon should already be seleced. 5. onnec a volage sensor across he 100Ω resisor, and connec i o he compuer via an analog por on he PASO inerface. 6. Inform he sofware which analog por you plugged he volage sensor ino by selecing he volage sensor icon and dragging i o he appropriae analog por. 7. onnec a volage sensor across he 330µF capacior, and connec i o he compuer via an analog por on he PASO inerface. 8. Inform he sofware which analog por you plugged he volage sensor ino by selecing he volage sensor icon and dragging i o he appropriae analog por. 9. lick he Sar/Sop Opions buon, and selec he Auomaic Sop ab. Selec he ime opion and inpu 6 seconds. lick OK. 10. reae a graph of he Volage across he apacior vs. ime. 11. Double click somewhere wihin he body of he acual graph, or click on he Graph Seings buon locaed on he graph oolbar o open he Graph Seings window. Selec he layou ab, and under Group Measuremens selec he Do No Group opion. Nex selec he Tools ab. Under he Smar Tools, se he Daa Poin Graviy o 0. Selec OK. 8M 9

10 Par II. ollecion of Daa and Analysis of he Time onsan 1. lick he Sar buon. The sofware will collec daa for 6 seconds and auomaically sop. 2. lick he auoscale buon locaed on he graph ool bar. 3. Maximize he graph o fill he screen. 4. Magnify a secion of he graph where he capacior is charging. Srech he ime scale if necessary. 5. Deermine he amoun of ime i akes o reach one half of he maximum volage using he smar ool. lick on he Smar Tool buon. When he smar ool curser has he following appearance, i can be moved o any locaion on he graph. Place he smar curser a he locaion where he capacior begins o charge. By moving he mouse slighly o he 2 nd quadran of he smar ool curser, you should be able o change he curser o he dela curser. When he dela curser is presen, lef click and drag he curser o he poin where he volage is one half of he maximum. The difference in ime beween hese poins is he ime a half-max, or T 1/2. 6. lick he auoscale buon on he graph. 7. Magnify a secion of he graph where he capacior is discharging. 8. Deermine he amoun of ime i akes o reach one half of he maximum volage using he smar ool in a similar manner as sep lick on he smar ool buon so ha he smar ool is off. Par III. ompleion of Sofware Se-up and Daa Analysis 1. Add a graph of he Volage Across he esisor vs. Time o he exising graph. lick and drag he icon ha represens he Volage Across he esisor o he exising graph. When he enire graph is boxed in a doed line, drop he icon. 2. epea sep one for he Oupu Volage of he signal generaor. 3. Make sure ha he align x-axis lock buon has been seleced. 4. You should now have hree graphs ha look like Figure 8 in he Foreword of he experimen. However, he graphs will be in a differen order. 5. lick on each graph one a a ime and urn on he smar ool. 6. Make he Volage Across he apacior vs. ime graph acive by clicking on i. 8M 10

11 7. lick he auoscale buon on he graph. 8. Magnify a secion of he graph in an area where he volage of he apacior is increasing. 9. Pick a poin where he volage of he apacior is increasing and deermine he volage of he apacior using he smar ool. Also deermine he volage of he esisor and Oupu Volage for he same ime. Prin he graph wih he smar ool daa visible. ommen on he resul. 10. epea seps 6 and 9 for a poin where he volage of he apacior is remaining sable. Prin he graph wih he smar ool daa visible. Make sure you commen on he resuls. 11. epea seps 6 and 9 for a poin where he volage of he apacior is decreasing. Prin he graph wih he smar ool daa visible. Make sure you commen on he resuls. 8M 11

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13 NAM STION DAT DATA AND ALULATION SUMMAY harging apacior T sar = s Time when he volage across he charging capacior sars o increase. T half max = s Time when he volage across he charging capacior reaches he value V = / 2. T 1/2 = s Time for he volage across he charging capacior o rise o half is maximum value. (T 1/2 = T half max T sar ) τ = s Time consan for he circui. (τ = T 1/ 2 / In 2) = Ω esisance of he resisor. = F apaciance of he capacior. τ = s Time consan for he circui. (τ = ) Discharging apacior T sar = s Time when he volage across he discharging capaciy sars o decrease. T half max = s Time when he volage across he discharging capacior reaches he value V = / 2. T I/2 = s Time for he volage across he discharging capacior o fall o half is maximum value. (T 1/2 = T half max T sar ) τ = s Time consan for he circui. (τ = T 1/ 2 / In 2) 8M 13

14 Volages and commens for a ime when he volage across he capacior is increasing: = vols V vols V = vols Volages and commens for a ime when he volage across he capacior is sable: = vols V vols V = vols Volages and commens for a ime when he volage across he capacior is decreasing: = vols V vols V = vols Quesions 1. The ime o half-maximum volage is how long i akes he capacior o charge half-way. Based on your experimenal resuls, how long does i ake for he capacior o charge o 75% of is maximum? 2. Afer four "half-lives", o wha percenage of he maximum charge is he capacior charged? 3. Wha is he maximum charge for he capacior in his experimen? 4. Wha is he maximum curren flowing hrough he resisor in his experimen? 8M 14