RC Circuits RC 2 RC

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Justina Griffin
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1 ircuis a b a b q ( / ) q = 1 e q q = e /

2 esisor-capacior circuis Le s add a apacior o our simple circui ecall volage drop on? V = Wrie KVL: = d Use = Now eqn. has only : d KVL gives Differenial Equaion! d d = We will solve his laer. For now, look a qualiaive behavior

3 apaciors ircuis, ualiaive Basic principle: apacior resiss change in resiss changes in V harging (i akes ime o pu he final charge on) niially, he capacior behaves like a wire (ΔV =, since = ). As curren coninues o flow, charge builds up on he capacior i hen becomes more difficul o add more charge he curren slows down Afer a long ime, he capacior behaves like an open swich. Discharging niially, he capacior behaves like a baery. Afer a long ime, he capacior behaves like a wire.

4 UL1, AT 2 2A A = he swich is hrown from posiion b o posiion a in he circui shown: The capacior is iniially uncharged. Wha is he value of he curren + jus afer he swich is hrown? (a) + = (b) + = /2 (c) + = 2 / a b 2B Wha is he value of he curren afer a very long ime? (a) = (b) = /2 (c) > 2 /

5 UL1, AT 2 2A A = he swich is hrown from posiion b o posiion a in he circui shown: The capacior is iniially uncharged. Wha is he value of he curren + jus afer he swich is hrown? (a) + = (b) + = /2 (c) + = 2 / a b Jus afer he swich is hrown, he capacior sill has no charge, herefore he volage drop across he capacior =! Applying KVL o he loop a =+, = = /2

6 2A UL1, AT 2 A = he swich is hrown from posiion b o posiion a in he circui shown: The capacior is iniially uncharged. Wha is he value of he curren + jus afer he swich is hrown? (a) + = (b) + = /2 (c) + = 2 / a b 2B Wha is he value of he curren afer a very long ime? (a) = (b) = /2 (c) > 2 / The key here is o realize ha as he curren coninues o flow, he charge on he capacior coninues o grow. As he charge on he capacior coninues o grow, he volage across he capacior will increase. The volage across he capacior is limied o ; he curren goes o.

7 U11Pf11: The capacior is iniially uncharged, and he wo swiches are open. E 3) Wha is he volage across he capacior immediaely afer swich S 1 is closed? a) V c = b) V c = E c) V c = 1/2 E 4) Find he volage across he capacior afer he swich has been closed for a very long ime. a) V c = b) V c = E c) V c = 1/2 E niially: = V = = E/(2) = E =

8 U11Pf11: E 6) Afer being closed a long ime, swich 1 is opened and swich 2 is closed. Wha is he curren hrough he righ resisor immediaely afer he swich 2 is closed? a) = b) =E/(3) c) =E/(2) d) =E/ Now, he baery and he resisor 2 are disconneced from he circui, so we have a differen circui. Since is fully charged, V = E. niially, acs like a baery, and = V /.

9 ircuis (Time-varying currens, charging) harge capacior: iniially uncharged; connec swich o a a = alculae curren and charge as funcion of ime. a b Loop heorem = onver o differenial equaion for : Would i maer where is placed in he loop?? = d d d = + d

10 harging apacior harge capacior: d = + d Guess soluion: a b = (1 e ) heck ha i is a soluion: d d d d + = e / 1 / = + e + (1 e ) =! Noe ha his guess fis he boundary condiions: = = = =

11 harging apacior harge capacior: a = ( / 1 e ) urren is found from differeniaion: b d = = / e onclusion: d apacior reaches is final charge(= ) exponenially wih ime consan τ =. urren decays from max (= /) wih same ime consan.

12 harging apacior harge on ( / ) = 1 e Max = 63% Max a = 2 d = = d Max = / urren e / 37% Max a = /

13 Discharging apacior a d d + = Guess soluion: b = e e = / τ / heck ha i is a soluion: d d = / 1 e e + = / + e / =! d d Noe ha his guess fis he boundary condiions: = = = =

14 Discharging apacior Discharge capacior: a urren is found from differeniaion: = e = e d = = e d / τ / / Minus sign: urren is opposie o original definiion, i.e., charges flow away from capacior. b onclusion: apacior discharges exponenially wih ime consan τ = urren decays from iniial max value (= -/) wih same ime consan

15 harge on Discharging apacior 2 = / e Max = 37% Max a = zero urren d = = d Max = -/ e / 37% Max a = - /

16 U11Pf 11: The wo circuis shown below conain idenical fully charged capaciors a =. ircui 2 has wice as much resisance as circui 1. 8) ompare he charge on he wo capaciors a shor ime afer = a) 1 > 2 b) 1 = 2 c) 1 < 2 niially, he charges on he wo capaciors are he same. Bu he wo circuis have differen ime consans: τ 1 = and τ 2 = 2. Since τ 2 > τ 1 i akes circui 2 longer o discharge is capacior. Therefore, a any given ime, he charge on capacior 2 is bigger han ha on capacior 1.

17 Prefligh 11: The circui below conains a baery, a swich, a capacior and wo resisors 1) Find he curren hrough 1 afer he swich has been closed for a long ime. a) 1 = b) 1 = E/ 1 c) 1 = E/( ) Afer he swich is closed for a long ime.. The capacior will be fully charged, and 3 =. (The capacior acs like an open swich). So, 1 = 2, and we have a one-loop circui wih wo resisors in series, hence 1 = E/( )

18 harging Discharging 2 2 ( / 1 ) = e = / e / d = = d e / d = = e d / - /

19 Power disribuion sysems emark; houses ypically have wo single phase, 12v ho lines onvenional House wiring has 24/12V lines Which prong is ho??

20 Elecriciy oss Your elecric coss are charged in unis of energy, Kilowa-hour 1 Kilowa-Hour = 1 was x 36 sec = 3.6 million joules Oahu residenial charges are abou 2 cens per K-H. This can be read on your A Kilowa-Hour meer ouside your house. Elecriciy coss in Hawaii are high! Example 1W ligh bulb run for 1hrs/day for 3 days uses.1 KW x 1 x 3 = 3 Kilowa-Hours and coss $6..

RC (Resistor-Capacitor) Circuits. AP Physics C

RC (Resistor-Capacitor) Circuits. AP Physics C (Resisor-Capacior Circuis AP Physics C Circui Iniial Condiions An circui is one where you have a capacior and resisor in he same circui. Suppose we have he following circui: Iniially, he capacior is UNCHARGED