Chapter 28 - Circuits

Transcription

1 Physics 4B Lecure Noes Chaper 28 - Circuis Problem Se #7 - due: Ch 28 -, 9, 4, 7, 23, 38, 47, 53, 57, 66, 70, 75 Lecure Ouline. Kirchoff's ules 2. esisors in Series 3. esisors in Parallel 4. More Complex Circuis 5. Elecrical Meers 6. C Circuis n a circui, charges move from one place o anoher carrying energy. These charges can be hough of as buckes ha carry energy around a circui. The baery fills he buckes. The buckes are empied a various places around he circui, bu he buckes hemselves never disappear. They reurn o he baery o be refilled. These basic ideas are summarized in Kirchoff's ules and are applicable o even he mos complicaed circuis.. Kirchoff's ules The Juncion Theorem: "The curren ino any juncion is exacly eual o he curren ou of he juncion." This heorem is explained by he Law of Conservaion of Charge. The Loop Theorem: "The sum of all he volage drops around any loop in a circui mus be zero." This heorem is explained by The Law of Conservaion of Energy and he fac ha he elecric force is conservaive. 2. esisors in Series 2 N The loop heorem reuires: 2 L N = 0 = + 2 +L+ N Ohm's Law says: =, =, 2 = 2 2, L N = N N = L+ N N The juncion heorem means ha: = = 2 =L= N = + 2 +L+ N = + 2 +L+ N s = i esisors in Series 28-

2 Physics 4B Lecure Noes 3. esisors in Parallel 2 2 N N The juncion heorem means ha: = + 2 +L+ N Ohm's Law says: =, =, 2 = 2 2, L N = N N = L+ N N The loop heorem reuires: = 0, 2 = 0, L N = 0 = = 2 =L= N So, = + 2 +L+ N = + 2 +L+ N 4. More Complex Circuis p = i esisors in Parallel ) Circui elemens in series have he same curren, bu divide up a common volage. 2) Circui elemens in parallel have he same volage, bu divide up a common curren. Many resisor circuis are jus combinaions of series and parallel. They can be sudied using he series and parallel rules. This is called "circui reducion." Oher circuis are no combinaions of series and parallel. These circuis have o be examined wih Kirchoff's ules. Example : For he circui shown find (a)he euivalen resisance, (b)he curren provided by he baery and (c)he curren hrough and volage across each resisor (=90.0, =3.00kΩ, 2 =4.00kΩ, 3 =.00kΩ, 4 =2.00kΩ and 5 =6.00kΩ) (a)use he idea of circui reducion. 3 and 4 are in series. They can be replaced wih an euivalen resisor, s = = 3.00kΩ. 2 5 s s and 5 are in parallel. They can be replaced wih an euivalen resisor, p = s + 5 p = 2.00kΩ. 2 p 28-2

3 p and 2 are in series. They can be replaced wih an euivalen resisor, s = p + 2 = 6.00kΩ. Physics 4B Lecure Noes s and are in parallel. The euivalen resisor is, = s + = 2.00kΩ. (b)use Ohm's ule = = = kΩ = 45.0mA (c)follow he curren and volage drops around he circui using Kirchoff's ules and Ohm's ule. n summary, () (ma) (kω) Example 2: A 6.00 baery ( ) wih a 2.00Ω inernal resisance and a 3.00 baery wih a 3.00Ω inernal resisance is conneced o a 6.00Ω resisor as shown. Find he erminal volage for each baery and he curren hrough he 6.00Ω resisor. This circui is no a combinaion of series and parallel resisor, so we mus go back o he more basic ideas of Kirchoff's ules o solve he problem. Applying he juncion heorem a poin A i 2 = i + i. The loop heorem around he lower loop 2 i i 2 r 2 = 0 and around he upper loop + i i r = 0. This gives hree euaions for he hree unknown currens. Solving he loop heorem euaions for i and i 2, i 2 = 2 i and i r = + i. 2 r Subsiuing ino he juncion heorem euaion, 2 i = i + + i. r 2 r Solving for i, i = 2 r 2 r + r 2 + r 2 r = 0.333A he minus sign means ha we chose he direcion of his curren wrong. Subsiuing back for i and i 2, i = 2.00A and i 2 =.67A. s The erminal volage is he acual poenial difference across he erminals of he baery, = i r = 2.00 and 2 = 2 i 2 r 2 = 2.00 i r A i i 2 2 r

4 5. Elecrical Meers Physics 4B Lecure Noes The basic consiuen par of a volmeer, ammeer, or ohmmeer is a galvanomeer. Mechanical galvanomeers are made from a coil of wire and a magne. Modern galvanomeers are C chips. They all funcion in he same basic way. They respond linearly o small currens. The differen ypes of meers are consruced from a galvanomeer and properly placed resisors. The schemaic symbols for differen ypes of meers are shown a he righ. G G GA GΩ Galvanomeer olmeer Ammeer Ohmeer Example 3: A galvanomeer of inernal resisance 0.20Ω reads full scale when 5.0µA passes hrough i. Design an ammeer o read up o.00a wih his galvanomeer. f he enire.00a goes hrough he galvanomeer i will blow up. We mus provide an alernae pah for he curren. We can conrol he fracion of he curren ha goes hrough he galvanomeer wih a resisor called a "shun." Using he juncion heorem, = i G + i. The loop heorem reuires i G r i = 0. Solving for i and subsiuing ino he juncion heorem euaion, i = r i G = i G + r r ( ) = =.80µΩ. i G i i G r G Example 4: Use an idenical galvanomeer o build a volmeer o measure up o 0.0. f he enire 0.0 is across he galvanomeer he curren will be huge and i will cook. We mus cu down he curren ha goes hrough he galvanomeer wih a resisor in series. Using he loop heorem, = i G r + i G. Solving for, = i G r = 6.67x0 5 Ω. i G r G 0.0 Example 5: The ammeer from example 3 and he volmeer from example 4 are used o measure he resisance in he circui shown. Find he difference beween he raio of he meer readings and he rue resisance. Using he loop heorem, = i ( + A ) where is he volmeer reading and A is he resisance of he ammeer which can be found from he parallel rule, A = r + shun A = r shun r + shun shun =.80µΩ. Using as he raio of he meer readings he euaion from he loop heorem can be wrien, i i = i ( + A ) = + A = + shun. The difference beween he meer raio and he acual resisance is, = shun =.80µΩ. ε i i A i 28-4

5 Physics 4B Lecure Noes 6. C Circuis Kirchoff's ules are very widely applicable. This is no surprising considering hey come from he Laws of Conservaion of Energy and Conservaion of Charge. They can be used o analyze circuis wih boh resisors and capaciors. A ypical C circui is shown a he righ. When he swich connecs a and b curren flows and he baery begins o charge he capacior. When he capacior is charged curren can no longer flow. The uesion is, how long does his ake? A some inermediae ime he curren in he circui is i and he charge on he capacior is. Applying he loop heorem, i c = 0. From he definiion of curren, he C curren mus eual he rae a which he capacior charges, i = d. The euaion from he loop heorem d becomes, d d c d = 0. This euaion can be solved for () by solving for and inegraing, C d d d = C c C d C = d c C d d C du = 0 C 0 c C = C ln C u c C C = c C Solving for, = C e C Charging C Circui c a C c b d The graph of he charge on he capacior as a funcion of ime is shown. Noe: )The charge is iniially zero. 2)The charge grows exponenially. 3)The maximum charge occurs when he capacior volage maches he baery volage. =C When he swich connecs b and c he capacior discharges. The loop heorem reuires, i d + C = 0. The curren mus eual he rae a which he capacior discharges, i = d d. The euaion from he loop heorem becomes, d d = negraing, d = d d C C o d = 0 d d C d C. ln = C o d C = C o e d C = C o e C Discharging C Circui 28-5

6 Physics 4B Lecure Noes The graph of he charge on he capacior as a funcion of ime is shown. =C o Noe: )The charge is iniially he capaciance imes he iniial volage on he capacior. 2)The charge dies exponenially. Example 6: A 5.00µF capacior is charged o 0.0. A 0.0cm piece of 2.00mm diameer copper wire is used o shor i ou. Find he ime i akes for he capacior's volage o drop o 0.0m. For he discharge of a capacior, = C o e C C = o e C = o e C. Solving for he ime, = e C ln = o o C = Cln. o The resisance can be found from is definiion, = ρ l A = 0.00 (.7x0 8 ) π(0.0000) 2 = 5.4x0 4 Ω. Puing he numbers in, = Cln = (5.4x0 4 )(5.00x0 6 )ln o 0.0 =.9x0 8 s = 9ns. Chaper 28 - Summary Kirchoff's ules: The Juncion Theorem: "The curren ino any juncion is exacly eual o he curren ou of he juncion." The Loop Theorem: "The sum of all he volage drops around any loop in a circui mus be zero." esisors in Series s = esisors in Parallel p = i i Charging C Circui = C e C Discharging C Circui = C o e C 28-6