Direc Curren Circuis February 19, 2014 Physics for Scieniss & Engineers 2, Chaper 26 1
Ammeers and Volmeers! A device used o measure curren is called an ammeer! A device used o measure poenial difference is called a volmeer! To measure he curren, he ammeer mus be placed in he circui in series! To measure he poenial difference, he volmeer mus be wired in parallel wih he componen across which he poenial difference is o be measured Volmeer in parallel High resisance Ammeer in series Low resisance February 19, 2014 Physics for Scieniss & Engineers 2, Chaper 26 2
Real-life Volmeer in a Simple Circui! Consider a simple circui consising of a source of emf wih poenial difference V emf = 150. V and a resisor wih resisance R = 100 kω.! A volmeer wih resisance R V = 10.0 MΩ is conneced across he resisor. PROBLEM 1:! Wha is he curren in he circui before he volmeer is conneced? SOLUTION 1:! Ohm s Law allows us o calculae he curren before: i = V emf R = 150. V 100. 10 3 Ω =1.50 10 3 A =1.50 ma February 19, 2014 Chaper 26 6
Real-life Volmeer in a Simple Circui PROBLEM 2:! Wha is he curren in he circui afer he volmeer is conneced? SOLUTION 2:! The equivalen resisance of he resisor and he volmeer conneced in parallel is given by: 1 R eq = 1 R + 1 R V! The equivalen resisance is: R eq = ( )( 10.0 10 6 Ω) RR V = 100. 10 3 Ω R + R V 100. 10 3 Ω+10.0 10 6 Ω = 9.90 104 Ω = 99.0 kω February 19, 2014 Chaper 26 7
Real-life Volmeer in a Simple Circui! The curren afer connecing he volmeer is: i = V emf R = 150. V 99.0 10 4 Ω =1.52 10 3 A =1.52 ma! So he curren in he circui increases by 0.02 ma, when he volmeer is conneced because he parallel combinaion of he resisor and he volmeer has a lower resisance han ha of he resisor alone.! The effec is small.! The larger he resisance of he volmeer, he beer. February 19, 2014 Chaper 26 8
Circuis! So far we have deal wih circuis conaining sources of emf and resisors! The currens in hese circuis did no vary in ime! Now we will sudy circuis ha conain capaciors as well as sources of emf and resisors! These circuis have currens ha vary wih ime! Consider a circui wih a source of emf, V emf, a resisor R, a capacior C February 19, 2014 Physics for Scieniss & Engineers 2, Chaper 26 9
Charging a Capacior! We sar wih he capacior uncharged! We close he swich and curren begins o flow, building up opposie charges on he plaes of he capacior and creaing a poenial difference across he capacior! When he capacior is fully charged, no more curren flows! The poenial difference across he plaes is equal o he poenial difference provided by he source of emf! The magniude of he oal charge on each plae is q o = CV emf February 19, 2014 Physics for Scieniss & Engineers 2, Chaper 26 10
Charging a Capacior! While he capacior is charging, we can analyze he he curren flowing by applying Kirchhoff s Loop Rule ( ) V emf V R V C = V emf i( )R q C = 0! The change of he charge on he capacior is he curren i( ) = dq( ) d! We can rewrie our equaion as ( ) ( ) R dq + q d C = V emf or dq( ) + q( ) d = V emf R dq Differenial equaion relaing q and d The erm V c is negaive since he op plae of he capacior is conneced o he posiive - higher poenial - erminal of he baery. Thus analyzing couner-clockwise leads o a drop in volage across he capacior! February 19, 2014 Physics for Scieniss & Engineers 2, Chaper 26 11
Charging a Capacior! The soluion for his differenial equaion is ( ) = q max 1 e q! The consan τ is he ime consan given by τ =! The consan q max is given by q max = CV emf τ! So we can wrie he soluion o our differenial equaion ( ) = CV emf 1 e q Mah Reminder: February 19, 2014 Physics for Scieniss & Engineers 2, Chaper 26 12
! A = 0 we have ( ) = CV emf 1 e q 0! A = we have q Charging a Capacior ( ) = CV emf 1 e! The curren flowing hrough he circui is he ime derivaive of he charge ( ) i( )= dq d = V emf R! A = 0 we have i( 0) = V emf R 0! A = we have i( ) = V emf R e e = 0 = 0 = CV emf February 19, 2014 Physics for Scieniss & Engineers 2, Chaper 26 13
Charging a Capacior February 19, 2014 Physics for Scieniss & Engineers 2, Chaper 26 14
Discharging a Capacior! Now we consider a circui conaining only one resisor and a fully charged capacior! The charge on he capacior is q max! Now we shor he resisor across he capacior by moving he swich from posiion 1 o posiion 2! Curren flows unil he capacior is compleely discharged! While he capacior is discharging, Kirchhoff s Loop Rule gives us ( ) i( )R V C = i( )R q C = 0 February 19, 2014 Physics for Scieniss & Engineers 2, Chaper 26 15
Discharging a Capacior! Using he definiion of curren, we can rewrie our equaion as ( ) ( ) R dq d + q C = 0! The soluion o his differenial equaion is ( ) = q max e q! A = 0 we have q( 0) = q max e! A = we have ( ) = q max e q 0 = q max = 0 February 19, 2014 Physics for Scieniss & Engineers 2, Chaper 26 16
Discharging a Capacior! The curren flowing hrough he circui is he ime derivaive of he charge ( ) = dq( ) i d ( ) = q max! A = 0 we have ( ) = q max i 0! A = we have i ( ) ( ) ( ) = q max e e 0 e ( ) = q max = 0 February 19, 2014 Physics for Scieniss & Engineers 2, Chaper 26 17
Discharging a Capacior February 19, 2014 Physics for Scieniss & Engineers 2, Chaper 26 18
Time Required o charge a Capacior! Consider a circui consising of a 12.0 V baery, a 50.0 Ω resisor, and a 100.0 μf capacior wired in series! The capacior is iniially compleely discharged! PROBLEM! How long afer he circui is closed will i ake o charge he capacior o 90% of is maximum charge? SOLUTION! The charge on he capacior as a funcion of ime is given by ( ) = q max 1 e q February 19, 2014 Physics for Scieniss & Engineers 2, Chaper 26 20
Time Required o charge a Capacior! We wan o know he ime unil q()/q max = 0.90 1 e ( ) = q = 0.90 0.10=e q max! Taking he naural log of boh sides gives us ln( 0.10)=ln e! So he ime o discharge he capacior o 90% of is maximum charge is February 19, 2014 Physics for Scieniss & Engineers 2, Chaper 26 21 ln ( 0.10 )= = ln( 0.10) = ( 50.0 Ω) ( 100.0 10 6 F) ( 2.30) = 0.0115 s = 11.5 ms Mah Reminder: ln(e x )=x
! Charging a capacior:! Discharging a capacior: q = q e 0! Time consan: i Summary: Circui q () = q0 1 e τ i V emf dq = = e d R dq d q 0 = = τ = e Physics for Scieniss & Engineers 2 22
Rae of Energy Sorage in a Capacior! A resisor wih R = 2.50 MΩ and a capacior wih C = 1.25 μf are conneced in series wih a baery for which V emf = 12.0 V! A = 2.50 s afer he circui is closed, wha is he rae a which energy is being sored in he capacior? SOLUTION THINK! When he circui is closed, he capacior begins o charge! The rae a which energy is sored in he capacior is given by he ime derivaive of he amoun of energy sored in he capacior, which is a funcion of he charge on he capacior February 19, 2014 Physics for Scieniss & Engineers 2, Chaper 25 23
Rae of Energy Sorage in a Capacior SKETCH! The circui conains a baery, a resisor, and a capacior in series. February 19, 2014 Physics for Scieniss & Engineers 2, Chaper 25 24
Rae of Energy Sorage in a RESEAH Capacior! The charge on he capacior as a funcion of ime is ( ) = CV emf 1 e q! The energy sored in a capacior wih charge q is U = 1 2 q 2 C! The ime derivaive of he sored energy is du 1 q 2 ( ) dq( ) d = d d 2 C = q C d! The ime derivaive of he charge is he curren i( ) = dq( ) = V emf d R e February 19, 2014 Physics for Scieniss & Engineers 2, Chaper 25 25
Rae of Energy Sorage in a SIMPLIFY Capacior! Combining our equaions ( ) du d = q C i du d = V 2 emf R e CV ( ) emf 1 e = CALCULATE! We firs calculae he ime consan 1 e C V emf R e = ( 2.50 10 6 Ω) ( 1.25 10 6 F) = 3.125 s February 19, 2014 Physics for Scieniss & Engineers 2, Chaper 25 26
Rae of Energy Sorage in a Capacior! Now we calculae he rae of change of energy sored in he capacior du d = ( 12.0 V)2 2.50 s 2.50 10 6 Ω e 3.125 s 1 e 2.50 s 3.125 s = 1.425 10 5 W DOUBLE-CHECK (Energy Conservaion)! The curren a = 2.50 is i( ) = 12.0 V 2.50 10 6 Ω e 2.50 s 3.125 s = 2.16 10 6 A February 19, 2014 Physics for Scieniss & Engineers 2, Chaper 25 27
Rae of Energy Sorage in a Capacior! The rae of energy dissipaion a = 2.50 is P = du d = i2 R = 2.16 10 6 A ( ) 2 2.50 10 6 Ω ( ) = 1.16 10 5 W! The rae a which he baery delivers energy o he circui a = 2.50 s is P = du d = iv emf = 2.16 10 6 A ( ) 12.0 V ( ) = 2.59 10 5 W! The energy supplied by he baery is equal o he sum of he energy dissipaed in he resisor plus he energy sored in he capacior February 19, 2014 Physics for Scieniss & Engineers 2, Chaper 25 28