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Nobel Prize in Physics, 2011 The Nobel Prize in Physics 2011 has been awarded to Saul Perlmutter, Brian P Schmidt and Adam G Riess for discovering the accelerating expansion of the universe Nobel Prize in Physics, 2010 "for groundbreaking experiments regarding the two-dimensional material graphene" Since it is practically transparent and a good conductor, graphene is suitable for producing transparent touch screens, light panels, and maybe even solar cells. Graphene transistors are predicted to be substantially faster than today’s silicon transistors and result in more efficient computers. Andre Geim Konstantin Novoselov Geim and Novoselov extracted the graphene from a piece of graphit such as is found in ordinary pencils. Using regular adhesive tape they managed to obtain a flake of carbon with a thickness of just one atom. Lectures 10-11 (Ch. 26) R-C circuits I. Steady state regime (R and C in series, R and C in parallel) II. Transient regimes 1.Charging the capacitor a) b) initial conditions temporal dynamics 2.Discharging the capacitor a) b) initial conditions temporal dynamics III. Applications Steady state regime (t→∞) NB: In steady state regime current does not go through the capacitor: I=i(t→∞)=0! =0! =0! 1.C and R are in series Switch has been closed for a long time. I=i(t→∞)=0 →Vab=IR=0, Vbc =ε, q(t →∞) = Q=εC (capacitor is fully charged) R1 2. C and R are in parallel I1 a b I1 I=0 R2 Switch has been closed for a long time. c I=i(t→∞)=0 but I1≠0 ! I1= ε/(R1+R2) Vab= εR1 /(R1+R2) Vbc =ε R2 /(R1+R2) q(t →∞) = Q=CVbc =CεR2 /(R1+R2) (capacitor is fully charged) Example. (Problem 26.74) 1. Switch is open. In the steady state regime find: a)voltage at the points a and b; b) charge on each capacitor. Current does no go through the capacitor in the steady state regime. Hence there is no current in this circuit. a) Vb=0x3Ω=0, Va=18V-0x3Ω=18V b) Q1=3μFx18V=54 μC Q2=6μFx18V=108 μC V=0 2. Switch is closed. In the steady state regime find: a)voltage at the points a and b; b) charge on each capacitor. Current does not go through the capacitors but it does go through 6Ω and 3Ω resistors. I=18V/9Ω=2A , Va=Vb=2Ax3Ω=6V Q1=3μFx6V=18 μC Q2=6μFx12V=72 μC Example. 1. Switch is open. In the steady state regime find: a) voltage at the points a and b; b) charge on each capacitor. Current does not go through the capacitors but it does go through 14Ω and 10Ω resistors. V=24V 14Ω a 10Ω Q1 S 2 μF b Q2 2 μF I=24V/24Ω=1A a) Va=1Ax10Ω=10V Vb=24V/2=12V b) Q1=Q2=Q=24VxCeq=24Vx1μF=24μC V=0 2. Switch is closed. In the steady state regime find: a) voltage at the points a and b; b) charge on each capacitor. Current is the same as in the previous part: I=24V/24Ω=1A Va=1Ax10Ω=10V (also the same) Vb=Va=10V Q2=10Vx2μF=20μC; Q1=14Vx2μF=28μC Example. Switch is closed at t=0. Find the final charge on the capacitor at t→∞. i1+i2 S 12V i2 10 Ω i1 1Ω a 1.5μF 9Ω b 5Ω o Q CVab , Vab Vao Vbo i1 9 i2 5 12V 12V i1 1.2 A, i2 0.8 A 10 15 Vab 10.8V 4V 6.8V Q 10.2C Charging the capacitor I.C and R in series 1. Initial conditions Switch is closed at t=0. Capacitor is initially uncharged: q( t=0)= 0→ Vbc(t=0)=q( t=0)/C=0 → Vab (t= 0)=ε → i(t=0)=ε /R Current appears at t=0. at t=0 Example. All capacitors are initially uncharged. Switch is closed at t=0. a)Find Vab(t=0); b) Find Vab(t→∞). a S 12V a)t 0 : Vac 0 (becauseq 0), 3μF c 5Ω b 1Ω 4μF Vab Vac Vcb 12V Vcb 5 10V Vab 10V 6 Q b)t : Vcb 0 (becausei 0),Vac 3F 2μF Q 12VC eq , 1 1 1 , Ceq 3F 6F 24C Ceq 2F , Vac 8V 3F Example. Problem 26.72 . The capacitor is initially uncharged. The switch is closed at t=0. a) Immediately after the switch is closed, what is a current through each resistor? b) what are the final currents and final charge ? 8Ω i1+i2 S i2 3Ω 6Ω 42V i1 4μF 42V a)t 0 : Vcb 0 Req 10 i1 i2 4.2 A 10 i1 6 i2 3 i2 2i1 3i1 4.2 A i1 1.4 A, i2 2.8 A 42V b)t : i2 0, i1 i2 i1 3A 14 Q 18V 4F 72C 2. Transient regime iR q / C 0, i dq dt dq C q dq dt dt RC q C CR 0 0 q t q C dt q C ln e CR C CR C 0 t t t q Q (1 e ); Q C , CR QV CV [ ] [ ] 1s VI VA t t t dq Q C i e e i Ie ; I dt CR CR R Example. Verify an energy balance in the process of charging the capacitor. U bat Pout dt idt I 0 0 0 0 0 t e dt I U R PR dt R i 2 dt RI 2 e 0 2t RC R 2C RI 2 R 2 RC 2C dt 2 2 2R 2 Q 2 2C 2 2C UC 2C 2C 2 One half of the energy provided by the battery is stored in capacitor, another half is dissipated in the resistor. • Example. Which fraction of the maximum possible energy was stored in capacitor by the moment when current droped 3 times? t* t* I 1 i (t*) Ie e 3 3 t* 2 q(t*) Q(1 e ) Q 3 4U C max q 2 (t*) 4Q 2 U C (t*) 2C 9 2C 9 Example. Switch is closed at t=0. a)Find Vab(t*) if q(t*)=Q/3, where Q is the charge of the capacitor when t→∞; b) Find characteristic time constant of this circuit. a S 12V 3μF c 5Ω b 1Ω 4μF 2μF Vab Vac Vcb t t Q 2 q (t*) Q (1 e ) e 3 3 t 2 i (t*) Ie I 3 2 Q Vab I 5 3 3 3F Q 12VC eq , Ceq 2 F , Q 24C 12V I 2A 6 Charging the capacitor. 2. C and R in parallel 1. Initial conditions. Capacitor is initially uncharged. Switch is closed at t=0. Current i1 appears at t=0. q( t=0)= 0→ R1 i1+i2 R2 b Vbc(t=0)=q( t=0)/C=0 →i2 (t=0)=0 Vab(t=0)= ε → i1+i2 (t=0)=i1=ε /R1 a i1 c i2 Example. Capacitor is initially uncharged. At t=0 the switch is closed. Find a)the initial and the final current; b)the final charge. S 10V 20Ω c 0.5μF 40Ω b 10V a )t 0 : Vcb 0, i 0.5 A 20 10V t i 0.17 A 60 b)Q Vcb (t ) 0.5F 0.17 A 40 0.5F 3.3C 2. Transient regime. Charge increases, current, i1-i2 decreases till steady state regime considered above is established at t →∞ . q 1 q i1R1 0 i1 ( ) C R1 C q i1R1 i2 R2 0 i2 CR2 CReq qR1 dq q i1 i2 dt R1 CReq CReq R1 i1 R2 b i1-i2 1 1 1 R1 R2 Req Req R1 R2 R1 R2 t qR1 CReq dq dt t , ln( ) 0 qR1 CReq 0 CReq CReq CReq q t q Q(1 e ), Q CR2 , CReq R1 R2 a c i2 Discharging the capacitor. I.C and R in series 1.Initial conditions. Capacitor is initially charged and disconnected from the battery. Switch is closed at t=0. Current appears at t=0. q( t=0)= Q→ Vbc(t=0)=Q/C=0 → Vab (t= 0)=-Q/C → i(t=0)=-Q/CR at t=0 2.Transient regime iR q / C 0, i dq dt dq q dq dt dt RC q CR Q 0 q t t q t ln q Qe , CR Q t dq Q i e dt Example. Verify an energy balance in the process of discharging the capacitor. Q2 UC 2C U R PR dt R i 2 dt R 0 0 0 Q2 2 e 2t Q2R Q2 dt 2 2C Example. The switch was closed for a long time so that the capacitor was fully charged (Q=72 μ C, see solution of this part of the problem above).Then the switch is opened at t=0. a) Immediately after the switch is opened, what is a current through each resistor? b) what are the currents through each resistor when charge decreased twice? 8Ω S 42V 6Ω 3Ω 4μF a )t 0 : through8 , c urrent 0; 72C through6 and 3 : I 2A 4 F 9 b)q (t*) Qe i (t*) Ie t Q/2e t I / 2 1A t 1/ 2 Discharging the capacitor. 2. C and R in parallel 1. Initial conditions. Capacitor is initially charged and disconnected from the battery. Switch is closed at t=0. Current appears at t=0. q( t=0)= Q→ Vbc(t=0)=Q/C→i2 (t=0)=Q/CR2 Vba(t=0)= Q/C → i1 (t=0)=Q /CR1 2. Transient regime. Charge decreases , currents also decrease to zero. q dq q i1 R1 i2 R2 ; (i1 i2 ) C dt ReqC t 1 1 1 ; q Qe , ReqC Req R1 R2 t t t Qe Qe Qe i1 , i2 , i1 i2 CR1 CR2 CReq R1 i1 a R2 b i1+i2 c i2 Example. Switch was closed for a long time. Then it was opened at t=0. At which moment of time charge decreases up to 5% of its value at t=0. S 12V 10 Ω 1Ω 1.5μF 9Ω 5Ω t* q (t*) Qe 0.05Q e t* ln 20, CReq t* 0.05 1 1 1 Req 6.16 Req 11 14 Applications of R-C circuits 1. Sawtooth voltage (pacemakers, turn signals, windshield wipes, etc.) V Gas filled tube 2. Frequency filtration HF output,LF filtr (ώ<1/RC do not go through) ε(t) ~ LF output,HF filtr (ώ>1/RC) do not go through) T t 1. Example. You need to built a pacemaker with 72beats/min. You have capacitor with C=2.5μF and it is known that the discharge starts at v=0.45 ε. Which resistance you should select? V ε Gas filled tube T t T 1 60s 0.83s f 72 T T V (t ) (1 e ) 0.45 e 0.55 T 1 T 0.83s ln RC R 0.2M 6 0.55 ln 1.82 2.5 10 F ln 1.82