1. 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

2. 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.

3. 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)initial conditions b)temporal dynamics 2. Discharging the capacitor a)initial conditions b)temporal dynamics III. Applications

4. 2. C and R are in parallel Switch has been closed for a long time. R1R1 R2R2 I=i(t→∞)=0 but I 1 ≠0 ! I 1 = ε/(R 1+ R 2 ) V ab = εR 1 /(R 1+ R 2 ) V bc =ε R 2 /(R 1+ R 2 ) q(t →∞) = Q=CV bc =CεR 2 /(R 1+ R 2 ) (capacitor is fully charged) I1I1 I1I1 I=0 a bc Steady state regime (t→∞) =0! NB: In steady state regime current does not go through the capacitor: I=i(t→∞)=0! 1.C and R are in series Switch has been closed for a long time. I=i(t→∞)=0 →V ab =IR=0, V bc =ε, q(t →∞) = Q=εC (capacitor is fully charged)

5. 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. V=0 Current does no go through the capacitor in the steady state regime. Hence there is no current in this circuit. a) V b =0x3Ω=0, V a =18V-0x3Ω=18V b) Q 1 =3μFx18V=54 μC Q 2 =6μFx18V=108 μC 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, V a =V b =2Ax3Ω=6V Q 1 =3μFx6V=18 μC Q 2 =6μFx12V=72 μC

6. Example. 1.Switch is open. In the steady state regime find: a) voltage at the points a and b; b) charge on each capacitor. V=0 Current does not go through the capacitors but it does go through 14Ω and 10Ω resistors. I=24V/24Ω=1A a)V a =1Ax10Ω=10V V b =24V/2=12V b) Q 1 =Q 2 =Q=24VxC eq =24Vx1μF=24μC 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 V a =1Ax10Ω=10V (also the same) V b =V a =10V Q 2 =10Vx2μF=20μC; Q 1 =14Vx2μF=28μC V=24V S 14Ω 10Ω 2 μF a b Q1Q1 Q2Q2

7. Example. Switch is closed at t=0. Find the final charge on the capacitor at t→∞. 12V S 1Ω1Ω 10 Ω 9Ω9Ω 5Ω5Ω 1.5μF a b i 1 +i 2 i1i1 i2i2 o

8. Charging the capacitor at t=0 q( t=0)= 0→ V bc (t=0)=q( t=0)/C=0 → V ab (t= 0)=ε → i(t=0)=ε /R I. C and R in series Current appears at t=0. 1. Initial conditions Switch is closed at t=0. Capacitor is initially uncharged:

9. Example. All capacitors are initially uncharged. Switch is closed at t=0. a)Find Vab(t=0); b) Find Vab(t→∞). 12V 2μF2μF 5Ω5Ω 4μF4μF 3μF3μF 1Ω1Ω a b c S

10. 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 ? 42V 8Ω8Ω 4μF4μF 6Ω6Ω 3Ω3Ω S i1i1 i2i2 i 1 +i 2

11. 2. Transient regime

12. Example. Verify an energy balance in the process of charging the capacitor. One half of the energy provided by the battery is stored in capacitor, another half is dissipated in the resistor.

13. Example. Which fraction of the maximum possible energy was stored in capacitor by the moment when current droped 3 times?

14. 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. 12V 2μF2μF 5Ω5Ω 4μF4μF 3μF3μF 1Ω1Ω a b c S

15. Charging the capacitor. 1. Initial conditions. Capacitor is initially uncharged. Switch is closed at t=0. Current i 1 appears at t=0. R1R1 R2R2 q( t=0)= 0→ V bc (t=0)=q( t=0)/C=0 → i 2 (t=0)=0 V ab (t=0)= ε → i 1+ i 2 (t=0)=i 1 =ε /R 1 i 1+ i 2 i2i2 b c 2. C and R in parallel a i1i1

16. Example. Capacitor is initially uncharged. At t=0 the switch is closed. Find a)the initial and the final current; b)the final charge. 10V S 20Ω 40Ω 0.5μF c b

17. R1R1 R2R2 i1i1 i2i2 b c a i 1 -i 2 2. Transient regime. Charge increases, current, i 1 -i 2 decreases till steady state regime considered above is established at t →∞.

18. Discharging the capacitor. at t=0 q( t=0)= Q→ V bc (t=0)=Q/C=0 → V ab (t= 0)=-Q/C → i(t=0)=-Q/CR 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.

19. 2.Transient regime

20. Example. Verify an energy balance in the process of discharging the capacitor.

21. 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 ? 42V 8Ω8Ω 4μF4μF 6Ω6Ω 3Ω3Ω S

22. Discharging the capacitor. 1.Initial conditions. Capacitor is initially charged and disconnected from the battery. Switch is closed at t=0. Current appears at t=0. R1R1 R2R2 q( t=0)= Q→ V bc (t=0)=Q/C→ i 2 (t=0)=Q/CR 2 V ba (t=0)= Q/C → i 1 (t=0)=Q /CR 1 i1i1 i2i2 bc 2. C and R in parallel a i 1 +i 2 2. Transient regime. Charge decreases, currents also decrease to zero.

23. 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. 12V S 1Ω1Ω 10 Ω 9Ω9Ω 5Ω5Ω 1.5μF

24. Applications of R-C circuits 2. Frequency filtration ~ ε(t) HF output,LF filtr ( ώ <1/RC do not go through) LF output,HF filtr ( ώ >1/RC) do not go through) 1. Sawtooth voltage (pacemakers, turn signals, windshield wipes, etc.) Gas filled tube V t T

25. 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? Gas filled tube V t T ε