2. a b  R C I I t q RC 2 RC 0 CC C a b + --  R + I I RC Circuits q RC2RC 0 t CC

3. Today… Calculate Discharging of Capacitor through a Resistor Calculate Charging of Capacitor through a Resistor

4. Review: Behavior of Capacitors Capacitors resist change in charge and voltage Charging –Initially, the capacitor behaves like a wire. –After a long time, the capacitor behaves like an open switch. Discharging –Initially, the capacitor behaves like a battery. –After a long time, the capacitor behaves like a wire.

5. 3) What is the voltage across the capacitor immediately after switch S 1 is closed? a) V c = 0 b) V c = E c) V c = 1/2 E E 4) Find the voltage across the capacitor after the switch has been closed for a very long time. a) V c = 0 b) V c = E c) V c = 1/2 E The capacitor is initially uncharged, and the two switches are open. Preflight 11: Initially: Q = 0 V C = 0 I = E/(2R) Q = E CI = 0

6. 6) After being closed a long time, switch 1 is opened and switch 2 is closed. What is the current through the right resistor immediately after the switch 2 is closed? E a) I R = 0 b) I R =E/(3R) c) I R =E/(2R) d) I R =E/R Preflight 11: Now, the battery and the resistor 2R are disconnected from the circuit, so we have a different circuit. Since C is fully charged, V C = E. Initially, C acts like a battery, and I = V C /R.

7. RC Circuits (Time-varying currents -- discharging) Discharge capacitor: C initially charged with Q = Q 0 = C  Connect switch to b at t = 0. Calculate current and charge as function of time. Convert to differential equation for Q : C a b ++ --  R I I  Loop theorem  Note: Although we know the current is flowing off the cap., we define it as shown so that …

8. Discharging Capacitor Guess solution: Check that it is a solution: Note that this “guess” incorporates the boundary conditions: C a b ++ --  R I I  ! R dQ dt Q C ee tRCt    // 0

9. Conclusion: Capacitor discharges exponentially with time constant  = RC Current decays from initial max value (= -  / R ) with same time constant Discharge capacitor: a C b + --  R + I I  Current is found from differentiation: Minus sign: Current is opposite to original definition, i.e., charges flow away from capacitor. Discharging Capacitor

10. Charge on C Max = C  37% Max at t = RC Current “Max” = -  /R 37% Max at t = RC t Q 0 CC RC 2 RC 0 -  / R I t zero

11. The two circuits shown below contain identical fully charged capacitors at t=0. Circuit 2 has twice as much resistance as circuit 1. 8) Compare the charge on the two capacitors a short time after t = 0 a) Q 1 > Q 2 b) Q 1 = Q 2 c) Q 1 < Q 2 Preflight 11: Initially, the charges on the two capacitors are the same. But the two circuits have different time constants:  1 = RC and  2 = 2RC. Since  2 >  1 it takes circuit 2 longer to discharge its capacitor. Therefore, at any given time, the charge on capacitor 2 is bigger than that on capacitor 1.

12. Lecture 11, ACT 1 The capacitor in the circuit shown is initially charged to Q = Q 0. At t = 0 the switch is connected to position a. At t = t 0 the switch is immediately flipped from position a to position b. (a) (b) (c) C a b R 3R3R Which of the following graphs best represents the time dependence of the charge on C ? 1A Which of the following correctly relates the value of t 0 to the time constant  while the switch is at a ? 1B (a) t 0 <  (b) t 0 =  (c) t 0 > 

13. Lecture 11, ACT 1 The capacitor in the circuit shown is initially charged to Q = Q 0. At t = 0 the switch is connected to position a. At t = t 0 the switch is immediately flipped from position a to position b. For 0 < t < t 0, the capacitor is discharging with time constant  = RC For t > t 0, the capacitor is discharging with time constant  = 3 RC, i.e., much more slowly. Therefore, the answer is (a) (b) has equal discharging rates (c) has faster discharging after t 0 than before. (a) (b) (c) C a b R 3R3R Which of the following graphs best represents the time dependence of the charge on C ? 1A

14. Lecture 11, ACT 1 The capacitor in the circuit shown is initially charged to Q = Q 0. At t = 0 the switch is connected to position a. At t = t 0 the switch is immediately flipped from position a to position b. C a b R 3R3R Which of the following correctly relates the value of t 0 to the time constant  while the switch is at a ? 1B (a) t 0 <  (b) t 0 =  (c) t 0 >  We know that for t = , the value of the charge is e -1 = 0.37 of the value at t = 0. Since the curve shows Q(t 0 ) ~ 0.6 Q 0, t 0 must be less than .

15. RC Circuits (Time-varying currents, charging) Charge capacitor: C initially uncharged; connect switch to a at t=0 Loop theorem  Convert to differential equation for Q : a b  R C I I Calculate current and charge as function of time.  Would it matter where R is placed in the loop??

16. Guess solution: Check that it is a solution: Note that this “guess” incorporates the boundary conditions: a b  R C I I Charge capacitor:  ! Charging Capacitor

17. Current is found from differentiation: Charge capacitor: a b  R C I I  Conclusion: Capacitor reaches its final charge( Q = C  ) exponentially with time constant  = RC. Current decays from max (=  / R ) with same time constant.

18. Charging Capacitor Q 0 CC Charge on C Max = C  63% Max at t = RC t RC 2 RC I 0 t  R Current Max =  / R 37% Max at t = RC

19. Lecture 11, ACT 2 At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. –At time t = t 1 = , the charge Q 1 on the capacitor is (1-1/ e ) of its asymptotic charge Q f = C . –What is the relation between Q 1 and Q 2, the charge on the capacitor at time t = t 2 = 2  ? a b  R C II R (a) Q 2 < 2 Q 1 (b) Q 2 = 2 Q 1 (c) Q 2 > 2 Q 1

20. Lecture 11, ACT 2 a b  R C II R (a) Q 2 < 2 Q 1 (b) Q 2 = 2 Q 1 (c) Q 2 > 2 Q 1 From the graph at the right, it is clear that the charge increase is not as fast as linear. In fact the rate of increase is just proportional to the current (d Q /dt) which decreases with time. Therefore, Q 2 < 2 Q 1.  Q Q2Q2 2Q12Q1 Q1Q1  The point of this ACT is to test your understanding of the exact time dependence of the charging of the capacitor. So the question is: how does this charge increase differ from a linear increase? Charge increases according to: At t=0 the switch is thrown from position b to position a in the circuit shown: The capacitor is initially uncharged. –At time t = t 1 = , the charge Q 1 on the capacitor is (1-1/ e ) of its asymptotic charge Q f = C . –What is the relation between Q 1 and Q 2, the charge on the capacitor at time t = t 2 = 2  ?

21. Charging Discharging RC 2RC t t Q 0 CC I 0  / R RC t 2RC 0 -  / R I t Q 0 CC

22. A very interesting RC circuit I1I1 I3I3 I2I2  R2R2 C R1R1 First consider the short and long term behavior of this circuit. Short term behavior: Initially the capacitor acts like an ideal wire. Hence, and Long term behavior: Exercise for the student!!

23. 10) Find the current through R 1 after the switch has been closed for a long time. a) I 1 = 0 b) I 1 = E/R 1 c) I 1 = E/(R 1 + R 2 ) The circuit below contains a battery, a switch, a capacitor and two resistors Preflight 11: After the switch is closed for a long time ….. The capacitor will be fully charged, and I 3 = 0. (The capacitor acts like an open switch). So, I 1 = I 2, and we have a one-loop circuit with two resistors in series, hence I 1 = E/(R 1 +R 2 )

24. Lecture 11, ACT 3 At t = 0 the switch is closed in the circuit shown. The initially uncharged capacitor then begins to charge. I1I1 I3I3 I2I2  R2R2 C R1R1 3B (a) (b) (c) 3A What will be the voltage across the capacitor a long time after the switch is closed? (a) V C = 0 (b) V C =  R 2 /(R 1 + R 2 ) (c) V C =  What is the charging time constant  ?

25. Lecture 11, ACT 3 At t = 0 the switch is closed in the circuit shown. The initially uncharged capacitor then begins to charge. I1I1 I3I3 I2I2  R2R2 C R1R1 3A What will be the voltage across the capacitor a long time after the switch is closed? (a) V C = 0 (b) V C =  R 2 /(R 1 + R 2 ) (c) V C =  After a long time the capacitor is completely charged, so no current flows through it. The circuit is then equivalent to a battery with two resistors in series. The voltage across the capacitor equals the voltage across R 2 (since C and R 2 are in parallel). Either from direct calculation, or remembering the “Voltage Divider Circuit”, V C = V R2 =  R 2 /(R 1 + R 2 ).

26. Lecture 11, ACT 3 (a) (b) (c) An ideal voltage source contributes no resistance or capacitance  time constant is entirely determined by C, R 1, and R 2. I personally find it easier to think about the circuit as if C was discharging than charging; I imagine that the capacitor is charged, and that the battery is replaced by a wire (which also has no resistance or capacitance). Since the battery supplies a constant voltage, it doesn’t affect the time constant. We simply need to find the effective resistance R eff through which the capacitor (dis)charges. Looking at the new circuit, it is clear that the capacitor would be (dis)charging through both R 1 and R 2, which are in parallel  their effective resistance is R eff = R 1 R 2 /(R 1 + R 2 ) and  = R eff C. At t = 0 the switch is closed in the circuit shown. The initially uncharged capacitor then begins to charge. –What is the charging time constant  ? I1I1 I3I3 I2I2  R2R2 C R1R1 3B

27. Very interesting RC circuit, detailed I1I1 I3I3 I2I2  R2R2 C R1R1 Loop 1 Loop 2 Node: Loop 1: Loop 2 : Eliminate I 1 in L 1 and L 2 using Node equation: Loop 1: Loop 2: eliminate I 2 from this Final differential eqn:

28. Very interesting RC circuit continued Try solution of the form: –and plug into ODE to get parameters A and τ Final differential eqn: time constant:  parallel combination of R 1 and R 2 Obtain results that agree with initial and final conditions: I1I1 I3I3 I2I2  R2R2 C R1R1 Loop 1 Loop 2

29. Very interesting RC circuit continued What happens when we discharge the capacitor? –Open the switch... Loop 1 and Loop 2 do not exist! I 2 is only current only one loop –start at x marks the spot... I2I2  R2R2 C R1R1 Different time constant for discharging I1I1 I3I3 I2I2  R2R2 C R1R1 Loop 1 Loop 2

30. Summary Kirchoff’s Laws apply to time dependent circuits  they give differential equations! Exponential solutions –from form of differential equation time constant  = RC –what R, what C ?? You must analyze the problem! series RC charging solution series RC discharging solution Next time: Start Magnetism