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I. I. I. a. I. a. R. R. b. b. +. +. C. C. e. -. -. e. RC Circuits. RC. 2 RC. C e. RC. 2 RC. C e. q. q. 0. 0. t. t. Text Reference: Chapter 26.6 . Examples: 26.17,18 and 19. Today…. Calculate Charging of Capacitor through a Resistor

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2 RC

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  1. I I I a I a R R b b + + C C e - - e RC Circuits RC 2RC Ce RC 2RC Ce q q 0 0 t t

  2. Text Reference: Chapter 26.6 Examples: 26.17,18 and 19 Today… • Calculate Charging of Capacitor through a Resistor • Calculate Discharging of Capacitor through a Resistor

  3. Last time--Behavior of Capacitors(from Lect. 10) • 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.

  4. Preflight 11: E The capacitor is initially uncharged, and the two switches are open. 3) What is the voltage across the capacitor immediately after switch S1 is closed? a) Vc = 0 b) Vc = E c) Vc = 1/2 E 4) Find the voltage across the capacitor after the switch has been closed for a very long time. a) Vc = 0 b) Vc = E c) Vc = 1/2 E

  5. Initially:Q = 0 VC = 0 I = E/(2R) After a long time: VC = EQ = E C I = 0

  6. Preflight 11: E 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? a) IR= 0 b) IR=E/(3R) c) IR=E/(2R) d) IR=E/R

  7. After C is fully charged, S1 is opened and S2 is closed. Now, the battery and the resistor 2R are disconnected from the circuit. So we now have a different circuit. Since C is fully charged, VC = E. Initially, C acts like a battery, and I = VC/R.

  8. I I a R b C e Would it matter where R is placed in the loop??  RC Circuits(Time-varying currents) • Charge capacitor: • Cinitially uncharged; connect switch toaat t=0 Calculate current and charge as function of time. • • Loop theorem Þ • Convert to differential equation for Q: No!

  9. I I a R b C e Note that this “guess” incorporates the boundary conditions: !  RC Circuits(Time-varying currents) • Charge capacitor: • • Guess solution: • Check that it is a solution:

  10. I • Charge capacitor: I a R b C e Þ • Conclusion: • Capacitor reaches its final charge(Q=Ce ) exponentially with time constant t = RC. • Current decays from max (=e /R) with same time constant. RC Circuits(Time-varying currents) • Current is found from differentiation:

  11. Charge on C Max = Ce 63% Max at t=RC Current Max =e /R 37% Max at t=RC Charging Capacitor RC 2RC Ce Q 0 t e /R I 0 t

  12. (c) Q2 > 2Q1 (b) Q2 = 2Q1 (a) Q2 < 2Q1 • The point of this ACT is to test your understanding of the exact time dependence of the charging of the capacitor. • Charge increases according to: • So the question is: how does this charge increase differ from a linear increase? 2Q1 Q2 • 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(dQ/dt)which decreases with time. • Therefore, Q2 < 2Q1. Q1 Q t 2t Lecture 11, ACT 1 a I I • At t=0 the switch is thrown from positionb to position ain the circuit shown: The capacitor is initially uncharged. • At time t=t1=t, the charge Q1 on the capacitor is (1-1/e) of its asymptotic charge Qf=Ce. • What is the relation between Q1 and Q2 , the charge on the capacitor at timet=t2=2t ? R b e C R

  13. I I a R b + + C e - -  RC Circuits (Time-varying currents) • Discharge capacitor: • C initially charged with Q=Ce • Connect switch to batt=0. • Calculate current and charge as function of time. • Loop theorem  • Convert to differential equation forQ:

  14. I I • Discharge capacitor: a R b + + C • Guess solution: e - - e -t/RC Q = C e Note that this “guess” incorporates the boundary conditions: Þ ! dQ Q - - t / RC t / RC e e + = - e + e = R 0 dt C RC Circuits(Time-varying currents) • Check that it is a solution:

  15. I I • Discharge capacitor: a -t/RC R e e Q = C b + + C e - - • Current is found from differentiation: Þ • Conclusion: • Capacitor discharges exponentially with time constant t = RC • Current decays from initial max value(= -e/R)with same time constant Minus sign: original definition of current “I” direction RC Circuits(Time-varying currents)

  16. e -t/RC Q = C e zero 0 Current Max= -e/R 37%Max att=RC I -e /R Discharging Capacitor RC 2RC Ce Charge on C Max = Ce 37% Max at t=RC Q 0 t t

  17. Preflight 11: The two circuits shown below contain identical fully charged capacitors att=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) Q1 > Q2 b) Q1 = Q2 c) Q1 < Q2

  18. Initially, the charges on the two capacitors are the same. But the two circuits have different time constants: t1 = RC and t2 = 2RC.Since t2 > t1it takes circuit 2 longer to discharge its capacitor. Therefore, at any given time, the charge on capacitor is bigger than that on capacitor 1.

  19. (c) (b) (a) Lecture 11, ACT 2 • At t=0 the switch is connected to positiona in the circuit shown: The capacitor is initially uncharged. • At t = t0, the switch is thrown from position a to position b. • Which of the following graphs best represents the time dependence of the charge on C? a b R 2R C e • For 0 < t < t0, the capacitor is charging with time constantt = RC • Fort > t0,the capacitor is discharging with time constantt = 2RC • (a) has equal charging and discharging time constants • (b) has a larger dischargingtthan a chargingt • (c) has a smallerdischargingtthan a charging t

  20. e -t/RC Q = C e 0 I -e /R Charging Discharging RC RC 2RC 2RC Ce Ce Q Q 0 0 t t e /R I 0 t t

  21. I1 I2 I3 A very interesting RC circuit e C R2 R1 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!!

  22. Preflight 11: The circuit below contains a battery, a switch, a capacitor and two resistors 10) Find the current through R1 after the switch has been closed for a long time. a) I1 = 0 b) I1 = E/R1 c) I1 = E/(R1+ R2)

  23. After the switch is closed for a long time ….. The capacitor will be fully charged, and I3 = 0. (The capacitor acts like an open switch). So, I1 = I2, and we have a one-loop circuit with two resistors in series, hence I1 = E/(R1+R2) What is voltage across C after a long time? Cis in parallel withR2 !! VC = I1R2 = E R2/(R1+R2) < E

  24. Loop 2 • Loop 1: I1 I2 I3 Loop 1 e C • Loop 2: R2 • Node: R1 • EliminateI1 inL1 and L2 using Node equation: eliminate I2 from this • Loop 1: • Loop 2: • Final differential eqn: Very interesting RC circuit continued

  25. Loop 2 • Final differential eqn: I1 I2 I3 Loop 1 e C R2 R1 time constant: t parallel combination of R1 and R2 Very interesting RC circuitcontinued • Try solution of the form: • and plug into ODE to get parameters A and τ • Obtain results that agree with initial and final conditions:

  26. Loop 2 I1 I2 I3 Loop 1 e C R2 R1 I2 e C R2 R1 Different time constant for discharging Very interesting RC circuitcontinued • What about discharging? • Open the switch... • Loop 1 and Loop 2 do not exist! • I2is only current • only one loop • start atxmarks the spot...

  27. Reading assignment: Ch. 28.1-2, 28.4 Examples: 28.1,4,5 and 6 Summary • Kirchoff’s Laws apply to time dependent circuitsthey give differential equations! • Exponential solutions • from form of differential equation • time constantt = RC • whatR, whatC??You must analyze the problem! • seriesRC charging solution • seriesRCdischarging solution e e -t/RC Q = C Next time: Start Magnetism

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