Physics 212 Lecture 11 RC Circuits Change in schedule Exam 2 will be on Thursday, July 12 from 8 – 9:30 AM.

RC Circuit Charging Capacitor uncharged, switch is moved to position “a” Kirchoff’s Voltage Rule Initially (q = q0 = 0) Long Term (Ic =0) a R Vbattery C b R Vbattery C a b I In general: Draw switch being closed Draw current at t = 0 Draw VC = 0 Reveal short term equation After long term, indicate IC = 0 implies VR = 0 11

Checkpoint 1a & Checkpoint 1b Q = 0 A circuit is wired up as shown below. The capacitor is initially uncharged and switches S1 and S2 are initially open. Checkpoint 1a & Checkpoint 1b V1 = V V2 = V B) V1 = 0 V2 = V Close S1, V1 = voltage across C immediately after V2 = voltage across C a long time after C) V1 = 0 V2 = 0 D) V1 = V V2 = 0 Immediately after the switch S1 is closed: Q = 0 After the switch S1 has been closed for a long time I = 0 V1 = 0 V = Q/C V2 = V VR = 0 13

I=0 I R Close S1 at t=0 (leave S2 open) V C 2R S1 S2 For t  V C R VC = V At t = 0 I VC = Q/C = 0 15

RC Circuit (Discharging) Capacitor has q0 = CV, switch is moved to position “b” Kirchoff’s Voltage Rule Initially (q=q0) Long Term (Ic =0) R Vbattery C a b R Vbattery C a b I + - In general: V Counterclockwise current -I 19

A circuit is wired up as shown below A circuit is wired up as shown below. The capacitor is initially uncharged and switches S1 and S2 are initially open. Checkpoint 1c IR + - 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 switch 2 is closed? A. IR = 0 B. IR = V/3R C. IR = V/2R D. IR = V/R A B C D 22

A circuit is wired up as shown below A circuit is wired up as shown below. The capacitor is initially uncharged and switches S1 and S2 are initially open. Checkpoint 1c IR + - 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 switch 2 is closed? A. IR = 0 B. IR = V/3R C. IR = V/2R D. IR = V/R A B C D 2R C V I V 22

A circuit is wired up as shown below A circuit is wired up as shown below. The capacitor is initially uncharged and switches S1 and S2 are initially open. Checkpoint 1d Now suppose both switches are closed. What is the voltage across the capacitor after a very long time? A. VC = 0 B. VC = V C. VC = 2V/3 A B C 26

Checkpoint 1d After both switches have been closed for a long time A circuit is wired up as shown below. The capacitor is initially uncharged and switches S1 and S2 are initially open. Checkpoint 1d Now suppose both switches are closed. What is the voltage across the capacitor after a very long time? A. VC = 0 B. VC = V C. VC = 2V/3 A B C After both switches have been closed for a long time The current through the capacitor is zero The current through R = current through 2R Vcapacitor = V2R V2R = 2/3 V 26

Close both S1 and S2 and wait a long time… R Close both S1 and S2 and wait a long time… C V 2R S1 S2 I 2R C R V No current flows through the capacitor after a long time. This will always be the case if the sources of EMF don’t change with time. VC Can do a lot of writing onthis one if wanted.. VC = (2/3)V I = V/(3R) V2R = I(2R) = (2/3)V = VC 27

DEMO – ACT 1 What will happen after I close the switch? Bulb 2 S R V Bulb 1 R C What will happen after I close the switch? Both bulbs come on and stay on. Both bulbs come on but then bulb 2 fades out. Both bulbs come on but then bulb 1 fades out. Both bulbs come on and then both fade out. The purpose of this Check is to jog the students minds back to when they studied work and potential energy in their intro mechanics class. No initial charge on capacitor V(bulb 1) = V(bulb 2) = V Both bulbs light No final current through capacitor V(bulb 2) = 0 30

DEMO – ACT 2 Suppose the switch has been closed a long time. Bulb 2 R S V Bulb 1 R C Suppose the switch has been closed a long time. Now what will happen after open the switch? Both bulbs come on and stay on. Both bulbs come on but then bulb 2 fades out. Both bulbs come on but then bulb 1 fades out. Both bulbs come on and then both fade out. The purpose of this Check is to jog the students minds back to when they studied work and potential energy in their intro mechanics class. Capacitor has charge (=CV) Capacitor discharges through both resistors 32

Calculation S In this circuit, assume V, C, and Ri are known. C initially uncharged and then switch S is closed. What is the voltage across the capacitor after a long time ? Circuit behavior described by Kirchhoff’s Rules: KVR: SVdrops = 0 KCR: SIin = Siout S closed and C charges to some voltage with some time constant Determine currents and voltages in circuit a long time after S closed The purpose of this Check is to jog the students minds back to when they studied work and potential energy in their intro mechanics class. 35

Calculation Immediately after S is closed: In this circuit, assume V, C, and Ri are known. C initially uncharged and then switch S is closed. What is the voltage across the capacitor after a long time ? R1 R2 R3 V C Immediately after S is closed: what is I2, the current through C what is VC, the voltage across C? (A) Only I2 = 0 (B) Only VC = 0 (C) Both I2 and VC = 0 (D) Neither I2 nor VC = 0 The purpose of this Check is to jog the students minds back to when they studied work and potential energy in their intro mechanics class. Why? We are told that C is initially uncharged (V = Q/C) I2 cannot be zero because charge must flow in order to charge C 37

Calculation I1 S In this circuit, assume V, C, and Ri are known. C initially uncharged and then switch S is closed. What is the voltage across the capacitor after a long time ? R1 R2 R3 V C Immediately after S is closed, what is I1, the current through R1 ? (A) (B) (C) (D) (E) The purpose of this Check is to jog the students minds back to when they studied work and potential energy in their intro mechanics class. Why? Draw circuit just after S closed (knowing VC = 0) V R1 R2 S R3 VC = 0 R1 is in series with the parallel combination of R2 and R3 39

Calculation S In this circuit, assume V, C, and Ri are known. C initially uncharged and then switch S is closed. What is the voltage across the capacitor after a long time ? R1 R2 R3 V C After S has been closed “for a long time”, what is IC, the current through C ? (A) (B) (C) The purpose of this Check is to jog the students minds back to when they studied work and potential energy in their intro mechanics class. V R1 R3 IC = 0 VC I Why? After a long time in a static circuit, the current through any capacitor approaches 0 ! This means we redraw circuit with open circuit in middle leg 41

Calculation S In this circuit, assume V, C, and Ri are known. C initially uncharged and then switch S is closed. What is the voltage across the capacitor after a long time ? R1 R2 R3 V C After S has been closed “for a long time”, what is VC, the voltage across C ? (A) (B) (C) (D) (E) The purpose of this Check is to jog the students minds back to when they studied work and potential energy in their intro mechanics class. V R1 R3 VC I VC = V3 = IR3 = (V/(R1+R3))R3 Why?? 43

Challenge What is tc, the charging time constant? Strategy We get: In this circuit, assume V, C, and Ri are known. C initially uncharged and then switch S is closed. S R1 R2 R3 V C What is tc, the charging time constant? Strategy Write down KVR and KCR for the circuit when S is closed 2 loop equations and 1 node equation Use I2 = dQ2/dt to obtain one equation that looks like simple charging RC circuit ( (Q/”C”) + “R”(dQ/dt) – “V” = 0 ) Make correspondence: “R” = ?, and “C” = ?, then t = “R” ”C” R2 C R3 The purpose of this Check is to jog the students minds back to when they studied work and potential energy in their intro mechanics class. We get: C

How do exponentials work? “Fraction of initial charge that remains” “How many time constants worth of time that have elapsed” 45

the longer it takes to get the same change… RC = 2 RC = 1 Time constant: t = RC The bigger t is, the longer it takes to get the same change… 47

Checkpoint 2a t = RequivC Which circuit has the largest time constant? The two circuits shown below contain identical capacitors that hold the same charge at t = 0. Circuit 2 has twice as much resistance as circuit 1. Checkpoint 2a RC = 1 RC = 2 Which circuit has the largest time constant? Circuit 1 Circuit 2 Same t = RequivC 49

The two circuits shown below contain identical capacitors that hold the same charge at t = 0. Circuit 2 has twice as much resistance as circuit 1. Checkpoint 2b Which of the following statements best describes the charge remaining on each of the the two capacitors for any time after t = 0? A. Q1 < Q2 B. Q1 > Q2 C. Q1 = Q2 D. Q1 < Q2 at first, then Q1 > Q2 after long time E. Q1 > Q2 at first, then Q1 < Q2 after long time 50

The two circuits shown below contain identical capacitors that hold the same charge at t = 0. Circuit 2 has twice as much resistance as circuit 1. Checkpoint 2b Which of the following statements best describes the charge remaining on each of the the two capacitors for any time after t = 0? A. Q1 < Q2 B. Q1 > Q2 C. Q1 = Q2 D. Q1 < Q2 at first, then Q1 > Q2 after long time E. Q1 > Q2 at first, then Q1 < Q2 after long time 50

Checkpoint 2b Checkpoint 2b Look at plot !!! Q = Q0e-t/RC The two circuits shown below contain identical capacitors that hold the same charge at t = 0. Circuit 2 has twice as much resistance as circuit 1. Checkpoint 2b Checkpoint 2b Which of the following statements best describes the charge remaining on each of the the two capacitors for any time after t = 0? A. Q1 < Q2 B. Q1 > Q2 C. Q1 = Q2 D. Q1 < Q2 at first, then Q1 > Q2 after long time E. Q1 > Q2 at first, then Q1 < Q2 after long time Look at plot !!! RC = 1 RC = 2 Q = Q0e-t/RC

“Dynamic” random access memory Charge capacitor – store a logical “1” Discharge capacitor – store a logical “0” Capacitor discharges through resistance between plates. Only holds Q for < 1 msec. Charge Q must be “refreshed” constantly, So memory is called dynamic.

Situation once things stop changing. 1. Capacitor is an open circuit for dc (direct current). VC = Q/C and IC = dQ/dt. If IC is flowing then Q is changing, so VC is changing. But in a dc circuit, nothing changes with time, so we must have IC = 0. R +Q V C IC -Q For t  V C R IC = 0 VC = V Situation once things stop changing.