RC Circuits - University Of Illinois · · 2012-06-28Physics 212 Lecture 11, Slide 1 Physics 212...

Physics 212 Lecture 11, Slide 1

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

• Kirchoff’s Voltage Rule

• Initially (q = q0 = 0)

• Long Term (Ic =0)

R Vbattery

C a

b battery

q-V + +IR = 0

C

00 0batteryV I R

battery

0

VI =

R

0 0battery

qV R

C

batteryq CV

In general:

0battery

q dqV R

C dt

/( ) (1 )t RCq t q e

/

0( ) t RCI t I e

11

a

R Vbattery

C

b

• Capacitor uncharged, switch is moved to position “a”

I

Physics 212 Lecture 11, Slide 3 13

Checkpoint 1a

Checkpoint 1b &

Close S1, V1 = voltage across C immediately after V2 = voltage across C a long time after

A) V1 = V V2 = V

B) V1 = 0 V2 = V

C) V1 = 0 V2 = 0

D) V1 = V V2 = 0

Immediately after the switch S1 is closed:

Q = 0 V1 = 0 V = Q/C

After the switch S1 has been closed for a long time

I = 0 V2 = V VR = 0

A circuit is wired up as shown below. The capacitor is initially uncharged and switches

S1 and S2 are initially open.


V 2R C

R

S1 S2

Close S1 at t=0 (leave S2 open)

For t

V C

R

I=0

VC = V

At t = 0

V C

R

I

VC = Q/C = 0

15


RC Circuit (Discharging)

• Kirchoff’s Voltage Rule

• Initially (q=q0)

• Long Term (Ic =0)

R Vbattery

C a

b 0q

IRC

00 0

qI R

C

0

0

qI

RC

0 0q

RC

0q

In general:

0q dq

RC dt

/

0( ) t RCq t q e

/0( ) t RCdq qI t e

dt RC

V

19

• Capacitor has q0 = CV, switch is moved to position “b”

R Vbattery

C a

b

I + -

-I


A

B

C

D

22

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


V

2R C V

I

A

B

C

D

22

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

26

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

• The current through the capacitor is zero

• The current through R = current through 2R

• Vcapacitor = V2R

• V2R = 2/3 V



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


2R C

R

S1 S2

Close both S1 and S2 and wait a long time…

V

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

I

I = V/(3R)

27

V2R = I(2R) = (2/3)V = VC VC = (2/3)V


DEMO – ACT 1

V C

S

Bulb 1

Bulb 2

What will happen after I close the switch? A) Both bulbs come on and stay on. B) Both bulbs come on but then bulb 2 fades out. C) Both bulbs come on but then bulb 1 fades out. D) Both bulbs come on and then both fade out.

R

R

30

No initial charge on capacitor V(bulb 1) = V(bulb 2) = V Both bulbs light

No final current through capacitor

V(bulb 2) = 0


DEMO – ACT 2

V C

S

Bulb 1

Bulb 2

Suppose the switch has been closed a long time. Now what will happen after open the switch? A) Both bulbs come on and stay on. B) Both bulbs come on but then bulb 2 fades out. C) Both bulbs come on but then bulb 1 fades out. D) Both bulbs come on and then both fade out.

R

R

32

Capacitor has charge (=CV) Capacitor discharges through both resistors


Calculation

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

V

R1 R2

C

R3

S

35


Calculation

V

R1 R2

C

R3

S

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

• 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

V

R1 R2

C

R3

S

• Why? – Draw circuit just after S closed

(knowing VC = 0)

• Immediately after S is closed, what is I1, the current through R1 ?

V

R1 R2

S

R3

VC = 0

I1

– R1 is in series with the parallel combination of R2 and R3


39

(A) (B) (C) (D) (E)

1 3

V

R R1

V

R1 2 3

V

R R R 1 2 3

1 2 2 3 1 3

R R RV

R R R R R R

32

321

RR

RRR

V


Calculation

V

R1 R2

C

R3

S

• 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

After S has been closed “for a long time”, what is IC, the current through C ?

(A) (B) (C)

1

V

R 0

V

R1 R3 IC = 0 VC

I


2

V

R

41


Calculation

V

R1 R2

C

R3

S

• Why??

After S has been closed “for a long time”, what is VC, the voltage across C ?

(A) (B) (C) (D) (E)

V3

1 3

RV

R R 0

2

2 31

2 3

RV

R RR

R R

2

1 2

RV

R R

– VC = V3 = IR3 = (V/(R1+R3))R3


V

R1 R3 VC

I I

43


Challenge In this circuit, assume V, C, and Ri are known. C initially uncharged and then switch S is closed.

V

R1 R2

C

R3

S

• 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”

C

What is tc, the charging time constant?

R2

C R3

2 3

2 3

1

2 ( )

R Rt C

R R

We get:

CRR

RRRc

31

312t


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 1 2 3 4 5 6 7 8 9 10

0

Q t

Q

0

t

RCQ t Q e

t

RC

“Fraction of initial charge that remains”

“How many time constants worth of time that have elapsed”

45

How do exponentials work?


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 1 2 3 4 5 6 7 8 9 10

0

Q t

Q

t

RC = 1

RC = 2

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

0

t

RCQ t Q e

47


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 1 2 3 4 5 6 7 8 9 10

RC = 1

RC = 2

Which circuit has the largest time constant?

A) Circuit 1

B) Circuit 2

C) Same

49

t = RequivC

Checkpoint 2a

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


Checkpoint 2b





A. Q1 < Q2 B. Q1 > Q2 C. Q1 = Q2




A. Q1 < Q2 B. Q1 > Q2 C. Q1 = Q2



Checkpoint 2b

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 1 2 3 4 5 6 7 8 9 10

RC = 1

RC = 2

Q = Q0e-t/RC

Look at plot !!!



Checkpoint 2b


Charge capacitor – store a logical “1”

Discharge capacitor – store a logical “0”

Capacitor discharges through resistance

between plates.

Only holds Q for < 1 msec.

“Dynamic” random access memory

Charge Q must be “refreshed” constantly,

So memory is called dynamic.


1. Capacitor is an open circuit for dc (direct current).

R

+Q

V C

IC

-Q

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.

For t

V C

R

IC = 0

VC = V

Situation once things stop changing.

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RC Circuits - University Of Illinois · · 2012-06-28Physics 212 Lecture 11, Slide 1 Physics 212...

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