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

a

b

R

C

II

q C e t RC 1 /

t

q

RC 2RC

0

C

C

a

b+

- -

R+

I I

RC Circuits

q

RC 2RC

0t

q C e t RC /

C

Today…

• Calculate Discharging of Capacitor through a Resistor

• Calculate Charging of Capacitor through a Resistor

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.

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

E

4) Find the voltage across the capacitor after the switch has beenclosed for a very long time.

a) Vc = 0 b) Vc = E

c) Vc = 1/2 E

The capacitor is initially uncharged, and the two switches are open.

Preflight 11:

Initially: Q = 0 VC = 0 I = E/(2R)

Q = E C I = 0

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) IR= 0

b) IR=E/(3R)

c) IR=E/(2R)

d) IR=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, VC = E. Initially, C acts like a battery, and I = VC/R.

RC Circuits (Time-varying currents -- discharging)

• Discharge capacitor:

C initially charged with Q = Q0 = 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

dQI

dt 0

C

Q

dt

dQR

0C

QIR• Loop theorem

Note: Although we know the current is flowing off the cap., we define it as shown so that …

Discharging Capacitor

• Guess solution:

• Check that it is a solution:

/ 1e t RCdQ

Cdt RC

CQt 00 Qt

Note that this “guess” incorporates the

boundary conditions:

C

a

b+ +

- -

R0dQ Q

Rdt C

I I

!RdQdt

QC

e et RC t RC / / 0

/ /0e et t RCQ Q C

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:

/t RCdQI e

dt R

Minus sign:Current is opposite to original definition,i.e., charges flow away from capacitor.

/ /0e et t RCQ Q C



Charge on C

Max = C

37% Max at t = RC

/t RCdQ

I edt R

Current

“Max” = -/R

37% Max at t = RC

t

Q

0

C RC 2RC

0

-/R

I

t

zero

/e t RCQ C

1

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) Q1 > Q2

b) Q1 = Q2

c) Q1 < Q2

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.

Lecture 11, ACT 1• The capacitor in the circuit shown is

initially charged to Q = Q0. At t = 0 the switch is connected to position a.

• At t = t0 the switch is immediately flipped from position a to position b.

(a) (b) (c)

time

QQ

0

t0time

QQ

0

t0time

QQ

0

t0

C

a b

R 3R

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 t0 to the

time constant while the switch is at a?1B

(a) t0 < (b) t0 = (c) t0 >




• For 0 < t < t0, the capacitor is discharging with time constant = RC• For t > t0, the capacitor is discharging with time constant = 3RC, i.e., much more slowly. Therefore, the answer is (a)

• (b) has equal discharging rates• (c) has faster discharging after t0 than before.

(a) (b) (c)

time

QQ

0

t0time

QQ

0

t0time

QQ

0

t0

C

a b

R 3R

Which of the following graphs best represents the time dependence of the charge on C?

1A




time

QQ

0

t0

C

a b

R 3R

Which of the following correctly relates the value of t0 to the

time constant while the switch is at a?1B

(a) t0 < (b) t0 = (c) t0 >

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(t0) ~ 0.6 Q0, t0 must be less than .

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

II

Calculate current and

charge as function of time.

dt

dQI

C

Q

dt

dQR

Would it matter where R is placed in the loop??0

QIR

C

• Guess solution:

•Check that it is a solution:

00 Qt

CQt

Note that this “guess” incorporates the

boundary conditions:

a

b

R

C

II

dQ QR

dt C

• Charge capacitor:

/ 1t RCdQC e

dt RC

)1(/ RC

tRCt ee

C

Q

dt

dQR !

(1 )t

RCQ C e

Charging Capacitor

Charging Capacitor

• Current is found from differentiation:

• Charge capacitor: a

b

R

C

II

/1 t RCQ C e

/t RCdQI e

dt R

Conclusion:

• Capacitor reaches its final charge(Q=C) exponentially with time constant = RC.

• Current decays from max (=/R) with same time constant.

Charging Capacitor

Q

0

C Charge on C

Max = C

63% Max at t = RC

/1 t RCQ C e

t

RC 2RC

I

0t

R /t RCdQ

I edt R

Current

Max =/R

37% Max at t = RC 2

http://webug.physics.uiuc.edu/courses/phys112/fall97/acts/lect11/act11b/sld001.htm

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 = t1 =, the charge Q1 on the capacitor is (1-1/e) of its asymptotic charge Qf = C.

– What is the relation between Q1 and Q2 , the

charge on the capacitor at time t = t2 = 2 ?

a

b

R

C

II

R

(a) Q2 < 2Q1 (b) Q2 = 2Q1 (c) Q2 > 2Q1

Lecture 11, ACT 2a

b

R

C

II

R

(a) Q2 < 2Q1 (b) Q2 = 2Q1 (c) Q2 > 2Q1

•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.

Q

Q2

2Q1

Q1

• 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?

)1( 2RCt

eCQ

• 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 = t1 =, the charge Q1 on the capacitor is (1-1/e) of its asymptotic charge Qf = C.

– What is the relation between Q1 and Q2 , the

charge on the capacitor at time t = t2 = 2 ?

Charging DischargingRC 2RC

t

t

Q

0

C

I

0

/R

/1 t RCQ C e

/t RCdQI e

dt R

RC

t

2RC

0

-/R

I

t

Q

0

C

/t RCdQI e

dt R

/e t RCQ C

A very interesting RC circuitI1

I3

I2

R2C

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!!

3

http://webug.physics.uiuc.edu/courses/phys112/fall97/acts/lect16/act16c/sld001.htm

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)

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 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)

Lecture 11, ACT 3• At t = 0 the switch is closed in

the circuit shown. The initially uncharged capacitor then begins to charge.

I1

I3

I2

R2C

R1

3B

(a) (b) (c) 1 2

1 2

R RC

R R

1 2( )R R C 1R C

3A • What will be the voltage across the capacitor a long time after the switch is closed?

(a) VC = 0 (b) VC = R2/(R1+ R2)

(c) VC =

•What is the charging time constant ?

Lecture 11, ACT 3• At t = 0 the switch is closed in

the circuit shown. The initially uncharged capacitor then begins to charge.

I1

I3

I2

R2C

R1

3A • What will be the voltage across the capacitor a long time after the switch is closed?

(a) VC = 0 (b) VC = R2/(R1+ R2)

(c) VC =

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 R2 (since C and R2 are in

parallel). Either from direct calculation, or remembering the

“Voltage Divider Circuit”, VC = VR2 = R2/(R1+ R2).

Lecture 11, ACT 3

(a) (b) (c)

• An ideal voltage source contributes no resistance or capacitance time constant is entirely determined by C, R1, and R2.

• 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 Reff through which the capacitor (dis)charges. Looking at the new circuit, it is clear that the capacitor would be (dis)charging through both R1 and R2, which are in parallel their effective resistance is Reff = R1R2/(R1 + R2) and = ReffC.

• 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 ?

I1

I3

I2

R2C

R13B

1 2

1 2

R RC

R R

1 2( )R R C 1R C

Very interesting RC circuit , detailed

I1

I3

I2

R2C

R1

Loop 1

Loop 2

• Node:

• Loop 1: 1 1 0Q

I RC

• Loop 2: 2 2 1 1 0I R I R

321 III

• Eliminate I1 in L1 and L2 using Node equation:

• Loop 1: 1 2 0Q dQ

R IC dt

• Loop 2: 2 2 1 2 0dQ

I R R Idt

eliminate I2 from this

• Final differential eqn:1 1 2

1 2

dQ Q

R dt R RC

R R

Very interesting RC circuit continued

• Try solution of the form:

– and plug into ODE to get parameters A and τ

• Final differential eqn:

1

21

21 RC

RRRR

Q

dt

dQ

time constant: parallel combination

of R1 and R2

/1)( teAtQ

• Obtain results that agree with initial and final conditions:

CRR

RR

21

212

1 2

RA C

R R

I1

I3

I2

R2C

R1

Loop 1

Loop 2

Very interesting RC circuit continued

• What happens when we

discharge the capacitor?– Open the switch...

• Loop 1 and Loop 2 do not exist!

• I2 is only current

• only one loop– start at x marks the spot...

I2 R2

C

R1

2 2 20Q dQ

I R but IC dt

Different time constant for discharging

I1

I3

I2

R2C

R1

Loop 1

Loop 2

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

/( ) 1 e t RCQ Q t

/( 0) e t RCQ Q t

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

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