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# 2RC - PowerPoint PPT Presentation

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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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

Examples: 26.17,18 and 19

Today…

• Calculate Charging of Capacitor through a Resistor

• Calculate Discharging of Capacitor through a Resistor

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.

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

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

After a long time:

VC = EQ = E C I = 0

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

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.

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!

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:

• 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:

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

(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

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:

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:

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)

-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

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

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.

(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

-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

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!!

• 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)

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

• 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

• 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:

I1

I2

I3

Loop 1

e

C

R2

R1

I2

e

C

R2

R1

Different time constant for discharging

Very interesting RC circuitcontinued

• Open the switch...

• Loop 1 and Loop 2 do not exist!

• I2is only current

• only one loop

• start atxmarks the spot...

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