2rc
This presentation is the property of its rightful owner.
Sponsored Links
1 / 27

2 RC PowerPoint PPT Presentation


  • 99 Views
  • Uploaded on
  • Presentation posted in: General

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

Download Presentation

2 RC

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


2rc

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


Today

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


Last time behavior of capacitors from lect 10

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.


2rc

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


2rc

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

After a long time:

VC = EQ = E CI = 0


2rc

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


2rc

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.


Rc circuits time varying currents

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!


Rc circuits time varying currents1

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:


Rc circuits time varying currents2

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:


Charging capacitor

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


Lecture 11 act 1

(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


Rc circuits time varying currents3

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:


Rc circuits time varying currents4

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:


Rc circuits time varying currents5

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)


Discharging capacitor

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


2rc

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


2rc

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.


Lecture 11 act 2

(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


Charging discharging

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


A very interesting rc circuit

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!!


  • 2rc

    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)


    2rc

    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


    Very interesting rc circuit continued

    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


    Very interesting rc circuit continued1

    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:


    Very interesting rc circuit continued2

    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...


    Summary

    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


  • Login