induction1 time varying circuits 2008 induction2 a look into the future we have one more week after...

Induction 1

Time Varying CircuitsTime Varying Circuits

20082008

Induction 2

A look into the futureA look into the future

We have one more week after today (+ one We have one more week after today (+ one day)day) Time Varying Circuits Including ACTime Varying Circuits Including AC Some additional topics leading to wavesSome additional topics leading to waves A bit of review if there is time.A bit of review if there is time.

There will be one more Friday morning quiz.There will be one more Friday morning quiz. I hope to be able to return the exams on I hope to be able to return the exams on

Monday at which time we will briefly review Monday at which time we will briefly review the solutions.the solutions.

Induction Induction 33

The Final ExamThe Final Exam

8-10 Problems 8-10 Problems similar to (or similar to (or exactly) Web-exactly) Web-AssignmentsAssignments

Covers the entire Covers the entire semester’s worksemester’s work

May contain some May contain some short answer short answer questions.questions.

Induction 4

Max Current Rate ofincrease = max emfVR=iR

~current

Induction 5

constant) (time

)1( /

R

L

eR

Ei LRt

We Solved the lo

op equation.

Induction 6

We also showed that

2

0

2

0

2

1

2

1

E

B

capacitor

inductor

u

u

Induction 7

At t=0, the charged capacitor is now connected to the inductor. What would you expect to happen??

Induction 8

The math …For an RLC circuit with no driving potential (AC or DC source):

2/12

2max

2

2

2

1

)cos(

:

0

0

L

R

LC

where

teQQ

Solutiondt

QdL

C

Q

dt

dQR

dt

diL

C

QiR

d

dL

Rt

Induction 9

The Graph of that LR (no emf) circuit ..

L

Rt

e 2

L

Rt

e 2

I

Induction 10

Induction 11

Mass on a Spring Result

Energy will swap back and forth. Add friction

Oscillation will slow down Not a perfect analogy

Induction 12

Induction 13

LC Circuit

High

Q/CLow

Low

High

Induction 14

The Math Solution (R=0):

LC

Induction 15

New Feature of Circuits with L and C

These circuits produce oscillations in the currents and voltages

Without a resistance, the oscillations would continue in an un-driven circuit.

With resistance, the current would eventually die out.

Induction 16

-1.5

-1

-0.5

0

0.5

1

1.5

0 1 2 3 4 5 6 7 8 9 10

Time

Vo

lts

Variable Emf Applied

emf

Sinusoidal

DC

Induction 17

Sinusoidal Stuff

)sin( tAemf

“Angle”

Phase Angle

Induction 18

Same Frequencywith

PHASE SHIFT

Induction 19

Different Frequencies

Induction 20

Note – Power is delivered to our homes as an oscillating source (AC)

Induction 21

Producing AC Generator

x x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x x

Induction 22

The Real World

Induction 23

A

Induction 24

Induction 25

The Flux:

tAR

emfi

tBAemf

t

BA

bulb

sin

sin

cos

AB

Induction 26

problems …

Induction 27

14. Calculate the resistance in an RL circuit in which L = 2.50 H and the current increases to 90.0% of its final value in 3.00 s.

Induction 28

18. In the circuit shown in Figure P32.17, let L = 7.00 H, R = 9.00 Ω, and ε = 120 V. What is the self-induced emf 0.200 s after the switch is closed?

Induction 29

32. At t = 0, an emf of 500 V is applied to a coil that has an inductance of 0.800 H and a resistance of 30.0 Ω. (a) Find the energy stored in the magnetic field when the current reaches half its maximum value. (b) After the emf is connected, how long does it take the current to reach this value?

Induction 30

16. Show that I = I0 e – t/τ is a solution of the differential equation where τ = L/R and I0 is the current at t = 0.

Induction 31

17. Consider the circuit in Figure P32.17, taking ε = 6.00 V, L = 8.00 mH, and R = 4.00 Ω. (a) What is the inductive time constant of the circuit? (b) Calculate the current in the circuit 250 μs after the switch is closed. (c) What is the value of the final steady-state current? (d) How long does it take the current to reach 80.0% of its maximum value?

Induction 32

27. A 140-mH inductor and a 4.90-Ω resistor are connected with a switch to a 6.00-V battery as shown in Figure P32.27. (a) If the switch is thrown to the left (connecting the battery), how much time elapses before the current reaches 220 mA? (b) What is the current in the inductor 10.0 s after the switch is closed? (c) Now the switch is quickly thrown from a to b. How much time elapses before the current falls to 160 mA?

Induction 33

52. The switch in Figure P32.52 is connected to point a for a long time. After the switch is thrown to point b, what are (a) the frequency of oscillation of the LC circuit, (b) the maximum charge that appears on the capacitor, (c) the maximum current in the inductor, and (d) the total energy the circuit possesses at t = 3.00 s?

Induction 34

Source Voltage:

)sin(0 tVVemf

Induction 35

Average value of anything:

Area under the curve = area under in the average box

T

T

dttfT

h

dttfTh

0

0

)(1

)(

T

h

Induction 36

Average Value

T

dttVT

V0

)(1

0sin1

0

0 T

dttVT

V

For AC:

Induction 37

So …

Average value of current will be zero. Power is proportional to i2R and is ONLY

dissipated in the resistor, The average value of i2 is NOT zero because

it is always POSITIVE

Induction 38

Average Value

0)(1

0

T

dttVT

V

2VVrms

Induction 39

RMS

2

2)(

2

2)

2(

2

1

)2

(1

0

02

0

20

0

20

0

20

220

VV

VdSin

VV

tT

dtT

SinT

TVV

dttT

SinT

VtSinVV

rms

rms

T

rms

T

rms

Induction 40

Usually Written as:

2

2

rmspeak

peakrms

VV

VV

Induction 41

Example: What Is the RMS AVERAGE of the power delivered to the resistor in the circuit:

E

R

~

Induction 42

Power

tR

VRt

R

VRitP

tR

V

R

Vi

tVV

22

0

2

02

0

0

sin)sin()(

)sin(

)sin(

Induction 43

More Power - Details

R

VVV

RR

VP

R

VdSin

R

VP

tdtSinR

VP

dttSinTR

VP

tSinR

VtSin

R

VP

rms

T

T

200

20

20

2

0

22

0

0

22

0

0

22

0

22

022

0

22

1

2

1

2

1)(

2

1

)(1

2

)(1

Induction 44

Resistive Circuit

We apply an AC voltage to the circuit. Ohm’s Law Applies

Induction 45

Con

sid

er

this

cir

cuit

CURRENT ANDVOLTAGE IN PHASE

R

emfi

iRe

Induction 46

Induction 47

Alternating Current Circuits

is the angular frequency (angular speed) [radians per second].

Sometimes instead of we use the frequency f [cycles per second]

Frequency f [cycles per second, or Hertz (Hz)] f

V = VP sin (t -v ) I = IP sin (t -I )

An “AC” circuit is one in which the driving voltage andhence the current are sinusoidal in time.

v

V(t)

t

Vp

-Vp

Induction 48

v

V(t)

t

Vp

-Vp

V = VP sin (t - v )Phase Term

Induction 49

Vp and Ip are the peak current and voltage. We also use the

“root-mean-square” values: Vrms = Vp / and Irms=Ip /

v and I are called phase differences (these determine whenV and I are zero). Usually we’re free to set v=0 (but not I).

2 2

Alternating Current Circuits

V = VP sin (t -v ) I = IP sin (t -I )

v

V(t)

t

Vp

-Vp

Vrms

I/

I(t)

t

Ip

-Ip

Irms

Induction 50

Example: household voltage

In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0.

Induction 51



This 120 V is the RMS amplitude: so Vp=Vrms = 170 V.2

Induction 52



This 120 V is the RMS amplitude: so Vp=Vrms = 170 V.This 60 Hz is the frequency f: so =2f=377 s -1.

2

Induction 53



This 120 V is the RMS amplitude: so Vp=Vrms = 170 V.This 60 Hz is the frequency f: so =2f=377 s -1.

So V(t) = 170 sin(377t + v).Choose v=0 so that V(t)=0 at t=0: V(t) = 170 sin(377t).

2

Induction 54

Review: Resistors in AC Circuits

ER

~EMF (and also voltage across resistor): V = VP sin (t)Hence by Ohm’s law, I=V/R:

I = (VP /R) sin(t) = IP sin(t) (with IP=VP/R)

V and I“In-phase”

V

t

I

Induction 55

This looks like IP=VP/R for a resistor (except for the phase change). So we call Xc = 1/(C) the Capacitive Reactance

Capacitors in AC Circuits

E

~C Start from: q = C V [V=Vpsin(t)]

Take derivative: dq/dt = C dV/dtSo I = C dV/dt = C VP cos (t)

I = C VP sin (t + /2)

The reactance is sort of like resistance in that IP=VP/Xc. Also, the current leads the voltage by 90o (phase difference).

V

t

I

V and I “out of phase” by 90º. I leads V by 90º.

Induction 56

I Leads V???What the **(&@ does that mean??

I

V

Current reaches it’s maximum at an earlier time than the voltage!

1

2

I = C VP sin (t +/2)

Phase=

-(/2)

Induction 57

Capacitor Example

E

~

CA 100 nF capacitor isconnected to an AC supply of peak voltage 170V and frequency 60 Hz.

What is the peak current?What is the phase of the current?

MX

f

C 65.2C

1

1077.3C

rad/sec 77.360227

Also, the current leads the voltage by 90o (phase difference).

I=V/XC

Induction 58

Again this looks like IP=VP/R for aresistor (except for the phase change).

So we call XL = L the Inductive Reactance

Inductors in AC Circuits

LV = VP sin (t)Loop law: V +VL= 0 where VL = -L dI/dtHence: dI/dt = (VP/L) sin(t).Integrate: I = - (VP / L cos (t)

or I = [VP /(L)] sin (t - /2)

~

Here the current lags the voltage by 90o.

V

t

I

V and I “out of phase” by 90º. I lags V by 90º.

Induction 59

Induction 60

Phasor Diagrams

Vp

Ipt

Resistor

A phasor is an arrow whose length represents the amplitude ofan AC voltage or current.

The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.

Phasor diagrams are useful in solving complex AC circuits.The “y component” is the actual voltage or current.

A phasor is an arrow whose length represents the amplitude ofan AC voltage or current.

The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.

Phasor diagrams are useful in solving complex AC circuits.The “y component” is the actual voltage or current.

Induction 61

Phasor Diagrams

Vp

Ipt

Vp

Ip

t

Resistor Capacitor

A phasor is an arrow whose length represents the amplitude ofan AC voltage or current.The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.Phasor diagrams are useful in solving complex AC circuits.The “y component” is the actual voltage or current.


Induction 62

Phasor Diagrams

Vp

Ipt

Vp

IpVp Ip

Resistor Capacitor Inductor



Induction 63

Steady State Solution for AC Current (2)

• Expand sin & cos expressions

• Collect sindt & cosdt terms separately

• These equations can be solved for Im and (next slide)

1/ cos sin 0

1/ sin cos

d d

m d d m m

L C R

I L C I R

sin sin cos cos sin

cos cos cos sin sin

d d d

d d d

t t t

t t t

High school trig!

cosdt terms

sindt terms

cos sin cos sinmm d d m d d m d

d

II L I R t t t

C

Induction 64


• Expand sin & cos expressions

• Collect sindt & cosdt terms separately

• These equations can be solved for Im and (next slide)

1/ cos sin 0

1/ sin cos

d d

m d d m m

L C R

I L C I R

sin sin cos cos sin

cos cos cos sin sin

d d d

d d d

t t t

t t t

High school trig!

cosdt terms

sindt terms

cos sin cos sinmm d d m d d m d

d

II L I R t t t

C

Induction 65

• Solve for and Im in terms of

• R, XL, XC and Z have dimensions of resistance

• Let’s try to understand this solution using “phasors”


1/tan d d L CL C X X

R R

m

mIZ

22L CZ R X X

L dX L

1/C dX CInductive “reactance”

Capacitive “reactance”

Total “impedance”

1/ cos sin 0

1/ sin cos

d d

m d d m m

L C R

I L C I R

Induction 66

REMEMBER Phasor Diagrams?

Vp

Ipt

Vp

Ip

t

Vp Ip

t


A phasor is an arrow whose length represents the amplitude ofan AC voltage or current.The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.Phasor diagrams are useful in solving complex AC circuits.

A phasor is an arrow whose length represents the amplitude ofan AC voltage or current.The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.Phasor diagrams are useful in solving complex AC circuits.

Induction 67

Reactance - Phasor Diagrams

Vp

Ipt

Vp

Ip

t

Vp Ip

t


Induction 68

“Impedance” of an AC Circuit

R

L

C~

The impedance, Z, of a circuit relates peakcurrent to peak voltage:

IV

Zpp (Units: OHMS)

Induction 69

“Impedance” of an AC Circuit

R

L

C~

The impedance, Z, of a circuit relates peakcurrent to peak voltage:

IV

Zpp (Units: OHMS)

(This is the AC equivalent of Ohm’s law.)

Induction 70

Impedance of an RLC Circuit

R

L

C~

As in DC circuits, we can use the loop method:

- VR - VC - VL = 0 I is same through all components.

Induction 71


R

L

C~

As in DC circuits, we can use the loop method:

- VR - VC - VL = 0 I is same through all components.

BUT: Voltages have different PHASES

they add as PHASORS.

Induction 72

Phasors for a Series RLC Circuit

Ip

VRp

(VCp- VLp)

VP

VCp

VLp

Induction 73


By Pythagoras’ theorem:

(VP )2 = [ (VRp )2 + (VCp - VLp)2 ]

Ip

VRp

(VCp- VLp)

VP

VCp

VLp

Induction 74


By Pythagoras’ theorem:

(VP )2 = [ (VRp )2 + (VCp - VLp)2 ]

= Ip2 R2 + (Ip XC - Ip XL)

2

Ip

VRp

(VCp- VLp)

VP

VCp

VLp

Induction 75


Solve for the current:

Ip Vp

R2 (Xc XL )2

Vp

Z

R

L

C~

Induction 76


Solve for the current:

Impedance:

Ip Vp

R2 (Xc XL )2

Vp

Z

Z R2 1

C L

2

R

L

C~

Induction 77

The circuit hits resonance when 1/C-L=0: r=1/When this happens the capacitor and inductor cancel each otherand the circuit behaves purely resistively: IP=VP/R.


Ip Vp

Z

Z R2 1

C L

2

The current’s magnitude depends onthe driving frequency. When Z is aminimum, the current is a maximum.This happens at a resonance frequency:

LC

The current dies awayat both low and highfrequencies.

IP

01 0

21 0

31 0

41 0

5

R = 1 0 0

R = 1 0

r

L=1mHC=10F

Induction 78

Phase in an RLC CircuitIp

VRp

(VCp- VLp)

VP

VCp

VLp

We can also find the phase:

tan = (VCp - VLp)/ VRp

or; tan = (XC-XL)/R.

or tan = (1/C - L) / R

Induction 79

Phase in an RLC Circuit

At resonance the phase goes to zero (when the circuit becomespurely resistive, the current and voltage are in phase).

IpVRp

(VCp- VLp)

VP

VCp

VLp

We can also find the phase:

tan = (VCp - VLp)/ VRp

or; tan = (XC-XL)/R.

or tan = (1/C - L) / R

More generally, in terms of impedance:

cos R/Z

Induction 80

Power in an AC Circuit

V(t) = VP sin (t)

I(t) = IP sin (t)

P(t) = IV = IP VP sin 2(t) Note this oscillates

twice as fast.

V

t

I

t

P

= 0

(This is for a purely resistive circuit.)

Induction 81

The power is P=IV. Since both I and V vary in time, sodoes the power: P is a function of time.


Use, V = VP sin (t) and I = IP sin (t+) :

P(t) = IpVpsin(t) sin (t+)

This wiggles in time, usually very fast. What we usually care about is the time average of this:

PT

P t dtT

10

( ) (T=1/f )

Induction 82


Now: sin( ) sin( )cos cos( )sin t t t

Induction 83


P t I V t tI V t t t

P P

P P

( ) sin( )sin( )sin ( )cos sin( )cos( )sin

2


Induction 84



P P

P P


2

sin ( )

sin( ) cos( )

2 1

2

0

t

t t

Use:

and:

So P I VP P1

2cos


Induction 85



P P

P P


2

sin ( )

sin( ) cos( )

2 1

2

0

t

t t

Use:

and:

So P I VP P1

2cos

Now:

which we usually write as P I Vrms rms cos

sin( ) sin( )cos cos( )sin t t t

Induction 86


P I Vrms rms cos goes from -900 to 900, so the average power is positive)

cos( is called the power factor.

For a purely resistive circuit the power factor is 1.When R=0, cos()=0 (energy is traded but not dissipated).Usually the power factor depends on frequency.

Induction 87

16. Show that I = I0 e – t/τ is a solution of the differential equation where τ = L/R and I0 is the current at t = 0.

Induction 88

17. Consider the circuit in Figure P32.17, taking ε = 6.00 V, L = 8.00 mH, and R = 4.00 Ω. (a) What is the inductive time constant of the circuit? (b) Calculate the current in the circuit 250 μs after the switch is closed. (c) What is the value of the final steady-state current? (d) How long does it take the current to reach 80.0% of its maximum value?

Induction 89

27. A 140-mH inductor and a 4.90-Ω resistor are connected with a switch to a 6.00-V battery as shown in Figure P32.27. (a) If the switch is thrown to the left (connecting the battery), how much time elapses before the current reaches 220 mA? (b) What is the current in the inductor 10.0 s after the switch is closed? (c) Now the switch is quickly thrown from a to b. How much time elapses before the current falls to 160 mA?

Induction 90

52. The switch in Figure P32.52 is connected to point a for a long time. After the switch is thrown to point b, what are (a) the frequency of oscillation of the LC circuit, (b) the maximum charge that appears on the capacitor, (c) the maximum current in the inductor, and (d) the total energy the circuit possesses at t = 3.00 s?

induction1 time varying circuits 2008 induction2 a look into the future we have one more week after...

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current slide

phase shift slide

different frequencies

real world slide

perfect analogy slide

emf sinusoidal dc slide

oscillating source ac

emf circuit