Induction - Spring 20061 Time Varying Circuits April 10, 2006

Induction - Spring 2006 1

Time Varying CircuitsTime Varying Circuits

April 10, 2006April 10, 2006


What is going on?What is going on?

There are only 7 more classes and There are only 7 more classes and the final is 3 weeks away.the final is 3 weeks away. Scotty, beam me somewhere else!Scotty, beam me somewhere else!

Exam IssuesExam Issues Look Ashamed!Look Ashamed!

Inductor CircuitsInductor Circuits Quiz FridayQuiz Friday AC Next week & Following MondayAC Next week & Following Monday


Second Problem

a a

h

Both Currents are going intothe page.

B

OtherWire

ah

stuff

hahh

dh

d

ha

h

r

h

rr

Sin

Sinr

ISintotal

(max)0)2(

~

~1

~)(

~

)(2

2)(2

22

22

0

B

B

BB


First Problem

V

8/

8)2(2

)(22

2

2

2

1

22

222222

2

2

mM

eRReM

consteRV

RqB

Vq

RBqm

Vq

mqBR

v

qBRm

m

Vqv

R

mvqvB

mvVq


The Last two Problems were similar to WebAssigns that were also reviewed in class.Circular Arc – Easy Biot-Savart

Moving Rod


Question

What about these problems was “unfair”?

Why so many blank or completely wrong pages?


And Now …..

From the past


Max Current Rate ofincrease = max emfVR=iR

~current


constant) (time

)1( /

R

L

eR

Ei LRt

Solve the lo

op equation.


We also showed that

2

0

2

0

2

1

2

1

E

B

capacitor

inductor

u

u


LR Circuit

i

then0,ELet

0

equationcapacitor

theas form same

0

:0 drops voltageof sum

dt

dqR

C

qE

dt

diLiRE

Steady Source


Time Dependent Result:

R

L

eR

Ei LRt

constant time

)1( /


R

L


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


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


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

L

Rt

e 2

L

Rt

e 2



Mass on a Spring Result

Energy will swap back and forth. Add friction

Oscillation will slow down Not a perfect analogy



LC Circuit

High

Q/CLow

Low

High


The Math Solution (R=0):

LC


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.


-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


Sinusoidal Stuff

)sin( tAemf

“Angle”

Phase Angle


Same Frequencywith

PHASE SHIFT


Different Frequencies


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


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


The Real World


A



The Flux:

tAR

emfi

tBAemf

t

BA

bulb

sin

sin

cos

AB


April 12, 2006April 12, 2006


ScheduleSchedule

TodayToday Finish InductorsFinish Inductors

FridayFriday Quiz on this weeks materialQuiz on this weeks material Some problems and then AC circuitsSome problems and then AC circuits

MondayMonday Last FULL week of classesLast FULL week of classes Following Monday is last day of classFollowing Monday is last day of class

FINAL IS LOOMING!FINAL IS LOOMING!


Some ProblemsSome Problems


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.


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.


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?


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?


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?


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?


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?


Back to Variable Back to Variable SourcesSources


Source Voltage:

)sin(0 tVVemf


Average value of anything:

Area under the curve = area under in the average box

T

T

dttfT

h

dttfTh

0

0

)(1

)(

T

h


Average Value

T

dttVT

V0

)(1

0sin1

0

0 T

dttVT

V

For AC:


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


Average Value

0)(1

0

T

dttVT

V

2VVrms


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


Usually Written as:

2

2

rmspeak

peakrms

VV

VV


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

E

R

~


Power

tR

VRt

R

VRitP

tR

V

R

Vi

tVV

22

0

2

02

0

0

sin)sin()(

)sin(

)sin(


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


AC CircuitsAC Circuits

April 17, 2006April 17, 2006


Last Days …Last Days …

If you need to, file your taxes TODAY!If you need to, file your taxes TODAY! Due at midnight.Due at midnight.

Note: File on Web has been updated.Note: File on Web has been updated. This weekThis week

Monday & Wednesday – AC Circuits Monday & Wednesday – AC Circuits followed by problem based reviewfollowed by problem based review

Friday – Review problems Next Week Friday – Review problems Next Week Monday – Complete Problem review.Monday – Complete Problem review.


Final ExaminationFinal Examination

Will contain 8-10 problems. One will Will contain 8-10 problems. One will probably be a collection of multiple choice probably be a collection of multiple choice questions.questions.

Problems will be Problems will be similarsimilar to WebAssign to WebAssign problems. Class problems may also be a problems. Class problems may also be a source.source.

You have 3 hours for the examination.You have 3 hours for the examination. SCHEDULE: MONDAY, MAY 1 @ 10:00 AMSCHEDULE: MONDAY, MAY 1 @ 10:00 AM

http://pegasus.cc.ucf.edu/%7Eenrsvc/registrar/http://pegasus.cc.ucf.edu/%7Eenrsvc/registrar/calendar/exam/calendar/exam/

http://pegasus.cc.ucf.edu/%7Eenrsvc/registrar/calendar/exam/

http://pegasus.cc.ucf.edu/%7Eenrsvc/registrar/calendar/exam/


Back to ACBack to AC


Resistive Circuit

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


Con

sid

er

this

cir

cuit

CURRENT ANDVOLTAGE IN PHASE

R

emfi

iRe



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


v

V(t)

t

Vp

-Vp

V = VP sin (wt - v )Phase Term


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


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.




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




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




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


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


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.


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)


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


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.



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.


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.



Phasor Diagrams

Vp

Ipt

Vp

IpVp Ip

Resistor Capacitor Inductor




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



• 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


• 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


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.


Reactance - Phasor Diagrams

Vp

Ipt

Vp

Ip

t

Vp Ip

t



“Impedance” of an AC Circuit

R

L

C~

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

IV

Zpp (Units: OHMS)


“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.)


Impedance of an RLC Circuit

R

L

C~E

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

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



R

L

C~E

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

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

BUT: Voltages have different PHASES

they add as PHASORS.


Phasors for a Series RLC Circuit

Ip

VRp

(VCp- VLp)

VP

VCp

VLp



By Pythagoras’ theorem:

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

Ip

VRp

(VCp- VLp)

VP

VCp

VLp



By Pythagoras’ theorem:

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

= Ip2 R2 + (Ip XC - Ip XL)

2

Ip

VRp

(VCp- VLp)

VP

VCp

VLp



Solve for the current:

Ip Vp

R2 (Xc XL )2

Vp

Z

R

L

C~



Solve for the current:

Impedance:

Ip Vp

R2 (Xc XL )2

Vp

Z

Z R2 1

C L

2

R

L

C~


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


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


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


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.)


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 )



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



P t I V t tI V t t t

P P

P P

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

2





P P

P P


2

sin ( )

sin( ) cos( )

2 1

2

0

t

t t

Use:

and:

So P I VP P1

2cos





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



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.

Documents

Induction - Spring 20061 Time Varying Circuits April 10, 2006