induction1 time varying circuits 2009 induction 2 the final exam approacheth 8-10 problems similar...

Induction 1

Time Varying CircuitsTime Varying Circuits

20092009

Induction Induction 22

The Final Exam ApproachethThe Final Exam Approacheth

8-10 Problems 8-10 Problems similar to Web-similar to Web-AssignmentsAssignments

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

May contain some May contain some short answer short answer (multiple choice) (multiple choice) questions.questions.

Induction Induction 33

Spring 2009 Final Exam ScheduleTuesday, April 28 - Monday, May 4No Exams on Sunday

EXAMTIMES

Class Meeting Times

EXAM DAY 1TUES 4/28

EXAM DAY 2WED 4/29

EXAM DAY 3THURS 4/30

EXAM DAY 4FRI 5/1

EXAM DAY 5SAT 5/2

EXAM DAY 6MON 5/4

7:00 a.m. -9:50 a.m.

7:30-10:20 T 9:00-10:15 TR (all a.m.)

7:30-10:20 W 8:30-9:20 MWF 9:00-10:15 MW (all a.m.)

7:30-8:45 TR 7:30-10:20 R (all a.m.)

7:30-10:20 F 9:00-10:15 M/7:30-8:45 F 9:30-10:20 MWF (all a.m.)

Finals For Saturday Classes Are Held During Regular Class Meeting Times

7:30-8:20 MWF 7:30-8:45 MW 7:30-10:20 M (all a.m.)

10:00 a.m. -12:50 p.m.

10:30-11:45 TR 10:30-1:20 T

10:30-1:20 W 11:30-12:20 MWF 12:00-1:15 MW 12:00-1:15 WF

10:30-1:20 R 12:00-1:15 TR

10:30-1:20 F 12:30-1:20 MWF 12:00-1:15 M/10:30-11:45 F

Finals For Saturday Classes Are Held During Regular Class Meeting Times

10:30-11:20 MWF 10:30-11:45 MW 10:30-1:20 M

1:00 p.m. -3:50 p.m.

1:30-2:45 TR 1:30-4:20 T

1:30-4:20 W 2:30-3:20 MWF 3:00-4:15 WF 3:00-4:15 MW

1:30-4:20 R 3:00-4:15 TR

1:30-4:20 F 3:30-4:20 MWF 3:00-4:15 M/1:30-2:45 F

FREE PERIOD1:30-2:20 MWF 1:30-2:45 MW 1:30-4:20 M

4:00 p.m. -6:50 p.m.

6:00-7:15 TRand Alternate Time

6:00-7:15 MWand Alternate Time

4:30-5:45 TRand Alternate Time

4:30-7:15 Fand Alternate Time

FREE PERIOD4:30-5:45 MWand Alternate Time

7:00 p.m. -9:50 p.m.

6:00-8:50 T 7:30-10:20 T (all p.m.)

6:00-8:50 W 7:00-7:50 MWF 7:30-8:45 MW 7:30-10:20 W (all p.m.)

6:00-8:50 R 7:30-8:45 TR 7:30-10:20 R (all p.m.)

6:00-8:50 F 8:00-8:50 MWF (all p.m.)

FREE PERIOD6:00-6:50 MWF 6:00-8:50 M 7:30-10:20 M (all p.m.)

HowjaDo??

Induction 4

A. I done goodB. I done okC. I done not so okD. I screwed up major

The Test Itself Was

Induction 5

A. FairB. Not so fair.C. Really Unfair.D. The worst kind of unfair in the entire

universe.

Back to Circuits for a bit ….

Sort of like RC circuit issues.

Definition

Current in loop produces a magnetic fieldin the coil and consequently a magnetic flux.

If we attempt to change the current, an emfwill be induced in the loops which will tend tooppose the change in current.

This this acts like a “resistor” for changes in current!

Remember Faraday’s Law

dt

ddVemf

sE

Lentz

Look at the following circuit:

Switch is open NO current flows in the circuit. All is at peace!

Close the circuit…

After the circuit has been close for a long time, the current settles down.

Since the current is constant, the flux through

the coil is constant and there is no Emf. Current is simply E/R (Ohm’s Law)


When switch is first closed, current begins to flow rapidly.

The flux through the inductor changes rapidly. An emf is created in the coil that opposes the

increase in current. The net potential difference across the resistor

is the battery emf opposed by the emf of the coil.


dt

demf

0

)(

dt

diRV

notationVEbattery

Moving right along …

0

solonoid, aFor

N. turns,ofnumber the toas wellas

current the toalproportion isflux The

0

)(

dt

diLiRV

dt

diL

dt

d

NLii

dt

diRV

notationVE

B

battery

Definition of Inductance L

i

NL B

UNIT of Inductance = 1 henry = 1 T- m2/A

is the flux near the center of one of the coilsmaking the inductor

Consider a Solenoid

n turns per unit lengthniB

or

nliBl

id enclosed

0

0

0

sBl

So….

AnlL

or

AlnL

ori

niAnl

i

nlBA

i

NL B

2

20

0

lengthunit

inductance/

Depends only on geometry just like C andis independent of current.

Inductive Circuit Switch to “a”. Inductor seems like a

short so current rises quickly.

Field increases in L and reverse emf is generated.

Eventually, i maxes out and back emf ceases.

Steady State Current after this.

i

THE BIG INDUCTION

As we begin to increase the current in the coil The current in the first coil produces a

magnetic field in the second coil Which tries to create a current which will

reduce the field it is experiences And so resists the increase in current.

Back to the real world…

i

0

equationcapacitor

theas form same

0

:0 drops voltageof sum

dt

dqR

C

qE

dt

diLiRE

Switch to “a”

Solution (See textbook)

R

L

eR

Ei LRt

constant time

)1( /

Switch position “b”

/

0

0

teR

Ei

iRdt

diL

E

Max Current Rate ofincrease = max emfVR=iR

~current

constant) (time

)1( /

R

L

eR

Ei LRt

Solve the lo

op equation.

IMPORTANT QUESTION

Switch closes. No emf Current flows for a while It flows through R Energy is conserved

(i2R)

WHERE DOES THE ENERGY COME FROM??

For an answerReturn to the Big C

We move a charge dq from the (-) plate to the (+) one.

The (-) plate becomes more (-)

The (+) plate becomes more (+).

dW=Fd=dq x E x d+q -q

E=0A/d

+dq

The calc

2

0

2

020

2

00

22

0

2

00

00

2

1

eunit volum

energy

2

1

2

1

2

1)(

2

2

)()()(

E

E

u

AdAdAd

AA

dW

or

q

A

dqdq

A

dW

dA

qdqddqEddqdW

The energy is inthe FIELD !!!

What about POWER??

Ridt

diLiiE

i

iRdt

diLE

2

:

powerto

circuit

powerdissipatedby resistor

Must be dWL/dt

So

2

2

2

12

1

CVW

LiidiLW

dt

diLi

dt

dW

C

L

L

Energystoredin the

Capacitor

WHERE is the energy??

l

Al

NiBA

l

Ni

niB

nilBll

id enclosed

0

0

0

0

0

B

or

0

sB

Remember the Inductor??

turn.onegh flux throu MagneticΦ

current.

inductorin turnsofNumber

i

Ni

NL

So …

l

AiN

l

NiANiW

l

NiA

iNi

NiLiW

L

Ni

i

NL

2220

0

0

0

22

2

1

2

1

2

1

2

1

2

1

2

0

2

0

22

0

0

2220

0

2

1

or

(volume) 2

1

2

1

B

:before From

2

1

BV

Wu

VBl

AlBW

l

Ni

l

AiNW

ENERGY IN THEFIELD TOO!

IMPORTANT CONCLUSION

A region of space that contains either a magnetic or an electric field contains electromagnetic energy.

The energy density of either is proportional to the square of the field strength.

Solution (From Before)

R

L

eR

Ei LRt

constant time

)1( /

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

Induction 35

The math …

Induction 36

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

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

The Graph of that LR (no emf) circuit

Induction 37

L

Rt

e 2

L

Rt

e 2

I

Induction 38

Mass on a Spring Result

Induction 39

Energy will swap back and forth. Add friction Oscillation will slow down Not a perfect analogy

Induction 40

LC Circuit

Induction 41

High

Q/CLow

Low

High

The Math Solution (R=0):

LC

Induction 42

New Feature of Circuits with L and C

Induction 43

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.

Variable Emf Applied

Induction 44

-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

emf

Sinusoidal

DC

Sinusoidal Stuff

Induction 45

)sin( tAemf

“Angle”

Phase Angle

Induction 46

Same Frequencywith

PHASE SHIFT

Different Frequencies

Induction 47

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

Induction 48

Producing AC Generator

Induction 49

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

Induction 50

Induction 51

A

Induction 52

The Flux:

Induction 53

tAR

emfi

tBAemf

t

BA

bulb

sin

sin

cos

AB

problems …

Induction 54

Induction 55

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 56

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 57

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 58

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 59

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 60

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 61

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?

Source Voltage:

Induction 62

)sin(0 tVVemf

Induction 63

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

Induction 64

0sin1

0

0 T

dttVT

V

T

dttVT

V0

)(1

For AC:

So …

Induction 65

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

Induction 66

0)(1

0

T

dttVT

V

2VVrms

RMS

Induction 67

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:

Induction 68

2

2

rmspeak

peakrms

VV

VV

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

Induction 69

E

R

~

Power

Induction 70

tR

VRt

R

VRitP

tR

V

R

Vi

tVV

22

0

2

02

0

0

sin)sin()(

)sin(

)sin(

More Power - Details

Induction 71

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

Resistive Circuit

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

Induction 72

Con

sid

er

this

cir

cuit

Induction 73

CURRENT ANDVOLTAGE IN PHASE

R

emfi

iRe

Induction 74

Induction 75

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 76

v

V(t)

t

Vp

-Vp

V = VP sin (t - v )Phase Term

Induction 77

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 78

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 79



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

Induction 80



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 81



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 82

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 83

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 84

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 85

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 86

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 87

Induction 88

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 89

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 90

Phasor Diagrams

Vp

Ipt

Vp

IpVp Ip

Resistor Capacitor Inductor



Induction 91

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 92


• 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 93

• 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 94

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 95

Reactance - Phasor Diagrams

Vp

Ipt

Vp

Ip

t

Vp Ip

t


Induction 96

“Impedance” of an AC Circuit

R

L

C~

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

IV

Zpp (Units: OHMS)

Induction 97

“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 98

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 99


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 100

Phasors for a Series RLC Circuit

Ip

VRp

(VCp- VLp)

VP

VCp

VLp

Induction 101


By Pythagoras’ theorem:

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

Ip

VRp

(VCp- VLp)

VP

VCp

VLp

Induction 102


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 103


Solve for the current:

Ip Vp

R2 (Xc XL )2

Vp

Z

R

L

C~

Induction 104


Solve for the current:

Impedance:

Ip Vp

R2 (Xc XL )2

Vp

Z

Z R2 1

C L

2

R

L

C~

Induction 105

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 106

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 107

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 108

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 109

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 110


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

Induction 111


P t I V t tI V t t t

P P

P P

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

2


Induction 112



P P

P P


2

sin ( )

sin( ) cos( )

2 1

2

0

t

t t

Use:

and:

So P I VP P1

2cos


Induction 113



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 114


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 115

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 116

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 117

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 118

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?

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