lecture 1410

8/2/2019 Lecture 1410

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Chapter 31

Electromagnetic Oscillations and AlternatingCurrent

In this chapter we will cover the following topics:

-Electromagnetic oscillations in an LC circuit

-Alternating current (AC) circuits with capacitors

-Resonance in RCL circuits

-Power in AC-circuits-Transformers, AC power transmission

(31 - 1)

8/2/2019 Lecture 1410


Suppose this page is perpendicular to a uniform magnetic field

and the magnetic flux through it is 5Wb. If the page is turned by

30 around an edge the flux through it will be:

A. 2.5Wb

B. 4.3Wb

C. 5Wb

D. 5.8Wb

E. 10Wb

8/2/2019 Lecture 1410


A car travels northward at 75 km/h along a straight road in a

region where Earth’s magnetic field has a vertical component of

0.50 × 10−4 T. The emf induced between the left and right

side, separated by 1.7m, is:

A. 0

B. 1.8mV

C. 3.6mVD. 6.4mV

E. 13mV

8/2/2019 Lecture 1410


L

C The circuit shown in the figure consists of a capacitor

and an inductor . We give the capacitor an initial

chanrge and then abserve what happens. The capacitor will discharge th

C

L

Q

LC Oscillations

rough the inductor resulting in a time

dependent current .i

We will show that the charge on the capacitor plates as well as the current

1in the inductor oscillate with constant amplitude at an angular frequency

The total energy in the circuit is t

q i

LC

U

ω =

2 2

he sum of the energy stored in the electric field

of the capacitor and the magnetic field of the inductor. .2 2

The total energy of the circuit does not change with time. Thus

E B

q LiU U U

C dU

= + = +

2

2

2

2

0

0.1

0

dt

dU q dq di dq di d q Li i

dt C d

d q L

t dt dt q

dt dt dt C +

=

= + = = → → == (31 - 2)

8/2/2019 Lecture 1410


L

C

2

2

2

2

10 ( )

This is a homogeneous, second order, linear differential equation

which we have encountered previously. We used it to d

10

escribe

the simple harmonic oscillat

o

d q

L qdt C

d qq

dt LC

+ == →

+

eqs.1

22

20

with sol

r (SHO)

( )

ution: ( ) cos( )

d x x

dt

x t X t

ω

ω φ

+ =

= +

eqs.2

( )

If we compare eqs.1 with eqs.2 we find that the solution to the differential

equation that describes the LC-circuit (eqs.1) is:1

( ) cos where , and is the phase angle.

The current

q t Q t LC

ω φ ω φ = + =

( )sindq

i Q t dt

ω ω φ = = − +

( )( ) cosq t Q t ω φ = +

1 LC

ω =

(31 - 3)

8/2/2019 Lecture 1410


L

C

( )

( ) ( )

2 22

2 2 2 22 2

2

The energy stored in the electric field of the capacitor

cos2 2

The energy stored in the magnetic field of the inductor

sin sin2 2 2

The total energy

2

E

B

E B

q QU t

C C

Li L Q QU t t

C

U U U

QU

ω φ

ω ω φ ω φ

= = +

= = + = +

= +

= ( ) ( )2

2 2cos sin2

The total energy is constant;

Qt t C C

ω φ ω φ + + + =

energy is conserved

2

2

3The energy of the has a value of at 0, , , , ...2 2 2

3 5The energy of the has a value of at , , , ...

2 4 4 4

When is maximum is ze E B

Q T T t T C

Q T T T t

C

U U

=

=

electric field maximum

magnetic field maximum

Note : ro, and vice versa

(31 - 4)

8/2/2019 Lecture 1410


0t =

1

2

/ 8t T =

3

/ 4t T =

4

3 / 8t T =

5

5/ 2t T =

432

1

6

6

5 / 8t T =

3 / 4t T =

7 / 8t T =

7

8

7

8

(31 - 5)

8/2/2019 Lecture 1410


2

2

If we add a resistor in an RL cicuit (see figure) we must

modify the energy equation because now energy is

being dissipated on the resistor.

2 E B

dU i Rdt

qU U U

C

= −

= + = +

Damped oscillations in an RCL circuit

22

2

Li dU q dq di Li i R

dt C dt dt → = + = −

( )

2

2

2

/ 2

2

2 2

1 0 This is the same equation as that

of the damped harmonics o 0 which hscillator:

The a

as the solution

( ) co ngul r f s a

:

bt mm

dq di d q d q dqi L R qdt dt dt dt d

d x dxm b kx

dt dt

x t x e t

t C

ω φ −

+ + =

′= +

= → = → + + =

( )

2

2

2

2

/ 2 1( )

requency

For the damped RCL circuit the solut

cos

ion is:

The angular fre que

4

ncy

4

Rt L Rq

k bm m

t Qe t

LC L

ω

ω φ ω − ′ ′= + = −

′ = −

(31 - 6)

8/2/2019 Lecture 1410


/ 2 Rt LQe

−

/ 2 Rt LQe

−

( )q t Q

Q−

( )q t ( )/ 2( ) cos Rt Lq t Qe t ω φ

− ′= +

2

2

1

4

R

LC Lω ′ = −

/ 2

2

2

The equations above describe a harmonic oscillator with an exponetially decaying

amplitude . The angular frequency of the damped oscillator

1is always smaller than the angular f

4

Rt LQe

R

LC Lω

−

′ = −

2

2

1requency of the

1undamped oscillator. If the term we can use the approximation

4

LC R

L LC

ω

ω ω

=

′ ≈=

(31 - 7)

8/2/2019 Lecture 1410


A battery for which the emf is constant generates

a current that has a constant direction. This type

of current is known as " " or " "

In chapter 30 we encountered a d

Alternating Current

dcdirect current

ifferent type

of sourse (see figure) whose emf is:

sin sin where , is the area of the generator

windings, is the number of the windings, is the angular frequency of the

rotation of the windings, and is the magnetic field.

m m NAB t t NAB A

N

B

ω ω ω ω

ω = = =E E E

This type of generator

is known as " " or " " because the emf as well as the current

change direction with a frequency 2 . In the US 60 Hz.

Almost all commercial electrical

f f πω = =acalternating current

power used today is ac even though the

analysis of ac circuits is more complicated than that of dc circuits.

The reasons why ac power was adapted will be discussed at the end of this

chapter.

(31 - 8)

sinm t ω =E E

8/2/2019 Lecture 1410


L

C

Our objective is to analyze the circuit shown in the

figure ( circuit). The discussion will be greatly

simplified if we examine what happens if we connect

each of the three elem

RCL

Three Simple Circuits

ents ( , , and ) separately

to an ac generator.

R C L

From now on we will use the standard notation for ac circuitanalysis. Lower case letters will be used to indicate the

values of ac quantities. Upper case letters

will be used

A convention

instantaneous

( )

to indicate the constant amplitudes of ac quantities.

Example: The capacitor charge in an LC circuit was written as:cos

The symbol is used for the instantaneous value of the charge

The symbol

q Q t

q

ω φ = +

is used for the constant amplitude of Q q

(31 - 9)

8/2/2019 Lecture 1410


In fig.a we show an ac generator connected to a resistor

From KLR we have: 0 sin

The current amplitude

The voltage across is equal to sin

The voltage

m

R R

m R

R m

R

i R i t R R

I R

v R t

ω

ω

− = → = =

=

A resistive load

EEE

E

E

amplitude is equal toThe relation between the voltage and

current amplitudes is:

In fig.b we plot the resistor current and the

resistor voltage as function of time t.Both quanti

m

R

R

R RV I R

i

v

=

E

ties reach their maximum values

at the same time. We say that voltage and

current are .in phase

(31 - 10)

R RV I R=

8/2/2019 Lecture 1410


A convenient method for the representation of ac

quantities is that of phasors

The resistor voltage and the resistor current are represented

by rotating vectors known as phasors using the following conventions:

Phasors rotate in the counterclockwise direction with angula

R Rv i

1. r speed

The length of each phasor is proportional to tha ac quantity amplitude

The projection of the phasor on the vertical axis gives the instantaneous

value of the ac quantity.

The rotation

ω

2.

3.

4. angle for each phasor is equal to the phase of the

ac quantity ( in this example)t ω (31 - 11)

8/2/2019 Lecture 1410


2 2

We define the "root mean square" (rms) value of as follows:

The equation looks the same

as in the DC case. This power appears a

s

2

heat on

rmsmrms

V

V P

R

R

V ≡ → ⟨ ⟩ =E

2

0

2

2

0

22

0

2

1( ) sin

1

1

si

1sin

n

22

T

m

T m

m

T

P P t dt P t T R

td

P tdt R

t T

T

P R

ω

ω

ω

⟨ ⟩ = = →

⟨

⟨ ⟩ =

⟩

=

=

→

∫

∫

∫

Average Power for R

E

E

E

2

2

m P R

⟨ ⟩ =E

2

2

mrmsV ≡

E

(31 - 12)

8/2/2019 Lecture 1410


( )

In fig.a we show an ac generator connected to a

capacitor

From KLR we have: 0

sin

cos sin 90

The voltage amplitude equal to

The current am

C

C m

C C m

C m

C

qC

q C C t

dqi C tdt t

dt

V

ω

ω ω ω

− = →

= =

= = = + °

A capacitive load

E

E E

E

E

plitude1/

The quantity is known as the

In fig.b we plot the capacitor current and the capacitor

voltage as function of time t

1

. The current the

v

/

C C

C

C

C

C

V I CV

C

i

v

X C ω

ω ω

= =

=

leads

capacitive reactance

oltage by a quarter of a period. The voltage and

current are .°out of phase by 90

O ω

C X

(31 - 13)

1

C

X C ω =

8/2/2019 Lecture 1410


2

2

2

0 0

sin cos

sin 22

1 1( ) = sin 2 0

2

A capacitor does not dissipate any power

on the average. In some parts o

2sin cos sin 2

h

f t

mC C

C

m

C

T T

m

C

P V I t t X

P t X

P P t dt tdt T X T

θ θ θ

ω ω

ω

ω

= =

→ =

⟨ ⟩ = =

=

∫ ∫

Average Power for C

Note :

E

E

E

e cycle it absorbes

energy from the ac generator but at the rest of the cycle

it gives the energy back so that on the average no

power is used!

0C P =

(31 - 14)

8/2/2019 Lecture 1410


( )

In fig.a we show an ac generator connected to an inductor

From KLR we have: 0 sin

sin cos sin 90

The voltage amplitude equa

m L L

m m m L L

L

L

di di L t

dt dt L L

i di tdt tdt t L L L

V

ω

ω ω ω ω ω

− = → = =

= = = − = − °∫ ∫

An inductive load

EEE

E E E

l to

The current amplitude

The quantity is known as the

In fig.b we plot the inductor current and the

inductor voltage as function of time t.

The current

m

L

L

L

L

L

V

I L

i

v

X L

ω

ω

==

E

inductive reactance

the voltage by a

quarter of a period. The voltage and

current are .°

lags behind

out of phase by 90

O

ω

L X

(31 - 15)

L X Lω =

8/2/2019 Lecture 1410


2

2

2

0 0

Power sin cos

sin22

1 1( ) = sin 2 0

2A inductor does not dissipate any power

on

2sin cos sin 2

the average. In some p rt

a s

m L L

L

m

L

T T

m

L

P V I t t

X

P t X

P P t dt tdt

T X T

ω ω

ω

ω

θ θ θ

= = −

→ = −

⟨ ⟩ ≡ − =

=

∫ ∫ Note :

E

E

E

Average Power for L

of the cycle it absorbes

energy from the ac generator but at the rest of the cycle

it gives the energy back so that on the average no

power is used!

0 L P =

(31 - 16)

8/2/2019 Lecture 1410


Circuitelement AveragePower Reactance Phase of current Voltage amplitude

Resistor

R

Current is in phase with the voltage

Capacitor

C

Current leads voltage by a quarter of a period

Inductor

L

Current lags behind

voltage by a quarter of a period

1C X

C ω =

L X Lω =

R RV I R=

C C C C

I V I X

C ω = =

L L L LV I X I Lω = =

R

SUMMARY

2

2

m R P

R⟨ ⟩ =

E

0C P ⟨ ⟩ =

0 L P ⟨ ⟩ =

(31 - 17)

8/2/2019 Lecture 1410


An ac generator with emf is connected to

an in series combination of a resistor , a capacitor

and an inductor , as shown in the figure. The phasor

for the ac genera

sinm

R C

L

t ω =The series RCL circuit

E E

( )

tor is given in fig.c. The current in

this circuit is described by the equation: sini I t ω φ = −( )sini I t ω φ = −

The current is for the resistor, the capacitor and the inductor

The phasor for the current is shown in fig.a. In fig.c we show the phasors for the

voltage across , the voltage across R C

i

v R v C

common

, and the voltage across .

The voltage is in phase with the current . The voltage lags behind

the current by 90 . The voltage leads ahead of the current by 90 .

L

R C

L

v L

v i v

i v i° °

(31 - 18)

8/2/2019 Lecture 1410


O

A B

Kirchhoff's loop rule (KLR) for the RCL circuit: . This equation

is represented in phasor form in fig.d. Because and have opposite directions

we combine the two in a single phasor

R C L

L C

L

v v v

V V

V

= + +E

( ) ( ) ( ) ( )

( )

( )

2 2 2 22 2 2 2

22

22

. From triangle OAB we have:

The denominator is known as the " "

of the circuit. The current amplitude

C

m R L C L C L C

m

L C

m L C

V

V V V IR IX IX I R X X

I Z R X X

Z R X X I

− = + − = + − = + − →

=+ −

= + − → =

E

E

E

impedance

2

2

1

m

Z

I

R L C ω ω

=

+ −

E

( )22 L C Z R X X = + −

m I Z

= E

( )sini I t ω φ = −

(31 - 19)

8/2/2019 Lecture 1410


O

A B

From triangle OAB we have: tan

We distinguish the following three cases depending on the relative values

of and .

0 The current phasor lags behind the generat

L C L C L C

R

L L

L C

V V IX IX X X V IR R

X X

X X

φ

φ

− − −= = =

> → >1. or phasor.The circuit is more inductive than capacitive

0 The current phasor leads ahead of the generator phasor

The circuit is more capacitive than inductive

0 The current phaso

C L

C L

X X

X X

φ

φ

> → <

= → =

2.

3. r and the generator phasor are in phase

( )sini I t ω φ = −

( )22 L C Z R X X = + −

L X Lω =

tan L C X X R

φ −=

1 C X

C ω =

(31 - 20)

8/2/2019 Lecture 1410


Fig.a and b: 0

The current phasor lags behindthe generator phasor. The circuit is more

inductive than capacitive

L C X X φ > → >1.

Fig.c and d: 0 The current phasor leads ahead of the generator

phasor. The circuit is more capacitive than inductive

Fig.e and f: 0 The current phasor and the generator ph

C L

C L

X X

X X

φ

φ

> → <

= → =

2.

3. asor are

in phase (31 - 21)

R

8/2/2019 Lecture 1410


In the RCL circuit shown in the figure assume that

the angular frequency of the ac generator can

be varied continuously. The current amplitude

in the circuit is given by the equation:

m I

ω

=

Resonance

E

2

2

The current amplitude1

1has a maximum when the term 0

1This occurs when

R LC

LC

LC

ω ω

ω ω

ω

+ −

− =

=

The equation above is the condition for resonance. When its is satisfied

A plot of the current amplitude as function of is shown in the lower figure.

This plot is known as a "

mres I

R

I ω

=E

resonance c "urve

mres I

R=

E1

LC ω =

(31 - 22)

2E

8/2/2019 Lecture 1410


2

We already have seen that the average power used by

a capacitor and an inductor is equal to zero. The

power on the average is consumed by the resistor.

The instantaneous power P i=

Power in an RCL ciruit

( )

( )

2

0

222 2

0

sin

1

The average power

1sin

2

cos

The term cos in the equation above is known as

the "

T

avg

T

avg rms

rmsavg rms rms rms rms rms rms rms

R I t R

P Pdt T

I R P I R t dt I R

T

R P I RI I R I I Z Z

ω φ

ω φ

φ

φ

= −

=

= − = =

= = = =

∫ ∫

E

E E

power fac " of the circuit. The average

power consumed by the circuit is maximum

when 0φ =

tor

2

avg rms P I R= cosavg rms rms P I φ = E (31 - 23)

8/2/2019 Lecture 1410


Power Station

Transmission lines

rms=735 kV , I rms = 500

A

home

110 V

T1T2

Step-up

transformer

Step-down

transformer

R = 220Ω

1000 km=l

2

The resistance of the power line . is fixed (220 in our example)

Heating of power lines This parameter is also fixed

( 55 MW in our exa

heat rms

R R A

P I R

ρ = Ω

=

Energy Transmission Requirements

l

mple)

Power transmitted (368 MW in our example)In our example is almost 15 % of and is acceptable

To keep we must keep as low as possible. The only wa

trans rms rms

heat trans

heat rms

P I P P

P I

=E

y to accomplish this

is by . In our example 735 kV. To do that we need a device

that can change the amplitude of any ac voltage (either increase or decrease)

rms rms =increasing E E

(31 - 24)

8/2/2019 Lecture 1410


The transformer is a device that can change

the voltage amplitude of any ac signal. It

consists of two coils with different number

of turns wound around a common iron core.

The transformer

The coil on which we apply the voltage to be changed is called the " " and

it has turns. The transformer output appears on the second coils which is knownas the "secondary" and has turns

P

S

N N

primary

. The role of the iron core is to insure that the

magnetic field lines from one coil also pass through the second. We assume that

if voltage equal to is applied across the primary then a voltag P V e appears

on the secondary coil. We also assume that the magnetic field through both coils

is equal to and that the iron core has cross sectional area A. The magnetic flux

through the primary

S

P

V

B

Φ ( )

The flux through the secondary ( )

P P P P

S

S S S S

d dB N BA V N A

dt dt

d dB N BA V N A

dt dt

Φ= → = − = −

ΦΦ = → = − = −

eqs.1

eqs.2

(31 - 25)

V V

8/2/2019 Lecture 1410


( )

( )

If we divide equation 2 by equation 1 we get:

P

S P

P P P P

S S S S S

S S S

P P S P P

d dB N BA V N A

dt dt

d dB N BA V N A

dt dt

dB N A

V N dt dBV N

N A

V

N

dt

V

N

ΦΦ = → = − = −

ΦΦ = → = − = −

−= =

−=→

eqs.1

eqs.2

The voltage on the secondary

If 1 We have what is known a " " transformer

If 1 We have what is known a " " transformer

Both types of tran

S S P

P

S S P S P

P

S S P S P

P

N V V

N

N N N V V

N

N N N V V

N

=

> → > → >

< → < → <

step up

step down

sformers are used in the transport of electric power over large

distances.

S P

S P

V V

N N =

(31 - 26)

V V

8/2/2019 Lecture 1410


P I S I

If we close switch S in the figure we have in addition to the primary current

a current in the secondary coil. We assume that the transformer is " "

i.e. it suffers no losses due to heating

P

S

I

I ideal

then we have: (eqs.2)

If we divide eqs.2 with eqs.1 we get:

In a step-up transformer ( ) we have that

In a step-down transformer ( )

P P S

P P S S

S S P P

P S S P

P S P

S

S P S P

S P

S

V I V I

V I V I

V N V N

N I I

N

N

I N I N

N I I

N N

=

= →

=

><

=

<we have that S P I I >

We have that:

(eqs.1)

S P

S P

S P P S

V V

N N

V N V N

=

→ =

S P

S P

V V

N N =

S S P P I N I N =

(31 - 27)

8/2/2019 Lecture 1410


Hitt

A generator supplies 100 V to the primary coil of a

transformer. The primary has 50 turns and the

secondary has 500 turns. The secondary voltage is:A. 1000 V

B. 500 V

C. 250 V

D. 100 V

E. 10V

8/2/2019 Lecture 1410


hitt

The main reason that alternating current replaced

direct current for general use is:

A. ac generators do not need slip ringsB. ac voltages may be conveniently transformed

C. electric clocks do not work on dc

D. a given ac current does not heat a power line asmuch as the same dc current

E. ac minimizes magnetic effects

lecture 1410

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