Alternating Currentv3

8/18/2019 Alternating Currentv3

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MagnetismAlternating-Current Circuits


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Alternating Current Generators

Alternating current motor

A coil of area A and N turns rotating with constant angular velocity in a uniform

magnetic field produces a sinusoidal emf

Instead of mechanically rotating, we can apply an ac potential difference

generated by other ac generator to the coil. This produces an ac current in the

coil, and the magnetic field exerts forces on the wires producing a torque that

rotates the coil.


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The magnetic field.Magnetic forces on moving charges.

Magnetic forces on a current element

Torques on current loops and magnets


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The Magnetic Field

Magnets

The Earth is a natural magnet with

magnetic poles near the north and

south geographic poles.


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The Magnetic Field

Magnets Does an isolated magnetic

pole exist?

The SI units of magnetic

field is the tesla [T]

Earth magnetic field isabout 10-4 T

Powerful laboratories

produce fields of 1 -2 T as a

maximum

1 Gauss [G] = 10-4 Tesla [T]


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The Magnetic Field


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The Magnetic Field. Magnetic force on a moving charge

A proton is moving in a region of crossed fields E =

2x105 N/C and B = 3000 G, as shown in the figure.

(a) What is the speed of the proton if it is not

deflected. (b) If the electric field is disconnected,

draw the path of the proton


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The Magnetic Field. Magnetic force on a current element

In the case of a straight segment

of length L


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The Magnetic Field. Torques on Current loops and

Magnets

A circular loop of radius 2 cm has 10 turns of wire and carries a current of 3 A. The

axis of the loop makes an angle of 30º with a magnetic field of 8000 G. Find the

magnitude of the torque on the loop.

B

A current-carrying loopexperiences no net force in a

uniform magnetic field, but it does

experience a torque that tends to

twist the loop


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Sources of Magnetic Field

The Magnetic Field of Moving Charges

The Magnetic Field of Currents. The Biot-

Savart Law

The Magnetic Field Due to a Current loop


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Moving Point Charges are the source of Magnetic Field


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A solenoid is a wire tightlywounded into a helix of

closely space turns . A

solenoid is used to produce a

strong, uniform magnetic

field in the region surrounded

by the loops

For a long

solenoid Find the magnetic field at the center of asolenoid of 600 turns, length 20 cm;

radius 1.4 cm that carries a current of 4 A


A solenoid

In this figure, the length is ten

times longer than the radius

n = N/L; N number of turns;

L : length of solenoid


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Sources of Magnetic Field: Solenoid and magnets

Left: Magnetic field lines of a solenoid; right Magnetic field lines of a bar magnet


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Magnetic Induction

Magnetic Flux

Induced EMF and Faraday´s Law

Lenz´s Law


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MAGNETIC INDUCTION

In 1830, Michel Faraday in England and Joseph Henry in the USA

independently discovered that in a changing magnetic field, a changing

magnetic flux through a surface bounded by a stationary loop of wire induces

a current in the wire: emf induced and induced current. This process is

known as induction.

In a static magnetic field, a changing magnetic flux through a surface bounded

by a moving loop of wire induces an emf in the wire: motional emf

Magnetic flux through a surface bounded by a loop of wire.

The SI unit for magnetic flux is

weber [Wb] 1Wb= 1 T·m2 Find the magnetic flux through a 40 cm longsolenoid with a 2.5 cm radius and 600 turns

carrying a current of 7.5 A.


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The induced emf is in such direction as to

oppose, or tend to oppose, the change that

produces it. Lenz´s Law

MAGNETIC INDUCTION


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The coil with many turns of wire gives

a large flux for a given current in thecircuit. Thus, when the current

changes, there is a large emf induced

in the coil opposing the change. This

self-induced emf is called a back emf

MAGNETIC INDUCTION


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Eddy Currents

Heat produced by eddy currents constitute a power loss in a

transformer. But eddy currents have some practical applications:

damping mechanical oscillations, magnetic braking system


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InductanceSelf-inductance The SI unit of

inductance is

the henry [H]

1 H = 1 Wb/A= 1 T.m2

.A-1

Find the self-inductance of a solenoid of length 10 cm,

area 5 cm2, and 100 turns


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Magnetic Energy


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Alternating Current Generators

Alternating current motor

A coil of area A and N turns rotating with constant angular velocity in a uniform


Instead of mechanically rotating, we can apply an ac potential difference

generated by other ac generator to the coil. This produces an ac current in the

coil, and the magnetic field exerts forces on the wires producing a torque that

rotaes the coil.


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Alternating Current GeneratorsA coil of area A and N turns rotating with constant angular velocity in a uniform


f f frequency

t

t NBAdt d

t NBA

t and NBA

peak

m

m

m

π ω

δ ω ε ε

δ ω ω φ ε

δ ω φ

δ ω θ θ φ

2;

)sin(

)sin(

)cos(

cos

=

+=

+=−=

+=+==


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Alternating Current Circuits:Alternating current in a Resistor

Inductors in Alternating CurrentsCapacitors in Alternating Currents


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Potential drop across the resistor, VRCurrent in the resistor I

Power dissipated in the resistor, P

Average power dissipated in the

resistor Paverage

Alternating currents in a Resistor

R

V I t

R

V I

t V t V

peak R

peak

peak R

peak R R

,,

,max

cos

coscos

=⇒=

===

ω

ω ω ε ε

Rt I R I P

Rt I R I P

av peak avav

peak

))(cos()(

)(cos

222

222

ω

ω

==

==


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Root-Mean-Square Values


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Inductors in Alternating Current Circuits

The potential drop across the inductor

led the current 90º (out of phase)

Instantaneous power delivered by the emf to the inductor is not zero

The average power delivered by the emf to the inductor is zero.

)2

cos(sincos

coscos

,,,

,max

π ω

ω ω

ω ω

ω ω ε ε

−==→=

====

t L

V t

L

V I t

L

V

dt

dI

t V t dt

dI LV

peak L peak L peak L

peak L L

INDUCTIVE

REACTANCE L

V I

L

V I

rms L

rms

peak L

peak ω ω

,,; ==


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Inductors in Alternating Current Circuits

Instantaneous power delivered by the emf to the inductor is not zero

The average power delivered by the emf to the inductor is zero.

INDUCTIVE

REACTANCE

L

V I

L

V I

rms L

rms

peak L

peak ω ω

,,; ==

The potential drop across a 40-mHinductor is sinusoidal with a peak

potential drop of 120 V. Find the

inductive reactance and the peak current

when the frequency is (a) 60 Hz, and

(b) 2000 Hz


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Capacitors in Alternating Current CircuitsThe

potential

drop lagsthe

current

by 90º

Power delivered by the emf in the capacitor: Instantaneous and average

CAPACITIVE

REACTANCE

)2

cos(1

sin

cos

coscos

,

,

,

,max

π ω

ω

ω ω

ω

ω ω ε ε

+

=→−==

=

⇒====

t

C

V I t CV

dt

dQ I

t CV Q

t V t C

QV

peak C

peak C

peak C

peak C C

=

=

C

V I

C

V I

rmsC

rms

peak C

peak

ω ω 1

;1

,,


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Driven RLC Circuits

Series RLC circuit The Kirchhoff´s rules governthe behavior of potentialdrops and current across the

circuit.

(a)When any closed-loop is

traversed, the algebraic sum

of the changes of potentialmust equal zero (loops rule)

(b)At any junction (branch

point) in a circuit where the

current can be divided, the

sum of the currents into the

junction must equal the sum

of the currents out of the

junction (junction rule)


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)cos(

cos

;cos

,

2

,

δ ω

ω

ω

−=

=++

=++=

t I I

t V C

Q R

dt

dQ

dt

Qd L

dt

dQ

I C

Q

IRdt

dI

Lt V

peak

peak app

peak app

Series RLC circuits


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Power delivered to the series RLC circuit

δ δ

δ

δ ε δ ε

π ω ω ε ε

coscos2

1

/cos/

coscos2

1

)2

cos(cos

,,

,

2

2

,

2

2

rmsrmsapp peak peak app

peak app peak

rmsapprmsav

rmsrms peak peak av

peak peak

I V I V P

Z V I and Z Ras Z

RV RI P

resistor theindissipated RI P

I I P

t I t I P

==

==

==

=

==

−==

Power factor:cosδ


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Phasors

Potential drop across a resistor can be

represented by a vector VR, which is called a

phasor. Then, the potential drop across the

resistor IR, is the x component of vector VR,

Potential drop across aseries RLC circuit

C

Q IR

dt

dI Lt V app ++=ω cos


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In the circuit shown in the figure, the

ac generator produces an rms voltageof 115 V when operated at 60 Hz. (a)

What is the rms current in the circuit

(b) What is the power delivered by

the ac generator (c) What is the rms

voltage across: points AB; points

BC; points CD; points AC; pointsBD?.

A certain electrical device draws 10 A rms and has an average power of 720 W

when connected to a 120-V rms 60-Hz power line. (a ) What is the impedance

of the device? (b) What series combination of resistance and reactance is thisdevice equivalent to? (c) If the current leads the emf, is the reactance inductive

or capacitive?


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The Transformer

Because of the iron core, there is a large

magnetic flux through each coil, even when

the magnetizing current Im in the primary

circuit is very small .

The primary circuit consists of an ac

generator and a pure inductance (we considera negligible resistance for the coil). Then the

average power dissipated in the primary coil

is zero. Why?: The magnetizing current in

the primary coil and the voltage drop across

the primary coil are out of phase by 90º1

1

2

222

11

V N

N V N V

N V

dt

d

dt

d

turn

turn

=→=

=

φ

φ

Secondary coil open circuit

The potential drop across the

primary coil is

If there is no flux leakage out of the ironcore, the flux through each turn is the

same for both coils, and then

A transformer is a device to raise or lower

the voltage in a circuit without an

appreciable loss of power. Power losses

arise from Joule heating in the smallresistances in both coils, or in currents

loops (eddy currents) within the iron core.

An ideal transformer is that in which these

losses do not occur, 100% efficiency.

Actual transformers reach 90-95%

efficiency


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The Transformer

rmsrmsrmsrms I V I V

I N I N

,2,2,1,1

2211

=

=

However, the potential drop in the primary is determined by the generator emfAccording to this, the total flux in the iron core must be the same as

when there is no load in the secondary. The primary coil thus draws an additional current

I1 to maintain the original fluxΦturn. The flux through each turn produced by this

additional current is proportional to N1I1. Since this flux equals – Φ́ turn, the additional

current I1 in the primary is related to the current I2 in the secondary by

These curents are 180 º out of phase and produce counteracting fluxes. Since I2 is in

phase with V2, the additional current I1 is in phase with the potential drop across the

primary. Then, if there are no losses

A resistance R, load resistance, in the

secondary circuit

A current I2 will be in the secondary coil, which is

in phase with the potential drop V2 across theresistance. This current sets up and additional flux

Φ´ turn through each turn, which is proportional to

N2I2. This flux opposes the original flux sets up

by the original magnetizing current Im in the

primary.

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