Alternating current source - University of Hawaiiplam/ph272_summer/L11/31_Lecture_Lam.pdf ·...

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley

PowerPoint® Lectures for

University Physics, Twelfth Edition

– Hugh D. Young and Roger A. Freedman

Lectures by James Pazun

Chapter 31

Alternating Current

Modified by P. Lam 8_8_2008


Topics for Chapter 31 & 32

• Alternating current source

• Voltage, current, power in AC circuit

• Complex impedance and its application to the L-

R-C series circuit

• AC Transformer


Alternating current (AC) source

An AC generator is a conducting coil rotating in a magnetic field.

The magnetic flux through the coil varies as a cosine function, hence

the induced emf varies as sine function.


Voltage and current relationship in AC circuit

Given : i(t) = Icos t

VR = iR = IRcos t

I & VR are "in phase"


VL = Ldi

dt= LI sin t

= LIcos t + /2( )

VL leads I by 900


VC =Q

C=1CI sin t

=1

CIcos t /2( )

VC lags I by 900


Resistance and “Reactance”


VR = iR = IRcos t

I & VR are "in phase"


VL = Ldi

dt= LI sin t

= LIcos t + /2( )

VL leads I by 900


VC =Q

C=1CI sin t

=1

CIcos t /2( )

VC lags I by 900

Compare the voltage equation for these three circuit elements.

What are the units for L and 1/( C)?

Answer : They have unit of resistance.

L XL = inductive reactance (an effective resistance of an inductor; it increases

with the frequency)

1/( C) XC = capacitive reactance (an effective resistance of a capacitor; it decreases

as the frequency increases)


The loudspeaker, a useful application

• The capacitor blocks the lowfrequencies and lets the highfrequencies go to thetweeter. XC=1/( C)

• The inductor blocks highfrequencies and lets lowfrequencies go to the woofer.XL= L


Resistance dissipates power but reactance doesn’t


VR = iR = IRcos t

I & VR are in phase


VL = Ldi

dt= LI sin t

= LIcos t + /2( )

VL leads I by 900


VC =Q

C=1CI sin t

=1

CIcos t /2( )

VC lags I by 900P(t) = i(t)VR (t)

= I2R(cos t)2

= always positive

Paverage = I2R < (cos t)2 >

= I2R1

2

Irms < [i(t)]2 >

= < I2(cos t)2 >

=I

2

Paverage = Irms2 R

P(t) = i(t)VL (t)

= LI2 cos t sin t

= positive and negative

Paverage = LI2 < cos t sin t >= 0

An inductor takes energy

from the source and then

gives it back, net power

dissipation is zero.

P(t) = i(t)VC (t)

=1

CI2 cos t sin t

= positive and negative

Paverage =1CI2 < cos t sin t >= 0

A capacitor takes energy

from the source and then

gives it back, net power

dissipation is zero.


The L-R-C circuit

Given : The source (t) = cos t, find i(t)

This is a hard problem. We will do an easier

problem first.


Find VR (t),VL (t),VC (t),and the source voltage (t)

Answer :

VR (t) = IRcos t, VL (t) = Ldi

dt= LI sin t, VC (t) =

Q

C=1

CI sin t

(t) = ?

Kirchoff 's rule :

VR VL VC = 0 (t) =VR +VL +VC

(t) = I Rcos t XL XC[ ]sin t{ }Combine cos t and sin t into a single function.

Let R

Z= cos ; Z = R2 + XL XC[ ]

2= impedance, and

XL XC[ ]Z

= sin

then, (t) = IZ cos cos t sin sin t{ } = IZ cos t +( )To answer our original question :

Given : (t) = cos t,

then i(t) =Zcos t( )


Geometric method to visualize Z and phase angle

R

XL= L

XC=1/( C)

ZXL- XC

Z = R2 + XL XC[ ]2

= impedance

cos =R

Z, sin =

XL XC[ ]Z

Note : If XL XC[ ] > 0 = positive voltage leads current

If XL XC[ ] < 0 = negative voltage lags current

Given : i(t) = Icos t then ( t) = IZ cos t +( )

Given : (t) = cos t then i( t) =Zcos t( )


Example: source frequency = natural frequency (resonance)

R

XL= L

XC=1/( C)

Z

Given : (t) =10cos t,R = 2 ,L = 2mH,C = 80μF

(1) What is the natural frequency ( o ) of this circuit?

(2) Suppose the frequency ( ) of the AC source = o ,

what the current i(t) through the circuit?

(3) What are the voltages across the R, L, and C?

Answer :

(1) o =1LC

=1

(2x10 3)(80x10 6)=14x104

rad

s= 2500

rad

s

(2)At resoance = o .

XL = oL =L

C= 5

XC =1

oC=

L

C= 5

Z = R2 + XL XC[ ]2

= R = 2 and = 0

i(t) =Zcos t( ) =

102cos2500t

(3)VR = iR =10cos2500t = (t)

VL = Ldi

dt= oLI sin ot = 25sin2500t

VC =Q

C=1

oCsin ot = +25sin2500t

Note(1) :VL cancels VCNote(2) :|VL | and |VC |>


Circuit behavior at resonance and off-resonance

• Minimum impedance occurs at resonance => maximum current(and =0)

• > o => >0

• < o => <0

R

XL= L

XC=1/( C)

ZXL- XC


Example: source frequency > natural frequency (resonance)

Given : (t) =10cos t,R = 2 ,L = 2mH,C = 80μF

(1) What is the natural frequency ( o ) of this circuit?

(2) Suppose the frequency ( ) of the AC source = 2 o ,

what the current i(t) through the circuit?

(3) What are the voltages across the R, L, and C?


Use impedance and phase angle method.

Given : (t) =10cos(120 t)

Find i(t),VR (t),VC (t)


Transformers - transfer energy from one coil to another

1 = N1d 1

dt

2 = N2

d 2

dt= magnetic flux per turn.

The iron core ensure that the magnetic

flux per turn is the same for coil 1 and coil 2

d 1

dt=d 2

dt2

1

=N2

N1

"Step up" transformer : N2

N1>1

"Step down" transformer : N2

N1<1


Transformers - energy conservation

Suppose we have a step up transformer

which steps up the 10 volts in the primary coil

into 1,000 volts in the secondary coil,

is there a violation of conservation of energy?

Answer is No. Suppose the secondary coil is connect to

a 1,000 resistor => I2 =1A,

then the primary coil must carry at least 100A to provide

the necessary power to generate the 1000 V in the secondary

coil.

In a step up transformer, the primary coil has fewer turns

but the conductor is much thicker to accommodate higher current.

Power in = Power out (if there are no losses)

V1I1 = V2I2

In real situation, there are losses

V1I1 > V2I2


Example of transformer energy loss

• Induced surface current on the the iron coil(eddy current) heats up the iron core =>energy is loss.

• Use laminated core to reduce eddy current.

Alternating current source - University of Hawaiiplam/ph272_summer/L11/31_Lecture_Lam.pdf ·...

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