3. Alternating Current

TOPICS

� Definition and Introduction

� AC Generator

Components of AC Circuits� Components of AC Circuits

� Series LRC Circuits

� Power in AC Circuits

� Transformers & AC Transmission

Introduction to AC�The electric power out of a home or office

power socket is in the form of alternating current (AC), as opposed to the direct current (DC) of a battery.

�Alternating current is used because it is easier to transport, and easier to “transform” easier to transport, and easier to “transform” from one voltage to another using a transformer.

� In Nigeria and UK, the frequency of oscillation of AC is 50 Hz. In the USA it is 60 Hz.

November 7, 2007

The AC GeneratorThe AC Generator

November 7, 2007

A coil of area A and N turns rotating with constantangular velocity in a uniform magnetic field producesa sinusoidal emf.

The slip rings and brushes allow the coil to rotatewithout twisting the connecting wires. Such a device iscalled a generator.

Alternating Current Generator Cont’d

December 5, 2007tdm ωεε sin= )sin( φω −= tIi d

It takes power to rotate the coil, butthat power can come from:

�moving water (a water turbine)

�air (windmill)

�gasoline motor (as in a car)

�steam (as in a nuclear power plant).November 7, 2007

tNBAd

tNBA

tandNBA

m

m

m

δωωφε

δωφδωθθφ

)sin(

)cos(

cos

+=−=

+=+==

November 7, 2007

fffrequency

t

tNBAdt

peak

m

πωδωεε

δωωε

2;

)sin(

)sin(

=

+=

+=−=

RLC Circuits with AC Power� When an RLC circuit is driven with an AC power

source, the “driving” frequency is thefrequency of the power source, while the circuitcan have a different “resonant” frequency.

� Let’s look at three different circuits driven by anAC EMF. The device connected to the EMF iscalled the “load.”

dω

2)2/(/1 LRLC −=′ω

December 5, 2007

called the “load.”

� What we are interested in is how the voltageoscillations across the load relate to the currentoscillations.

� We will find that the “phase” relationshipschange, depending on the type of load(resistive, capacitive, or inductive).

Leading & Lagging in Phasors� The figure shows, in (a), a

sine curve S(t) = sin(ωdt)and three other sinusoidalcurves A(t), B(t), and C(t),each of the form sin(ωdt–φ).

� (a) Rank the three curvesaccording to the value of φ,most positive first and mostmost positive first and mostnegative last.

Answer (a) C, B, A

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(b) Which curve corresponds to which phasor in part (b) of the figure? (c) Which curve leads the others?

November 7, 2007

(b) (c)

1 −>A A

2 −>B

3 −>S

4 −>C

A Resistive Load� Phasor Diagram: shows

the instantaneous phaseof either voltage orcurrent.

� For a resistor, the currentfollows the voltage, so thevoltage and current are inphase (φ = 0).

φ

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phase (φ = 0).

� If

� Then tR

VtIi d

RdRR ωω sinsin ==

tVv dRR ωsin=

Resistive Load Cont’d

November 7, 2007

Resistive Load

CURRENT

is in phase with is in phase with

VOLTAGE

November 7, 2007

Potential drop across the resistor,

AC Power in a Resistor

tVtV peakRR ,max coscos === ωωεεPotential drop across the resistor, VR Current in the resistor I Power dissipated in the resistor, P Average power dissipated in the resistor Paverage

R

VIt

R

VI

tVtV

peakRpeak

peakR

peakRR

,,

,max

cos

coscos

=⇒=

===

ω

ωωεε

RtIRIP

RtIRIP

avpeakavav

peak

))(cos()(

)(cos222

222

ω

ω

==

==

Root-Mean-Square Values

� For a capacitive load, the voltage across the capacitor is proportional to the charge

� But the current is the time derivative of the charge

A Capacitive Load

tC

Q

C

qv dC ωsin==

tCVdt

dqi dCdC ωω cos==

December 5, 2007

� In analogy to the resistance, which is the proportionality constant between current and voltage, we define the “capacitive reactance” as

� So that .

� The phase relationship is that φ = −90º, and current leads voltage.

tX

Vi d

C

CC ωcos= C

Xd

C ω1=

dt dCdC

Capacitive Load Cont’d

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Capacitors in Alternating Current CircuitsThepotentialdrop lagsthe currentby 90º

cos

coscos

,

,max

ω

ωωεε

=

⇒====

tCVQ

tVtC

QV

peakC

peakCC

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

CAPACITIVE REACTANCE

)2

cos(1

sin

cos

,,

,

πω

ω

ωω

ω

+

=→−==

=

t

C

VItCV

dt

dQI

tCVQ

peakCpeakC

peakC

=

=

C

VI

C

VI rmsC

rmspeakC

peak

ωω1

;1

,,

Capacitive Load

CURRENT

Leads Leads

VOLTAGE

By 90 degrees November 7, 2007

An Inductive Load� For an inductive load, the voltage across the inductor

is proportional to the time derivative of the current

� But the current is the time derivative of the charge

dt

diLv L

L =

tL

Vdtt

L

Vi d

d

Ld

LL ω

ωω cos sin

−== ∫

December 5, 2007

� Again in analogy to the resistance, which is the proportionality constant between current and voltage, we define the “inductive reactance” as

� So that .

� The phase relationship is that φ = +90º, and current

lags voltage.

tX

Vi d

L

LL ωcos−=

LX dL ω=

LL dω

∫

Inductive Load Cont’d

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Inductors in Alternating Current Circuits

The potential drop across theinductor led the current 90º (outof phase)

coscos ,max ωωεε ==== tVtdI

LV peakLL

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

πωω

ωω

ω

ωωεε

−==→=

====

tL

Vt

L

VIt

L

V

dt

dI

tVtdt

LV

peakLpeakLpeakL

peakLL

INDUCTIVE REACTANCE L

VI

L

VI rmsL

rmspeakL

peak ωω,, ; ==

Inductive Load

CURRENT

Lags Lags

VOLTAGE

By 90 degrees November 7, 2007

Summary Table

Circuit Element

Symbol Resistance or Reactance

Phase of Current

Phase Constant Amplitude Relation

Resistor R R In phase with vR

0º (0 rad) VR=IRR

ω − −π

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Capacitor C XC=1/ωdC Leads vR by 90º

−90º (−π/2) VC=ICXC

Inductor L XL=ωdL Lags vR by 90º

+90º (π/2) VL=ILXL

Driven RLC Series CircuitsThe Kirchhoff´s rules governthe behavior of potentialdrops and current across thecircuit.

(a) When any closed-loop istraversed, the algebraicsum of the changes ofpotential must equal zero(loops rule)

(a) At any junction (branchpoint) in a circuit wherethe current can bedivided, the sum of thecurrents into the junctionmust equal the sum ofthe currents out of thejunction (junction rule)

)cos(

cos

;cos

,

2

,

δω

ω

ω

−=

=++

=++=

tII

tVC

QR

dt

dQ

dt

QdL

dt

dQI

C

QIR

dt

dILtV

peak

peakapp

peakapp

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 a series RLC circuit C

QIR

dt

dILtVapp ++=ωcos

series RLC circuit

Power delivered to the series RLC circuit

δεδε

πωωεε

coscos2

1

)2

cos(cos

22

,2

2

rmsapprmsav

rmsrmspeakpeakav

peakpeak

Z

RVRIP

resistortheindissipatedRIP

IIP

tItIP

==

=

==

−==

Powerfactor: cosδ

δδ

δ

coscos2

1

/cos/

,,

,

2,

rmsrmsapppeakpeakapp

peakapppeak

rmsapprmsav

IVIVP

ZVIandZRasZ

VRIP

==

==

==

The Transformer

Because of the iron core, there is a large

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 small resistances 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

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 consider a 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 90º 1

1

2222

11

VN

NVNV

NV

dtd

dtd

turn

turn

=→=

=

φ

φ

Secondary coil open circuitThe potential drop across the primary coil is

If there is no flux leakage out of the iron core, the flux through each turn is the same for both coils, and then

The Transformer

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

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 the resistance. 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.

rmsrmsrmsrms IVIV

ININ

,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 I1to maintain the original flux Φturn. The flux through each turn produced by thisadditional current is proportional to N1I1. Since this flux equals – Φ´turn, theadditional 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

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November 7, 2007