4.0 ac circuits - part 1_3

8/13/2019 4.0 Ac Circuits - Part 1_3

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1Chapter 4 : AC Circuits

8/13/2019 4.0 Ac Circuits - Part 1_3


4.0 AC Circuits• 4.1 Inductors and capacitors

• 4.2 Generation, frequency, mean values

• 4.3 Phasor concepts• 4.4 Resistance, capacitance

and inductance in AC circuit• 4.5 Power in AC circuits

• 4.6 Three phase supply star and delta connections


8/13/2019 4.0 Ac Circuits - Part 1_3


The term or to give it its full description of

Alternating Current,

generally

refers

to

a

time

‐

varying waveform with the most common of all

being called a Sinusoid better known as a

Sinusoidal Waveform .


8/13/2019 4.0 Ac Circuits - Part 1_3


Sinusoidal waveforms are more generally called by

their short description as Sine Waves . Sine waves are by far one of the most important

types of

AC

waveform

used

in

electrical

engineering.


8/13/2019 4.0 Ac Circuits - Part 1_3


•

Direct Current or as it is more commonly called, is a form of current or voltage that flows

around an electrical circuit in one direction only,

making it a "Uni ‐directional" supply.

• Generally, both DC currents and voltages are

produced by power supplies, batteries, dynamos and solar cells to name a few.


8/13/2019 4.0 Ac Circuits - Part 1_3


• A DC voltage or current has a fixed magnitude

(amplitude) and

a

definite

direction

associated

with

it.

• For example, +12V represents 12 volts in the

positive direction, or ‐5V represents 5 volts in the

negative direction.


8/13/2019 4.0 Ac Circuits - Part 1_3


• We also know that DC power supplies do not change

their value with regards to time, they are a constant value flowing in a continuous steady state direction. In

other words, DC maintains the same value for all times

and a constant uni ‐directional DC supply never changes or becomes negative unless its connections are

physically reversed.

• An example of a simple DC or direct current

circuit is shown below.


8/13/2019 4.0 Ac Circuits - Part 1_3



8/13/2019 4.0 Ac Circuits - Part 1_3


• Inductors and capacitors are energy ‐storage elements

• They can store energy and later return it to

the circuit• They do not generate energy ‐only the energy

that has been put into these elements can be extracted

•

Said to

be


8/13/2019 4.0 Ac Circuits - Part 1_3


• Capacitors are constructed by separating two

sheets of conductor (usually metallic) by a thin

layer of insulating material

• The insulating material are called dielectric

can be

polyester,

polypropylene,

mica

or

a

variety of other material12Chapter 4 : AC Circuits

8/13/2019 4.0 Ac Circuits - Part 1_3


• In a way, a capacitor is a little like a battery.• Although they work in completely different ways,

capacitors and batteries both store electrical

energy .• A capacitor is much simpler than a battery, as it

can't produce new electrons ‐‐ it only stores them• In an ideal capacitor, the stored charge q is

proportional to the voltage between the plates


8/13/2019 4.0 Ac Circuits - Part 1_3


•

Capacitance, C has unit of farads (F) which is equivalent to coulombs per volt

• In most application, we deal with capacitances

in the range from a few picofarads up to

perhaps 0.01F


8/13/2019 4.0 Ac Circuits - Part 1_3


• Relationship between current and voltage

• As voltage increases, current flows through the

capacitance and charge accumulates on each plate• If voltage remain constant, the charge is constant

and the current is zero• Thus, capacitors act as open circuit for steady

state dc voltages


8/13/2019 4.0 Ac Circuits - Part 1_3


• An inductor is about as simple as an electronic component can get ‐‐ it is simply a coil of wire

•

Current flowing through the coil that links the coil

• Frequently, the coil form is composed of a

magnetic material such as iron or iron oxides

that increase the magnetic flux for a given

current16Chapter 4 : AC Circuits

8/13/2019 4.0 Ac Circuits - Part 1_3


• For an ideal inductor, the voltage is

proportional to the time rate of change of the current

• The polarity of voltage is such as to oppose

the change in current• Inductance (L)


8/13/2019 4.0 Ac Circuits - Part 1_3


• The voltage and current is related by

• Inductance has units of henries (H), which are equivalent to volt seconds per ampere


8/13/2019 4.0 Ac Circuits - Part 1_3



8/13/2019 4.0 Ac Circuits - Part 1_3


Types of Periodic Waveform


8/13/2019 4.0 Ac Circuits - Part 1_3


Basic Single Coil AC Generator


8/13/2019 4.0 Ac Circuits - Part 1_3


8/13/2019 4.0 Ac Circuits - Part 1_3


The points on the sinusoidal waveform areobtained by projecting across from thevarious positions of rotation between0oand 360 o to the ordinate of thewaveform that corresponds to theangle, θ and when the wire loop or coilrotates one complete revolution, or 360 o,

one full waveform is produced.23Chapter 4 : AC Circuits

8/13/2019 4.0 Ac Circuits - Part 1_3


From the plot of the sinusoidal waveform we can see

that when θ is equal to 0o, 180 o or 360 o, the

generated EMF (Electro ‐Motive Force) is zero as the coil cuts the minimum amount of lines of flux.


8/13/2019 4.0 Ac Circuits - Part 1_3



But when θ is equal to 90o

and 270o

the generated EMF is at its maximum value as the maximum

amount of flux is cut.

The sinusoidal waveform has a positive peak at

90o

and a negative peak at 270o

. Positions B, D, F and H generate a value of EMF

corresponding to the formula e = Vmax.sin θ .

8/13/2019 4.0 Ac Circuits - Part 1_3


THE GENERALISED FORMAT USED FOR ANALYSING AND CALCULATING THE VARIOUS VALUES OF

A SINUSOIDAL WAVEFORM IS AS FOLLOWS


8/13/2019 4.0 Ac Circuits - Part 1_3



8/13/2019 4.0 Ac Circuits - Part 1_3


Chapter 4 : AC Circuits 28

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8/13/2019 4.0 Ac Circuits - Part 1_3


T

π ω

2= f π ω 2=

( ) o

90cossin −= z z

Frequency

T f

1=

Angular frequency

To uniformity, the sinusoidal functions by using cosine

function rather than the sine function.

8/13/2019 4.0 Ac Circuits - Part 1_3


( )dt t vT

V T

2

0rms

1 ∫=

R

V P

2rms

avg =

( )dt t iT I T

2

0rms 1 ∫=

R I P 2rmsavg

=


8/13/2019 4.0 Ac Circuits - Part 1_3


2rms

mV V =

The rms value for a sinusoid is the peakvalue divided by the square root of two.

However, this is not true for other periodicwaveforms such as square waves or triangular

waves.


8/13/2019 4.0 Ac Circuits - Part 1_3



8/13/2019 4.0 Ac Circuits - Part 1_3


2 2 arctan b

A a b

a

= + ∠

|| A A

b a A j( )

cos j sin

cos jsin

A A A

A

ϕ ϕ

ϕ ϕ

= +

= +


8/13/2019 4.0 Ac Circuits - Part 1_3



8/13/2019 4.0 Ac Circuits - Part 1_3


• sinusoidal waveforms of the same frequency can havea Phase Difference between themselves whichrepresents the angular difference of the two sinusoidal

waveforms

• terms "lead" and "lag" as well as "in ‐phase" and "out ‐

of ‐phase" are used to indicate the relationship of onewaveform to the other with the generalized sinusoidalexpression given as: A (t) = Am sin(ω t ± Φ ) representingthe sinusoid in the time ‐domain form

• sinusoids can also be represented graphically in thespacial or phasor ‐domain form by a Phasor Diagram , andthis is achieved by using the rotating vector method


8/13/2019 4.0 Ac Circuits - Part 1_3



8/13/2019 4.0 Ac Circuits - Part 1_3


• The phasor diagram is drawn corresponding to time zero

( t = 0 ) on the horizontal axis.• The lengths of the phasors are proportional to the values of

the voltage, ( V ) and the current, ( I ) at the instant in time

that the phasor diagram is drawn.• The current phasor lags the voltage phasor by the angle, Φ , as

the two phasors rotate in an anticlockwise


8/13/2019 4.0 Ac Circuits - Part 1_3


If however, the waveforms are frozen at timet = 3 0 o, the corresponding phasor diagramwould look like the one shown below.

Once again the current phasor lags behind thevoltage phasor as the two waveforms are of

the same frequency.


8/13/2019 4.0 Ac Circuits - Part 1_3


Consider two AC voltages, V1 having a peak voltage of 20 volts,

and V2 having a peak voltage of 30 volts where V1 leads V2 by

60 o. The total voltage, VT of the two voltages can be found by

firstly drawing a phasor diagram representing the two vectors

and then constructing a parallelogram in which two of the

sides are the voltages, V1 and V2 as shown below.


8/13/2019 4.0 Ac Circuits - Part 1_3


How to find the total voltage, VT ? ‐‐‐‐‐‐ use the analytical method

called as Rectangular Form

In the

rectangular

form,

the

phasor is

divided

up

into

a

real

part,

x

and an imaginary part, y forming the generalisedexpression Z = x ± jy


8/13/2019 4.0 Ac Circuits - Part 1_3


Voltage, V2 of 30 volts points in the reference direction along the horizontal zero axis, then it has a horizontal component but no vertical component as

follows.Horizontal component = 30 cos 0 o = 30 volts• Vertical component = 30 sin 0o = 0 volts• This then gives us the rectangular expression for voltage V2 of: 30 + j0

Voltage, V1 of 20 volts leads voltage, V2 by 60 o, then it has both horizontal and vertical components as follows.• Horizontal component = 20 cos 60 o = 20 x 0.5 = 10 volts•

Vertical component = 20 sin 60o

= 20 x 0.866 = 17.32 volts• This then gives us the rectangular expression for voltage V1 of: 10 + j17.32

The resultant voltage, VT is found by adding together the horizontal and vertical

components as

follows.• VHorizontal = sum of real parts of V1 and V2 = 30 + 10 = 40 volts

• VVertical = sum of imaginary parts of V1 and V2 = 0 + 17.32 = 17.32 volts


8/13/2019 4.0 Ac Circuits - Part 1_3



8/13/2019 4.0 Ac Circuits - Part 1_3


Step 1: Determine the phasor for each termStep 2: Add the phasors using complex

arithmetic.Step 3: Convert the sum to polar form.Step 4: Write the result as a time function.


8/13/2019 4.0 Ac Circuits - Part 1_3


( ) ( )1 20cos 45 Vv t t ω = − o

( ) ( )2 10sin 60 Vv t t ω = + o

1 20 45 V= ∠−V o

2 10 30 V= ∠−V o

1 2Find ?

sv v v= + =

Solution:


8/13/2019 4.0 Ac Circuits - Part 1_3


4.0 ac circuits - part 1_3

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