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8/13/2019 4.0 Ac Circuits - Part 1_3
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1Chapter 4 : AC Circuits
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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
2Chapter 4 : AC Circuits
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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 .
3Chapter 4 : AC Circuits
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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.
4Chapter 4 : AC Circuits
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•
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.
5Chapter 4 : AC Circuits
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• 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.
6Chapter 4 : AC Circuits
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• 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.
7Chapter 4 : AC Circuits
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8Chapter 4 : AC Circuits
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9Chapter 4 : AC Circuits
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10Chapter 4 : AC Circuits
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• 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
11Chapter 4 : AC Circuits
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• 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
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• 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
13Chapter 4 : AC Circuits
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•
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
14Chapter 4 : AC Circuits
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• 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
15Chapter 4 : AC Circuits
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• 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
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• 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)
17Chapter 4 : AC Circuits
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• The voltage and current is related by
• Inductance has units of henries (H), which are equivalent to volt seconds per ampere
18Chapter 4 : AC Circuits
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19Chapter 4 : AC Circuits
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Types of Periodic Waveform
20Chapter 4 : AC Circuits
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Basic Single Coil AC Generator
21Chapter 4 : AC Circuits
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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
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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.
24Chapter 4 : AC Circuits
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25Chapter 4 : AC Circuits
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 .
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THE GENERALISED FORMAT USED FOR ANALYSING AND CALCULATING THE VARIOUS VALUES OF
A SINUSOIDAL WAVEFORM IS AS FOLLOWS
26Chapter 4 : AC Circuits
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27Chapter 4 : AC Circuits
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Chapter 4 : AC Circuits 28
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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.
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( )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
=
31Chapter 4 : AC Circuits
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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.
32Chapter 4 : AC Circuits
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33Chapter 4 : AC Circuits
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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
= +
= +
35Chapter 4 : AC Circuits
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36Chapter 4 : AC Circuits
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37Chapter 4 : AC Circuits
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39Chapter 4 : AC Circuits
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40Chapter 4 : AC Circuits
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• 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
41Chapter 4 : AC Circuits
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42Chapter 4 : AC Circuits
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43Chapter 4 : AC Circuits
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• 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
44Chapter 4 : AC Circuits
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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.
45Chapter 4 : AC Circuits
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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.
46Chapter 4 : AC Circuits
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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
47Chapter 4 : AC Circuits
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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
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49Chapter 4 : AC Circuits
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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.
51Chapter 4 : AC Circuits
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( ) ( )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:
52Chapter 4 : AC Circuits
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