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8/13/2019 4.0 Ac Circuits - Part 2
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8/13/2019 4.0 Ac Circuits - Part 2
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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 anddelta connections
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VR
IR
w t
VC
IC
w t
VL IL
w t
RESISTOR CAPACITOR INDUCTOR
A phasor is an arrow whose length represents the amplitude of an AC voltage or current.
Phasor diagrams are useful in solving complex AC circuits.
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R
L
C~
The impedance, Z, of a circuit relates max
current to max voltage:
Z V I m
m
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R
L
C~
As in DC circuits,
I is same through all components.
: Voltages have different PHASES
they add as PHASORS.
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Solve for the current:
Impedance:
Z V
X X R
V I m
Lc
mm
22 )(
Z R2
1
w C w L
2
R
L
C
~
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• STEP
1. Replace the time descriptions of V and I with
correct corresponding phasors.
2. Replace inductances and capacitances by their
complex impedances, where;
3. Analyze the circuit by using any techniques
learned before.
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• The instantaneous power dissipated in a
component is a product of the instantaneous
voltage and the instantaneous current
• In a the voltage and current are in
phase – calculation of p is straightforward
• In , there will normally be some
phase shift between v and i , and calculating the
power becomes more complicated
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Current, voltage and power versus
time for a purely resistive load
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.• Therefore, if a voltage is
applied across an inductance , the current will
be given by• Therefore
)2
2sin(
)sin(cos
sincos
)()()(
t I V
t t I V
t I t V
t it vt p
mm
mm
mm
w
w w
w w
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• Half of the time the power is positive, showing
energy is delivered to the inductance, where it
stored in magnetic field, and other half is
negative, showing inductance returns energy
to the source
• The average power is zero, we can say that
reactive power flows from source to load
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Current, voltage and power versus
time for a purely inductive load
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.
• Therefore, if a voltage is applied
across a capacitance , the current will be given by
• Then
)2
2sin(
)sin(cos
sincos
)()()(
t I V
t t I V
t I t V
t it vt p
mm
mm
mm
w
w w
w w
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• The average power is zero (reactive powerflow)
• The power for capacitances carries oppositesign from inductance
• Thus, reactive power is positive for inductanceand is negative for capacitance.
• If load contains both inductance andcapacitance with reactive power of equal
magnitude, the reactive powers cancel
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Current, voltage and power versus
time for a purely capacitive load
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• When a sinusoidal voltage is
applied across a circuit with resistance and
reactance, the current will be of the generalform
• Therefore, the instantaneous power, p is given
by
)cos(cos
)()()(
w w
t I t V
t it vt p
mm
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• By using trigonometric identity,
is reduced to
cos
2/
2/
cos2
1
rmsrms
mrms
mrms
mm
I V P
I I
V V
I V P
)cos(cos
)()()(
w w
t I t V
t it vt p
mm
cosrmsrms I V P
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• The dissipation given by
is termed the in the circuit and ismeasured in watts (W)
• The product of the voltage and current
is termed the . Toavoid confusion this is given the units of voltamperes (VA)
cosrmsrms I V P
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• In other words, the active power is the
apparent power times the cosine of the phaseangle.
•This cosine is referred to as the
coscos
S I V P rmsrms
factor Power amperes)volt(inpower Apparent
watts)(inpower Active
iv
S
P
cosfactor Power
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•When a circuit has resistive and reactive parts,the resultant power has 2 parts:
– The first is dissipated in the resistive element. Thisis the
– The second is stored and returned by the reactiveelement. This is the , which hasunits of
•
While reactive power is not dissipated it doeshave an effect on the system
– for example, it increases the current that must besupplied and increases losses with cables
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• Consider an
RL circuit
– the relationship
between the variousforms of power can
be illustrated using
a power triangle
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Active Power watts
Reactive Power var
Apparent Power S = V rmsIrms VA
S2 = P2 + Q2
cosrmsrms I V P
sinrmsrms I V Q
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– measures power directly using a single meter which
effectively multiplies instantaneous current and
voltage
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• R is the real part of the impedancethrough which the current flows
• X is the imaginary part of theimpedance through which the
current flows• Reactive power Q is positive for
inductive loads and negative forcapacitive loads
• Vrms
is the rms voltage across theresistance
• VXrms is the rms voltage across thereactance
R
V P
Rrms2
R I P rms2
X I Q rms2
X
V Q
Xrms2
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•
In resistive circuits the average power isequal to VI, where V and I are r.m.s. values
• In a the current leads the voltage
by 90 and the average power is zero
• In an the current lags the voltageby 90 and the average power is zero
•
In, the average power is VI cos
• The term is called the
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3-Phase-Definition
37
• AC system that involves only one voltage source
is called one phase system. For this type of
generator, the emf is generated using 1
conducting coil that is rotated and strike throughthe magnetic field (north-south pole).
• If the numbers of conducting coil increased, the
generator with multiple phase will be produced.
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3-Phase-Definition
38
• 3-phase generation is based on the generation of
one phase system. Therefore, 3 conducting coil
should be arranged and separated 120° among
each other. This will lead to the 3 differentvoltages with the θ of 120°.
• Normally, these 3 conducting coil are termed
‘red’, ‘blue’ and ‘yellow’.
• The rotation of the rotor in the generation of
voltage can be positive (anti-clockwise) or
negative (clock-wise).
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3-Phase-Definition
39
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3-Phase-Definition
40
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3-Phase-Definition
41
• Phasor form:
• The summation of VR, VB and VY =0
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42
• Since the generator involved with three different
elements, logically we need 6 wires to connect
these 3 elements.
• However, one can be arranged so that all the
wires can be connected correctly. It’s via the
delta or wye (star) connection. Via these
connection, there is only 4 wires needed, 3 foroutput and one being connected among
themselves. The latter is called neutral.
• This configuration is also important when we are
dealing with the load connection.
• Note that this connection is in wye/star
connection.
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43
3-Phase-Connection-Wye/Star
• There are two different voltages in 4 wire
y-connection: – Phase voltage-the voltage difference between appropriate
wire/line with neutral line (Vph)
– Line voltage-the voltage difference between line with other line
(VL)
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44
Vph?
VL?
Vph VRN, VBN, VYN
VL VRB, VRY, VYB
3-Phase-Connection-Wye/Star
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45
3-Phase-Connection-Wye/Star
VRN
VBN
VYN
-VYN
-VRN
-VBN
30° 30°
30°
VRY
VYB
VRB
VL= 3Vph
IL= Iph
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• A 415 V, 50 Hz, 3-phase supply is connected to
a star-connected balanced load. Each phase of
the load consists of a resistance of 25 Ω and
inductance 0.1 H, connected in series.
• Calculate a) phase voltage,
b) the line current drawn from the
supply
46
Example 4.6.1
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47
a) Phase voltage:
Vph=VL/√3= 415/√3=239.6 V
b) IL=Iph, so Iph= Vph/Zph
Zph
= ℎ
2 +
2
XL=2fL=2(50)(0.1)=31.42 Ω
Zph=(252+31.422)1/2 = 40.15
IL= 239.6/40.15=5.97 A
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48
• The star-connection is based on KVL principle
which involved voltage while delta-connection
is based on KCL which involved current.
3-Phase-Connection-Delta
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49
3-Phase-Connection-Delta
R
Y
B
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50
3-Phase-Connection-Delta
IB
IR
IY
IBR -IYB
30°
IRY-IRY
IYB-IBR
30°
30°
IL= 3Iph
VL= Vph
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• A balanced load of phase impedance 120
is connected in delta. When this load is
connected to a 600 V, 50 Hz, 3-phasesupply.
• Determine
a) the phase current, andb) the line current drawn
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• a) Vph = VL = 600 V;Iph=Vph/Zph=600/ 120=5A
•
b) IL= 3Iph = 3 x 5=8.66 A
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53
Power Dissipation in Star and Delta-connection Loads
• The power dissipation for both star and
delta connection are always the same.
The only difference is the value of VL andIL in the connection. The equation of
power dissipation is given by: Phase
power
factor
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• A balanced load of phase impedance 100
and power factor 0.8 is connected
a) in star, and
b) in delta, to a 400 V, 3-phase supply.
Calculate the power dissipation in each case.
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• a) IL=IpH=Vph/Zph;• Vph=VL/√3=400/√3
• IL=(400/√3)/100=4/√3
• P=√3ILVLcosф
= (√3)(4/√3)(400)(0.8)
= 1.28 kW
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Star/Delta Supplies and Loads
• The same equation can also be applied for a
combination of supplies and loads with the
same arrangement.
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• The star-connected stator of a three-phase, 50
Hz alternator supplies a balanced delta-
connected load. Each phase of the load
consists of a coil of resistance 15 Ω andinductance 36 mH, and the phase voltage
generated by the alternator is 231 V.
• Calculate (a) the phase and line currents, (b)the load power factor, and (c) the power
delivered to the load.
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59
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• a) Alternator: V1=Vph= 231 V;• V2=VL=√3Vph = (√3)(231)=400 V
• IL=Iph=I1
•
Load: VL=Vph=V2=400 V• XL= 2fL=2(50)(36 x 10-3)=11.3 Ω
• Zph=√(R2+XL2)=√152+11.32=18.78 Ω
• Iph=Vph/Zph=400/18.78=21.3 A
• IL=√3Iph = (√3)(21.3)=36.9 A; • So, I1=36.9 A
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• b) Power factor = cos ϕ = (R/Z) = 15/
18.78=0.8
• c) P = Iph2R = (21.3)2(15) = 6.81 kW for 1 phase
• For 3 phase = 6.81 x 3=20.4 kW
or P = √3VLILcos
ϕ=√3(400)(36.9)(0.8)=20.4
kW
61
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62
Conclusion
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