Ohms Law

1
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• ???
• ??.???????? ????????
• ???????????????????? ??????????????????????????
• ???????????????????????????????????????????
• ??????????????????????????????????????????????

2
Ohms Law

• The Ohms Law states that
• Where
• I Current (Amps)
• V Voltage (Volts)
• R Resistance (Ohms)

3
Ohms Law

• The formula for Ohms law only apply to ac
  circuits that are purely resistive
• However, many ac circuits have reactance (X) as
  well as resistance measured in ohms
• This reactance must be combined with the
  resistance to calculate the impedance (Z) and is
  also measured in ohms

4
Ohms Law

• Therefore,
• The result leads to

5
AC Power

• Alternating Current (AC) is an electrical current
  that periodically reverse its direction of flow
  because the polarity of the voltage source
  constantly changes
• When an AC circuit is connected to a resistance,
  the resulting alternating current follows the
  same alternating pattern as the voltage source
• This pattern is called a sine wave

6
AC Power

• A sine wave cycle is one variation of the sine
  wave from zero to maximum, back through zero, to
  minimum, and back to zero

7
AC Power

• The amplitude of the sine wave is expressed as
  either voltage or current
• The peak value is the current magnitude between
  zero and the highest point of the sine wave for
  one-half cycle

8
AC Power

• The current at any point on a sine wave is called
  the instantaneous current (i)
• It is also possible to determine the arithmetic
  value of the alternating current
• The average value is the arithmetical average of
  all values in the sine wave for one-half cycle

9
AC Power

• The most common method of measuring the amplitude
  of a sine wave is its root-mean-square (rms)
  value
• RMS is the square root of the average of all the
  instantaneous currents squared
• The rms value of a sine wave is readily
  determined by calculus but can perhaps be more
  easily understood by simple arithmetic

10
AC Power

• The RMS value is 0.707 times the zero-to-peak
  value of the sine wave

11
AC Power

• Average Value
• RMS Value

12
AC Power
13
AC Power

• The number of times this cycle occurs in one
  second is called frequency and is expressed in
  Hertz (Hz)
• Another parameter of interests is the length of
  time required for the sinusoidal function to pass
  through all its possible values

14
AC Power

• This time is referred to as the period of the
  function and is denoted as T which can be
  expressed as
• Omega (?) represents the angular frequency of the
  sinusoidal function and it is given as

15
Frequency

• A cycle per second is referred to as a hertz
• In north America, the frequency is 60 Hz
• However, in other countries may have different
  frequencies
• In Thailand, the frequency is 50 Hz

16
AC Power

• For the pure resistance circuit, both current and
  voltage are in the same phase

17
AC Power

• For the pure inductance circuit, current lags
  voltage with the angle of 90 degree

18
AC Power
Where
19
AC Power

• For the pure capacitance circuit, current leads
  voltage with the angle of 90 degree

20
AC Power
Where
21
AC Power

• For the series R-L circuit, current lags voltage
  with the angle of tan-1?L/R

22
Example

Let
Calculate The Voltage
23
Example
24
AC Power

• For the series R-C circuit, current leads voltage
  with the angle of tan-11/?CR

25
Example

• A sinusoidal voltage is given by
• Determine
• a) What is the period of the voltage?
• b) What is the frequency in hertz?
• c) What is the magnitude of v at t 2.778 ms
• d) What is the rms value of v

26
Example

• Solution
• a) Since ? 100p rad/s
• b)

27
Example

• c) At t 2.778 ms

Hence
d)
28
Phasor Diagram

• A diagram of phasor currents and voltages can be
  used to analyze the steady-state sinusoidal
  operation of a circuit
• A phasor diagram shows the magnitude and phase
  angle of each phasor quantity in the
  complex-number plane
• Phase angle are measured counterclockwise from
  the positive real axis and magnitude are measured
  from the origin of the axes

29
Phasor Diagram
30
Phasor Diagram

• For the pure resistance circuit, phasor current
  has the same phase (in phase) with the phase
  voltage

31
Phasor Diagram

• For the series R-L circuit, phasor current lags
  the phase voltage with the angle of ? degree

32
Phasor Diagram

• For the series R-C circuit, phasor current leads
  the phase voltage with the angle of ? degree

33
Phasor Diagram

• In this circuit, the parallel combination of R2
  and L2 represents a load on the output end of a
  distribution line
• The distribution line is modeled by the series
  combination of R1 and L1

R1
L1

Vs
R2
L2
VL
-
34
Phasor Diagram
Ib

• Because we are holding the amplitude of the load
  voltage constant, VL is chosen as the reference

R1
j?L1

I
Vs
R2
j?L2
VL
Ia
-
VL
35
Phasor Diagram

• We know that Ia is in phase with VL and its
  magnitude is
• We also know that Ib lag behind VL by 90 degree
  and its magnitude is

Ia
VL
VL
Ia
Ib
36
Phasor Diagram

• The line current I is equal to the sum of Ia and
  Ib
• The voltage drop across R1 is in phase with the
  line current, and the voltage drop across j?L1
  leads the line current by 90 degree

Ia
VL
Ib
I
37
Phasor Diagram

• The source voltage is the sum of the load voltage
  and the drop along the line

j?L1I
VL
Ia
R1I
Ib
I
38
Phasor Diagram
j?L1I
Vs
j?L1I
Ia
VL
R1I
Ib
R1I
I
39
Configuration

• Configuration is the number of phases and the
  types of connection between the utility and the
  user
• For the number of phases, they can be
• One Phase
• Two Phases
• Three Phases

40
Balanced Three-Phase

• A set of balanced three-phase voltages consists
  of three sinusoidal voltages that have identical
  amplitudes and frequency but are out of phase
  with each other by exactly 120 degree
• Because the phase voltages are out of phase by
  120 degree, two possible phase relationship can
  exist between a-phase voltage and the b- and
  c-phase voltage

41
Balanced Three-Phase

• One possibility is for the b-phase voltage to lag
  the a-phase voltage by 120 degree, in which case
  the c-phase voltage must lead the a-phase voltage
  by 120 degree
• This phase relationship is known as the abc phase
  sequence or positive phase sequence

42
Balanced Three-Phase
VC
VA
VB
Positive Phase Sequence
43
Balanced Three-Phase

• Another possibility is for the b-phase voltage to
  lead the a-phase voltage by 120 degree, in which
  case the c-phase voltage must lag the a-phase
  voltage by 120 degree
• This phase relationship is known as acb phase
  sequence or negative phase sequence

44
Balanced Three-Phase
VB
VA
VC
Negative Phase Sequence
45
Configuration

• A three-phase source can be either Wye-connected
  or Delta-connected and the three-phase loads can
  also be either Wye-connected or Delta-connected

Three-phase Voltage Source
Three-phase Load
Three-phase Line
46
Configuration

• Thus, the basic circuit in the above figure can
  take four different configurations
• Wye-Wye Connection
• Wye-Delta Connection
• Delta-Wye Connection
• Delta-Delta Connection

47
Wye-Wye Connection
A Three-Phase Y-Y System
48
Wye-Wye Connection

• Zga, Zgb, and Zgc represent the internal
  impedance associated with each phase winding of
  the voltage source
• Z1a, Z1b, and Z1c represent the impedance of each
  phase conductor of the line connecting the source
  to the load
• Zo is the impedance of the neutral conductor that
  connects the source neutral to the load neutral
• ZA, ZB, and ZC are the impedance of each phase of
  load

49
Wye-Wye Connection

• The circuit is a balanced three-phase circuit if
  it satisfies all the following criteria
• Van, Vbn, and Vcn form a set of balanced
  three-phase voltages
• Zga Zgb Zgc
• Z1a Z1b Z1c
• ZA ZB ZC

50
Wye-Wye Connection

• If the system is balanced, VN must be zero
• VN denote the node voltage between nodes N and n
• If VN is zero, there is no difference in
  potential between the source neutral, n, and the
  load neutral, N
• Consequently, the current in the neutral
  conductor is zero

51
Wye-Wye Connection

• When the system is balanced, three line currents
  are

52
Wye-Wye Connection

• In a balanced system, the three line currents
  form a balanced set of three-phase currents
• Thus, the current in each line is equal in
  amplitude and frequency and 120 degree out of
  phase with the other two line currents
• Hence, if the current IaA is calculated, the line
  currents IbB and IcC can be obtained without
  further computations

53
Wye-Wye Connection

• Once the line currents in the circuit are
  obtained, calculating any voltages of interest is
  relatively simple
• Of particular interest is the relationship
  between the line-to-line voltages and the
  line-to-neutral voltages
• Thus, line-to-line voltages at the load terminals
  in terms of the line-to-neutral load voltages are

54
Wye-Wye Connection
A
VAB
VAN
ZA
B
ZB
N
VCA
VBN
VBC
ZC
VCN
C
Line-to-Line and Line-to-Neutral Voltages
55
Wye-Wye Connection

• To show the relationship between the line-to-line
  voltages and the line-to-neutral voltages, a
  positive phase sequence is assumed and the
  line-to-neutral voltage of the a-phase is used as
  the reference

56
Wye-Wye Connection
VCN
VAN
VBN
Vp is Line-to-Neutral Voltage or Phase Voltage
57
Wye-Wye Connection

• From the above equations, it can be concluded
  that

58
Wye-Wye Connection

• The magnitude of the line-to-line voltage is
  times the magnitude of the line-to-neutral
  voltage
• The line-to-line voltages form a balanced
  three-phase set of voltages
• The set of line-to-line voltages leads the set of
  line-to-neutral voltages by 30 degree

59
Wye-Wye Connection
VCN
VAB
VCA
VAN
VBN
VBC
60
Wye-Delta Connection

• If the load in a three-phase circuit is connected
  in a delta, it can be transformed in to a wye by
  using the delta-wye-transformation

61
Wye-Delta Connection
Impedance Transformation from Delta-to-Wye
62
Wye-Delta Connection

• When the load is balanced, the impedance of each
  leg of the wye is one-third the impedance of each
  leg of the delta
• To determine the relationship between the phase
  currents and line currents, a positive phase
  sequence is used

63
Wye-Delta Connection
A
IaA
IAB
ICA
Z?
Z?
IbB
Z?
C
B
IBC
IcC
Relationship Between Line Currents and Phase
Currents in a Balanced Delta-Load
64
Wye-Delta Connection

• The previous figure shows that, in the delta
  configuration, the phase voltage is identical to
  the line voltage

ICA
IAB
IBC
Ip is the phase current
65
Wye-Delta Connection

• The line current in terms of the phase current
  can be written by Kirchhoffs law as following

66
Wye-Delta Connection

• By comparing the phase current and line current,
  it can be concluded that
• The magnitude of the line current is times
  the magnitude of phase current
• The set of line currents lags the set of phase
  current by 30 degree
• If a negative phase sequence is applied, line
  currents are larger than phase currents and
  lead the phase current by 30 degree

67
Delta-Wye Connection

• In the Delta-Wye three phase circuit, the source
  is delta-connected and the load is wye-connected
• The analysis of the delta-wye connection can be
  achieved by replacing the balanced
  delta-connected source with a wye equivalent
• The wye equivalent of the source by dividing the
  internal phase voltages of the delta-source by

68
Delta-Wye Connection

• If the phase sequence is positive, the phase
  angle of this set of three-phase voltages is
  shifted by the angle of -30 degree
• However, in the case of negative sequence, the
  phase angle of this set of three-phase voltages
  is shifted by the angle of 30 degree
• The internal impedance of the wye-equivalent is
  one third the internal impedance of the
  delta-source

69
The Y-equivalent of a balanced, three-phase,
delta-connected source (positive sequence)
70
Delta-Wye Connection

• For a positive phase sequence, the set of
  delta-source phase current (Iba, Icb, and Iac)
  lead the set of line currents IaA, IbB, and IcC)
  by 30 degree
• For a negative phase sequence, the phase currents
  in the source lag the line currents by 30 degree
• The magnitude of the phase current is times
  the magnitude of the line current

71
Delta-Delta Connection

• In the delta-delta circuit, both source and load
  are delta-connected
• By replacing both source and load with their
  wye-equivalent, the analysis of delta-delta
  connection can be achieved

72
Configuration

• Wye Connection
• Delta Connection

73
Example

• A three-phase, positive sequence, Y-connected
  generator has an impedance of 0.2 j0.5
  ohm/phase. The internal phase voltage of the
  generator is 120 V. The generator feeds a
  balanced, three-phase, Y-connected load having an
  impedance of 39 j28 ohm/phase.
• The impedance of line connecting the generator to
  the load is 0.8 j1.5 ohm/phase.
• The a-phase voltage of the generator is specified
  as the reference phasor

74
Example

• Calculate
• a) Line currents (IaA, IbB, and IcC)
• b) Line-to-Neutral voltages at the load (VAN,
  VBN, VCN)
• c) Line voltages at load (VAB, VBC, VCA)

75
Example
IaA
a
A
0.2 O
j0.5 O
0.8 O
j1.5 O
39 O
j28 O
N
76

• Solution
• a) The a-phase line current

IcC
IaA
IbB
For positive sequence
77

• b) The line-to-neutral voltages at load

VCN
For positive sequence
VAN
VBN
78

• c) The line-to-line voltages at load
• For a positive sequence, the line-to-line
  voltages lead the line-to-neutral voltages by 30
  degree

VAB
VCN
VCA
VAN
VBN
VBC
79
Configuration

• For the number of connections, they can be
• Two-wire connections
• Three-wire connections
• Four-wire connections
• Five-wire connections
• The electrical system between the generating
  source and the customers site is known as the
  distribution system

80
Configuration

• The wiring between the connection to the
  distribution system at the customers site and
  the equipment to be powered is called the
  premises wiring (system)
• The various loads that the electrical system is
  required to power have certain characteristics
  which affect the amount of power (watts) that is
  demanded by these loads

81
Configuration

• These characteristics are
• Resistance
• Inductance
• Capacitance
• Inductance and capacitance cause a difference
  between the real power and the apparent power
  required by the load in an AC circuit

82
Premises Wiring (System)
Branch Circuits
Distribution System
Main Feeder
Feeder
Loads
Medium Voltage Supply
Equipment Loads
Main Electrical Service Panel
Pole or Pad-mounted Transformer
Meter
Feeder Panel or Sub-Panel
Receptacles or Hard-wired
Building Service Transformer (if Required)
83
Capacity

• The capacity of a given system is the amperage
• This can be also be stated in volt-amperes (VA)
  or watts (W)
• Generally, in an AC circuit, the product of the
  measured rms value of the current and the
  measured rms value of the voltage equals the
  volt-amperes

84
Real and Reactive Power

• The VA product is only the Apparent power of
  the circuit

VA
VAR
Phase Angle
W
85
Real and Reactive Power

• To determine real or true power, one must know
  the amount (phase angle) the current leads or
  lags the voltage
• Real power associates with the purely resistive
• This phase angle can vary from 90 degree leading
  in a purely capacitive circuit through 0 degree
  in a purely resistive circuit, to 90 degree
  lagging in a purely inductive circuit

86
Real and Reactive Power
Where
87
Real and Reactive Power

• VAR (volt-amperes reactive) is defined two
  meanings
• The power absorbed by the reactive part of the
  load
• The power delivered by the capacitive part of the
  load
• Thus, VAR is simply referred as reactive power

88
Real and Reactive Power
Where
RF is Reactive Factor
89
Example

• An electrical load operates at 220 V, 15 A, and
  power factor 0.8. Calculate the real and
  reactive power.
• Solution
• Load VA 22015 3300 VA
• PF 0.8, cos? 0.8, ? 36.87o

3300 VA
Q
W
90
Example

• Real power
• Reactive power

91
Watthours

• Watts indicate the amount of power that is
  consumed by a circuit at any given time
• Typically, utility companies charge for power in
  kilowatt-hours (kWh)
• Hence, multiplying the number of watts consumed
  by the number of hours that the watts are being
  consumed produces watt-hours

92
Example

• ?????????????????????????????? (????????? 3
  ??????????????, ?????? 22 ?????????)
  ???????????????????????????????? TOU
  ???????????????????????????????????????????????
• ????????????????????
• On-Peak (?????? - ????? 09.00 - 22.00 ?.) 850 kW
• Off-Peak 1 (?????? - ????? 22.00 09.00 ?.) 600
  kW
• Off-Peak 2 (????? - ???????, ?????????????) 500
  kW

93
Example

• ????????????????
• On-Peak (?????? - ????? 09.00 - 22.00 ?.) 40,000
  ?????
• Off-Peak 1 (?????? - ????? 22.00 09.00
  ?.) 35,000 ?????
• Off-Peak 2 (????? ???????, ?????????????) 25,000
  ?????
• ?????????????
• ??????????????????????? 74.14 ???/?????????
• ??????????????? (Peak) 2.695 ???/?????

94
Example

• ??????????????? (Off-Peak) 1.1914 ???/?????
• ????????? 228.17 ???/?????
• ?????????????? 0.4328 ???/?????
• Solution
• ??????????????????????? 85074.17 63,044.50
  ???
• ??????????????? (400002.695)(350001.1914)(25
  0001.1914)
• 172,586.19 ???

95
Example

• ?????????????? 1000000.4328 43,280 ???
• ????????? 228.17 ???
• ??? (1) 279,138.86 ???
• ?????????????????? (7) (279138.36)(0.07)
  19,539.72
• ?????????????????? 298,678.58 ???

96
Complex power

• For circuit operating in sinusoidal-steady-state,
  real and reactive power conveniently calculated
  from complex power
• Complex power is the complex sum of real power
  and reactive power which can be defined as

97
Complex power

• Where
• S is Apparent Power (VA)
• P is Real Power (W)
• Q is Reactive power (VAR)

98
Complex power

• Thus, the apparent, real and reactive power in
  three-phase system are calculated as following
• In form of phase value

Vp is Line-to-Neutral Voltage or Phase Voltage
99
Complex power

• In form of line value

VL is Line-to-Line Voltage
100
Example

• The wye source with the rating of 100 kVA and 380
  V (line) supplies power the wye load. Load has a
  operating power factor of 0.8 lagging. Calculate
• Phase voltage at load
• Phase current at load
• Real power at load
• Reactive power at load

101
Example

• Solution
• a)
• b)

102
Example

• For wye-connection
• c)

103
Example

• d)

104
Power Factor

• Power factor is defined as the ratio of the real
  power to the apparent power
• Power factor will be leading or lagging
  depending on which way the load shifts the
  currents phase with respect to the voltages
  phase
• Inductive loads cause current to lag behind
  voltage, while capacitive loads cause current to
  lead voltage

105
Power Factor

• From the power triangle, the following equations
  can be expressed

S (VA)
Q (VAR)
W
106
Example

• An electrical load operates at 240 V rms. The
  load absorbs an average power of 8 kW at a
  lagging power factor of 0.8. Calculate the
  complex power of the load.
• Solution
• Since the power factor is described as lagging,
  it indicates that the load is inductive

107
Example
S
Q
P
108
Example

• A 50 kW fully-loaded motor is operating at 595
  volts, 3-phase drawing 88 A. Determine operating
  power factor.
• Solution

88 A
S
Q
Voltage Source 595 V, 3 phase
Motor 50 kW
P 50 kW
109
Example
Since motor is three-phase motor, thus the
apparent power is calculated from
Then, operating power can be calculated by
110
Example

• Alternatively, operating power factor can be
  calculated in just one step as following

111
Power Factor

• Low power factor results in
• Power loss in the network
• Higher transformer losses
• Increased voltage drop in power distribution
  networks
• Less power distributed via the network
• Increasing peak kVA

112
Power Factor Correction

• Power factor improvement can be obtained by
  adding capacitor into the system
• Capacitors produces the necessary leading
  reactive power to compensate the lagging reactive
  power

113
Power Factor Correction
VA1
VAR1
VA2
VAR2
W
Where Cos ?1 is power factor before power
factor correction Cos ?2 is power factor after
power factor correction VAR1 is reactive power
before power factor correction VAR2 is reactive
power after power factor correction
114
Power Factor Correction

• Calculation Method
• Suppose the power factor of both before and after
  power factor correction are known
• Thus

115
Power Factor Correction
Therefore, the required VAR can be calculated
from
116
Power Factor Correction

• Where to connect capacitor on the customers
  electrical system?
• Connect all together at any point past the
  utility metering (usually at the service entrance
  switchgear)
• Divide the total capacitor requirement in
  selected amounts and connect to the system about
  the load center of the feeder

117
Power Factor Correction

• Connect the capacitors as close to the loads,
  with which they will be trading VARs, as possible
• If a system or part of a system operates with a
  varying load, then connect the capacitors through
  a series of steps with each step switched
  through a contactor that is controlled by power
  factor controller

118
Example

• Suppose a factory currently has a peak demand at
  1250 kW with the operating power factor of 0.65.
  If this factory needs to improve its power factor
  to 0.95, calculate the require capacitor size.
• Solution

VA1
VAR1
PF1 0.65
1250 kW
Before Improving Power Factor
119
Example
Before Improving Power Factor
After Improving Power Factor
VA2
VAR2
1250 kW
120
Example
121
Example

• Suppose capacitor bank with a capacity of 50 kVar
  is designed for rated 240 V and 60 Hz. If this
  capacitor is applied for rated 220 V and 50 Hz,
  what is the actual capacity of this capacitor
  bank?

122
Example
123
Example
124
Transformer

• A schematic represents of a two-winding
  transformer with the phasor voltages E1 and E2
  across the windings

125
Transformer

• From figure the phasor current I1 entering
  winding 1, which has N1 turns
• The phasor current I2 entering winding 2, which
  has N2 turns
• Ohms law for the magnetic circuit states that
  the net magnetomotive force (mmf) equals the
  product of core reluctance (Rc) and the core flux
  (?)

126
Transformer

• For an ideal transformer, core is assumed to have
  infinite permeability, therefore, the core
  reluctance is zero

127
Transformer

• By applying the Faradays law, voltage at the
  primary side can be calculated as
• Assuming a sinusoidal-steady-state flux with
  constant frequency and replacing e(t) and
  by E and ?

128
Transformer

• Similarly, the voltage at secondary side is given
  as
• Therefore,

129
Transformer

• Turn ratio (a) is defined as follows
• Hence, the basic relation for an ideal
  single-phase two-winding transformers are

130
Transformer

• Two additional relations concerning complex power
  and impedance can be also derived
• The complex power entering winding 1

131
Transformer

• If an impedance Z2 is connected across winding 2
  of the ideal transformer, then
• This impedance, when measured from winding 1, is

132
Example

• A single-phase two-winding transformer is rated
  20 kVA, 480/120 V, 60 Hz. A source connected to
  the 480-V winding supplies an impedance load
  connected to the 120-V winding.
• The load absorbs 15 kVA at 0.8 p.f. lagging when
  the load voltage is 118 V. Assume that the
  transformer is an ideal. Calculate
• a) The voltage across the 480-V winding
• b) The load impedance
• c) The load impedance referred to the 480-V
  winding

133
Example

• Solution

Selecting the load voltage E2 as the reference
134
Example
S2 15 kVA
VAR
P
135
Example

• a) Turn Ratio
• The voltage across winding 1

136
Example

• b)
• Therefore,
• The load impedance is calculated from

137
Example

• c) The load impedance referred to the 480-V
  winding can be calculated by

138
Per-Unit System

• In power system a normalization of variable
  called per unit normalization is almost always
  used
• It is especially convenient if many transformers
  and voltage levels are involved
• One advantage of the per-unit system is to avoid
  the possibility of making serious calculation
  errors when referring quantities from one side of
  a transformer to the other

139
Per-Unit System

• The basic idea is to pick base values for
  quantities such as voltages, currents,
  impedances, and power
• Hence, the quantity in per unit is defined as the
  ratio of actual quantity to the base value
  quantity

140
Per-Unit System

• For Single Phase System

141
Per-Unit System

• For Three Phase System

142
Per-Unit System
Per Unit Value
143
Per-Unit System

• When only one component, such as a transformer,
  is considered, the nameplate ratings of that
  component are usually selected as base values
• When several components are involved, the system
  base values may be different from the nameplate
  ratings of any particular device
• It is then necessary to convert the per-unit
  impedance of a device from the old base to new
  base values

144
Per-Unit System

• To convert a per-unit impedance from old to
  new base use

145
Example

• Three zones of a single-phase circuit are
  connected by transformers T1 and T2 whose ratings
  are shown in figure. By using base values of 30
  kVA and 240 V in zone 1
• a) Draw the per-unit circuit and determine the
  per unit impedances and per-unit source voltage
• b) Calculate load current in both in per-unit and
  in amperes

146
Example

• Solution
• a) First the base values in each zone are
  determined.
• For base kVA will be 30 kVA for the entire
  network

Zone 1
Zone 2
Zone 3
Xline 2 O
Zload 0.9 j 0.2 O
30 kVA, 240/480 volts X0.1 pu
20 kVA, 460/115 volts X0.1 pu
147
Example

• Thus
• For base voltage in zone 1
• As moving across a transformer, the voltage base
  is changed to the transformer voltage ratings,
  hence

148
Example

• For base voltage in zone 3
• The base impedance in zone 2 and 3 are

149
Example

• The base current in zone 3 is calculated from
• Per-unit reactance of Transformer 1 is then
  calculated from

150
Example

• Per-unit reactance of line in zone 2 can be
  calculated from
• Per-unit reactance of transformer 2 must be
  converted from its nameplate rating to the system
  base (Vb2)

151
Example

• Alternatively, use Vb3
• Load in zone 3 has a per-unit impedance of

152
Example

• Finally, the per-unit source voltage is
  calculated from

Xpu(line)j0.2604 pu
Xpu(T1)j0.1 pu
Xpu(T2)j0.1378 pu
Zpu(load) 1.875j0.4167 pu
Vpu(source) 0.9167
153
Example

• The per-unit load current is then simply
  calculated from per-unit circuit

154
Example

• The actual load current is