UNESCO-NIGERIA
UNESCO
TECHNICAL &
VOCATIONAL EDUCATION
REVITALISATION PROJECT-PHASE
PROJECT
II
NATIONAL DIPLOMA IN
ELECTRICAL ENGINEERI
ENGINEERING
NG TECHNOLOGY
ELECTRICAL CIRCUITS (II
II)
COURSE CODE:
CODE EEC 249
YEAR III SEMESTER IV
THEORY
Version 1: December 2008
TABLE OF CONTENTS
Department
Electrical Engineering Technology
Subject
Electric Circuit Theory(I I)
Year
2
Semester
4
Course Code
EEC 249
Credit Hours
2
Theoretical
1
Practical
2
CHAPTER
•
1
: Power in A.C. Circuit
Assessment
CHAPTER
2
Assessment
•
Laboratory
•
Mid term Test1
3
Assessment
•
Laboratory 2
4
2
Weeks 4 – 7
: Time Domain Analysis
•
CHAPTER
Week 1-3
: Three Phase Systems
•
CHAPTER
1
3
Weeks 8 - 11
: Magnetic Coupling
•
Assessment
•
Laboratory 3
4
Weeks 12 –15
2
This Page is Intentionally Left Blank
3
Week 1
1.Power in A.C Circuit
At the end of this week, the students are expected to:
♦
Calculate power in a.c circuits containing:
♦
Resistance
♦
Inductance
♦
Capacitance
♦
Combination of the resistance, inductance and capacitance
1.1
POWER IN A.C CIRCUIT
1.1.1 Power in A.C circuit containing Resistance only
+
V mI m
i
P(t)
Average power
V mI m
2
+
e
R
_
t
0
i(t)
_
v(t)
Fig 1.1: (a) pure resistive circuit, (b) power wave form.
(b)
Inspection of the power waveform of fig 1.1 shows that its average value lies half way
between zero and its peak value of VmIm. That is
P = VmIm/2
Since V (the magnitude of the r.m.s value of voltage) is Vm/√2 and I (the magnitude
of the r.m.s value of current) is Im/√2, this can be written as P = VI. Thus, power
(average power) to a purely resistive load is P = VI (watts).
(1.1)
Example 1.1: Calculate the power dissipated by the circuit of fig 1.2
+
100V
I
R = 25Ω
_
Fig 1.2
4
Solution
I = 100V/25Ω = 4A
∴ P = VI = 100 x 4 = 400W
1.1.2 Power in A.C Circuit Containing Inductance only
For a purely inductive load as in fig 1.3(a), current lags voltage by 900. A sketch of P
versus time (obtained by multiplying V times i) then looks as shown in fig 1.3(b).
+
i
PL(t)
VI
i(t)
v(t)
+
PL
V
_
0
t
(a)
Energy
stored
_
Energy
released
Energy
stored
Energy
released
Fig 1.3: (a) pure inductive circuit (b) power waveform(b)
for a purely inductive circuit
Consider fig 1.3. Energy stored during each quarter-cycle is returned during the next
quarter cycle. Thus, the average power is zero. Consequently, the only power flowing
in the circuit is reactive power. This is given by
QL = VI (VAr)
(1.2)
Example 1.2: For the circuit of fig 1.4, determine the reactive power
+
100V
I
XL = 20Ω
_
Solution
I = 100V/20Ω = 5A
Fig 1.4
∴ QL = VI = 100 x 5 = 500VAr
5
1.1.3 Power in A.C circuit containing capacitance only
For a purely capacitive load current leads voltage by 900. Multiplications of V times i
yield the power curve of figure 1.5
+
PC(t)
i
v(t)
VI
i(t)
+
PC
_
0
V
(a)
-VI
t
(b)
Energy
released
Energy
released
Energy
stored
Energy
stored
(b)
Fig 1.5: (a) pure capacitive circuit (b) power waveform for a pure capacitive circuit.
Consider fig 1.5. Energy stored during each quarter-cycle is returned during the next
quarter cycle. Thus, the average power is zero. Consequently, the only power flowing
in the circuit is reactive power. This is given by
QC = I2XC
(1.3)
Example 1.3: With regard to fig 1.6, determine average and reactive power.
+
100V
I
XC = 40Ω
_
Fig 1.6
Solution
I = 100V/40Ω = 2.5A.
QC = VI = 100 x 2.5 = 250VAr.
P = 0W.
6
1.1.4 Calculations of power in A.C circuits containing R, L and C
Example 1.4: For the RL circuit of figure 1.7, I = 5A. Find the power and reactive power.
XL = 4Ω
R = 3Ω
P
5A
Q
Fig 1.7
Solution
P = I2R = (5)2 (3) = 75W
QL = I2XL = (5)2 (4) = 100VAr
Example 1.5: For the RC circuit of figure 1.8, determine the power consumed and the
reactive power.
+
40V
_
20Ω
R
80Ω
XC
Fig 1.8
Solution
P = V2/R = (40)2/20 = 80W
QL = V2/XC = (40)2/80 = 20VAr
Example 1.6: A 10Ω resistor, a 100µF capacitor and an inductor of 0.15H are connected in
series to a supply at 230V, 50Hz. Calculate the power consumed by the circuit.
Solution
XL = 2πf L = 2π x 50 x 0.15 = 47.1Ω
XC =
1 =
1
= 31.83Ω
2πfC 2π x 50 x 100 x 10-6
7
Z = √{R2 + (XL – XC)2} = √{102 + (47.1 – 31.83)2} = 18.28Ω
I = VS = 230 = 12.58A
Z
18.28
Power consumed, P = I2R = (12.58)2 x 10 = 1582W
8
Power in A.C Circuit
Week 2
At the end of this week, the students is expected to:
♦
Explain power factor and factors affecting its value
♦
Explain methods of power factor correction
1.2
POWER FACTOR AND FACTORS AFFECTING ITS VALUE
1.2.1 Power Factor
Power factor is the ratio of the real power dissipated in the load to the apparent power
of the load. Thus,
P.f = cosθ = P/S
(1.4)
Angle θ is the angle between voltage and current. For a pure resistance, therefore, θ =
00. For a pure inductance, θ = 900; for a pure capacitance, θ = -900. For a circuit
containing resistance and inductance, θ will be somewhere between 00 and 900; for a
circuit containing resistance and capacitance, θ will be somewhere between 00 and 900.
1.2.2 Factors Affecting the Value of Power Factor
1.
Increase in supply voltage
Due to increase in supply voltage, which usually occurs during low loads period, the
magnetizing current of inductive reactance increase and power factor of the plant as a
whole comes down.
2.
Improper Maintenance and Repairs
The power factor at which motors operates falls due to improper maintenance and
repairs of motors.
3.
Industrial Heating of Furnance
Induction furnances operates at a very low lagging power factor due to heating of the
furcance.
1.3
METHODS OF POWER FACTOR CORRECTION
1.
Synchronous Motors
These machines draw leading KVAr when they are over excited and, especially, when
they are running idle. They are employed for correcting of the power factor in bulk
and have the special advantage that the amount of correction can be varied by
changing their excitation.
9
2.
Static Capacitors
They are installed to improve the power factor of a group of a.c motors and are
practically loss-free. Since their capacitances are not variable, they tend to overcompensate on light loads, unless arrangements for automatic switching off the
capacitor bank are made.
3.
Phase Advancers
The power factor of induction motors is being improved by equipping the machines
with a phase advancer, which supply exciting current to the motor circuit. With the
arrangement, the phase angle between the supply voltage and current would be
decrease, thereby increasing the power factor.
10
Week 3
Power Calculation in A.C Circuit
At the end of this week, the students are expected to:
♦
Explain apparent power, reactive power and active power
♦
Solve problems on power factor, active power apparent power reactive power and
power factor
1.4
APPARENT
POWER,
REACTIVE
POWER
AND
ACTIVE
POWER
1.4.1 Apparent power
I
+
S
load
V
_
Fig 1.9: Apparent power S = VI
When a load has voltage V across it and current I through it as in figure 1.9, the power
that appears to flow to it is VI. However, if the load contains both resistance and
reactance, this product represents neither active power nor reactive power. Since VI
appears to represent power, it is called apparent power. Apparent power is given the
symbol S and units of volt-amperes (VA). Thus
S = VI (VA)
(1.5)
where V and I are the magnitude of the r.m.s voltage and current respectively.
1.4.2 Reactive Power
Consider figure 1.3. During the intervals that instantaneous power PL(t) is negative,
power is being returned from the load. (This can only happen if the load contains
reactive elements: L or C.) The portion of power that flows into the load then back out
is called reactive power. Reactive power is given by the symbol Q and units of voltampere reactive (VAr). Thus
Q = VIsinθ
(1.6)
1.4.3 Active power
Consider again figure 1.3. Since PL represents the power flowing to the load; its
average will be the average power to the load. Denote this average by the letter P. If P
has a positive value, it represents the power that is really dissipated by the load. For
11
this reason, P is called Real power. In modern terminology, real power is also called
active power. The unit of active power is in watt (W). Thus
P = VIcosθ (W)
1.5
(1.7)
SOLVED PROBLEMS ON POWER FACTOR, ACTIVE POWER,
APPARENT
POWER,
REACTIVE
POWER
AND
POWER
FACTOR CORRECTIONS
Example 1.7:
(a)
In figure 1.10, find the following:
The active power (b) the apparent power (c) the reactive power and (d) the
power factor.
R = 8Ω
XL = 7Ω
XC = 15Ω
V = 100
Fig 1.10
Solution
Z = √{R2 + (XC – XL)2} = √{82 + (15 – 7)2} = 11.314Ω
I = E = 100 = 8.84A
Z 11.314
From the phasor diagram of RLC series circuit shown in fig 1.11
VL
V
V L - VC
φ
VR
I
Fig 1.11
VC
φ = cos-1 (VR/V), where VR = IR = 8.84 x 8 = 70.72V
12
∴ φ = cos-1 (70.72/100) = 450
(a) P = VIcosφ = 100 x 8.84 x cos450 = 625W
(b) S = VI = 100 x 8.84 = 884VA
(c) Q = VIsinφ = 100 x 8.84 x sin450 = 625VAr
(d) P.f = cosφ = cos450 = 0.707
Example 1.8: A resistor of 40Ω is connected in parallel with a 67.6mH inductor, the
combination being supplied by a 120V, 50Hz supply. Calculate (a) the power factor
(b) the active power (c) the apparent power (d) the reactive volt-amperes consumed
Solution
IR = VS/R = 120/40 = A
120
= 5.65A
IL = VS =
-3
2πfL 2π x 50 x 67.6 x 10
I = √(IR2 + IL2) = √(32 + 5.652) = 6.397A
From the phasor diagram of RL parallel circuit (fig 1.12)
IR
VS
φ
IL
I
Fig 1.12
(a) P.f = cosφ = IR/I = 3/6.397 = 0.4690
(b) P = VIcosφ = 120 x 6.397 x 0.4690 = 360W
(c) S = VI = 120 x 6.397 = 767.64VA
(d) Q = VIsinφ, φ = cos-1(0.469) = 620
= 120 x 6.397 x sin620 = 677.8VAr
Example 1.9: A load of P = 1000KW with p.f = 0.5lagging is fed by a 5KV, 50Hz source. A
capacitor is added in parallel such that the power factor is improved to 0.8. Find the value of
the shunt capacitance needed to improve the power factor
13
Solution
The value of the shunt capacitance is given by
C = P(tanφ1 - tanφ2)
2πfV2
Before improvement.
P = 1000KW, cosφ1 = 0.5, ⇒ φ1 = cos-1(0.5) = 600
∴ Ptanφ1 = 1000K x tan600 = 1732.05KVAr
∴ C = 1732.05 – 750.82 ≅ 125µF
2π x 50 x (5000)2
14
Week 4
Three Phase Systems
At the end of this week, the students are expected to:
♦
Define polyphase system
♦
Explain the basic differences between single phase and three phase systems
♦
Explain phase sequence of a three phase system
♦
State the advantages of three phase circuits.
2.1
POLYPHASE SYSTEM
Circuits or system in which the ac sources operate at the same frequency but different
phases are known as polyphase systems.
2.2
BASIC DIFFERENCES BETWEEN SINGLE PHASE AND THREE
PHASE SYSTEMS
2.2.1 Single-Phase Systems
A single phase a.c power system consists of a generator connected through a pair of
wires (a transmission line) to a load. Figure 2.1 depicts a single phase two wire
system, where VP is the magnitude of the source voltage and φ is the phase. The
power in a single phase system is pulsating (not constant). For large motors, pulsating
power supply causes excessive vibration. Besides, it has neither delta nor star type of
connection.
Vp ∠Φ +_
ZL
Fig 2.1: single-phase system two-wire type.
2.2.2 Three Phase Systems
Figure 2.2 shows a three phase four wire systems. As distinct from a single phase
system, a three phase system is produced by a generator consisting of three sources
having the same amplitude and frequency but out of phase with each other by 1200.
When one phase or two phase inputs are required, they are taken from the three phase
system rather than generated independently. The instantaneous power in a three phase
system can be constant. This result in uniform power transmission and less vibration
of three phase machines. Generally, three-phase systems have two type of connection,
which are star and delta connections
15
Vp 00
_
+
Vp -12 00
_
+
a
A
ZL1
b
B
ZL2
c
C
ZL3
Vp +12 00
_
+
n
N
Fig 2.2: Three phase four wire-systems
2.3
PHASE SEQUENCE OF A THREE PHASE SYSTEM
Consider the waveform of three phase e.m.f shown in fig 2.3. From the waveform it is
seen that the e.m.f of phase A lead that of B by 1200, and in similar way, the e.m.f of
phase B lead that of C by 1200. Hence, the order in which the e.m.fs of phases A, B
and C attain their maximum values is ABC. It is called phase sequence ABC.
Therefore, Phase sequence is the order in which the three phases (A,B and C) reach
their maximum value.
e
A
B
C
ωt
120
0
2400
Fig 2.3: three phase waveforms
16
2.4
ADVANTAGES OF 3-PHASE CIRCUITS
1.
The instantaneous power in a three phase system can be held constant. This
results in uniform transmission.
2.
Three phase systems are more economical than single phase systems.
3.
Three phase systems are more efficient
17
Week 5
Three Phase Systems
At the end of this week, the students are expected to:
♦
Explain how 3-phase e.m.fs are produced
♦
Distinguish between star and delta three-phase system
♦
Derive the relationship between line and phase values of voltages and currents in a
star and delta connected windings
2.5
GENERATION OF THREE PHASE E.M.F’s
B1
A
C1
N
S
C
B
A1
Fig 2.4: Generation of 3-φ e.m.f’s
In figure 2.4, three similar coils (A,B, and C) are displaced from one another by 120
electrical degrees. If the coils are rotated within the magnetic field, emf would be
induced or generated in the three coils. It is evident that counterclockwise rotation
results in coil sides A, B, and C in the order A-B-C. The result for the three coils is as
shown in fig 2.3. Voltage B is 120 electrical degrees later than A, and C is 2400 later.
Changing the direction of rotation would result in A-C-B, which is called the ACB
phase sequence.
2.6
DIFFERENCE
BETWEEN
STAR
AND
DELTA
3–PHASE
SYSTEM
2.6.1 Star-Connected 3-Phase System
1.
Neutral wire is available
2.
Phase current = line current
3.
Phase voltage = line voltage
√3
4.
It can handle both lighting and power loads
18
2.6.2 Delta-Connected 3-Phase system
2.7
1.
Neutral wire is not present
2.
Phase current = line current
√3
3.
Phase voltage = line voltage
4.
It can handle power loads only
DERIVATION OF THE RELATIONSHIP BETWEEN LINE AND
PHASE VALUES OF VOLTAGES AND CURRENT IN A STAR
AND DELTA CONNECTED WINDINGS
2.7.1 Line voltages and phase voltages in a star connected windings
Line 1
VR
VRY
VB
VBR
Vy
Line2
VYB
Line3
(a)
Fig 2.5: (a) Star connection of a 3-Φ circuit
VRY
VR
-VY
-VB
300
VYB
VB
VY
-VR
VBR
19
Fig 2.6: phasor diagram of a star connected load
Consider the star connection of a three phase circuit shown in fig 2.5. It phasor
diagram is as shown in fig 2.6. To obtain the line voltages we proceed as under:
Let VRY = line voltage between red phase and yellow phase
VYB = line voltage between yellow phase and blue phase
VBR = line voltage between blue phase and red phase
VR = voltage across the red phase
VY = voltage across the yellow phase
VB = voltage across the blue phase
Thus, VYB, VBR and VYB are called line voltages, while VB, VR and VY are called
phase voltages.
The p.d between line 1 and 2 in (fig 2.5) is VRY. Hence, VRY is found by
compounding VR and VY reversed and its value is given by the diagonal of the
parallelogram of fig 2.6. Obviously, the angle between VR and VY reversed is 600.
The parallelogram is shown in fig 2.7 below.
VY
VRY
x
o
300
VR
Fig 2.7: parallelogram of fig 2.6
From fig 2.7, ox = ½ VRY
(2.1)
Also, ox = VR cos 300
(2.2)
Equating (2.2) to equation (2.1) gives
VRY = 2 (VRcos300)
= √3 VR = √3 Vph
Considering that the system is balanced,
∴ VYB = √3 Vph
also, VBR = √3 Vph
Now VRY = VYB = VBR = line voltage, say VL. Hence, in star connection
VL = √3Vph
(2.3)
20
2.7.2 Line currents and phase currents in a star connected windings
R
IR
N
IB
Iy
Y
B
Fig 2.8: star connection of three-phase circuit
Consider fig 2.8.
Let IR = current flowing through the red phase
IY = current flowing through the yellow phase and
IB = current flowing through the blue phase
IR, IY and IB are called phase current (Iph). It is seen that each of the phase currents is
equal to the current flowing through the respective lines. Thus, the current flowing
through the respective lines is known as the line current (IL)
Hence, in star connection, line current = phase current.
∴ IL = Iph
(2.4)
2.7.3 Line currents and phase currents in a delta connected windings
1
(IR – IB)
R
IB
IR
VRY
VBR
2
IY
(IY – IR)
Y
VYB
(IB – IY)
3 B
Fig 2.9 Delta connection of a three phase circuit
21
VRY
(IR – IB)
IR
-IY
Φ
-IB
(IB – Iy)
Φ
Φ
IY
IB
VYB
-IR
VBR
(IY – IR)
Fig 2.10 Phasor diagram of a delta connection of a three phase circuit
Consider the delta connection of a three phase circuit shown in fig 2.9. It phasor
diagram is as shown in fig 2.10. To obtain the line currents we proceed as under:
Let I1 = IR – IB
I2 = IY - IR
I3 = IB - IY
I1, I2 and I3 are called line currents
Line current I1 is also found by compounding IR and IB reversed as shown in fig 2.10.
Its value is given by diagonal of fig 2.10. The parallelogram of fig 2.10 is as shown in
fig 2.11.
IR
I1
x
o
300
IB
Fig 2.11: parallelogram of fig 2.10
From fig 2.11,
Ox = ½ I1
Also, ox = IB cos30
(2.5)
0
(2.6)
22
From equation (2.6) and (2.6), ½ I1 = √3/2 IB
If IR = IY = IB = phase current (Iph), then
Current in line no.1 is I1 = √3Iph
Current in line no.2 is I2 = √3Iph and
Current in line no.3 is I3 = √3 Iph
Since all the line currents are equal in magnitude, i.e I1 = I2 = I3 = IL
∴ IL = √3Iph
(2.7)
2.7.4 Line voltages and phase voltages in a delta connected windings
It is seen from fig 2.9 that there is only one phase winding completely included
between any pair of terminals. Hence, in delta connection, the voltage between any
pair of lines is equal to the phase voltage of the phase winding connected between the
two lines considered. Hence, for a balanced system, VRY = VYB = VBR = line voltage
VL. Then, it is seen that
VL = Vph
(2.8)
23
Week 6
Three Phase Systems
At the end of this week, the students are expected to:
♦
Derive an expression for power in a three-phase circuit (balanced and unbalanced)
♦
Explain the two wattmeter and single wattmeter methods of measuring three-phase
power
2.8
DERIVATION
FOR
POWER
IN
A
3-PHASE
CIRCUIT
(BALANCED AND UNBALANCED)
2.8.1 Power in a 3-Phase circuit (balanced-delta connection)
Consider the delta connection of a three phase circuit (fig 2.9). The total active power
in a delta connected circuit is the sum of the three phase powers. Hence,
Power per phase = VphIphcosφ
Total power = 3 x VphIph cos φ
However, Vph = VL and Iph = IL/√3. In terms of line values, the above expression for
power becomes
P = 3 x VL x IL x cosφ = √3 VLILcosφ
√3
(2.9)
Similarly, total reactive power is given by
Q = √3 VLILsinφ
(2.10)
and total apparent power of the three phase is
S = √3 VLIL
(2.11)
2.8.2 Power in a three phase circuit (unbalanced star connection)
IR
Vph
ZR
In
IY
Vph
ZY
N
ZB
Vph
IB
Fig 2.11: unbalanced three phase star connected load
Consider the unbalanced three phase star connected load, where it is assumed that the
phase voltages (Vph) are equal. The line currents are obtained from ohms law.
24
Power in phase R, PR = phase voltage x Phase current x cosφR, where φR is the angle
between phase voltage and phase current.
∴PR = Vph IRcosφR
(2.12)
Similarly, power in phase Y is obtained as
PY = Vph IY cosφY,
(2.13)
Power in phase B is
PB = Vph IB cosφB
(2.14)
Total power in the three phase is
PT = PR + PY + PB
= Vph IRcosφR + Vph IYcosφY + Vph IBcosφB
2.9
(2.15)
2-WATTMETER AND SINGLE-WATTMETER METHODS OF
MEASURING 3-PHASE POWER
2.9.1 2-Wattmeter Method
In this method, two wattmeters are connected in two phases and their potential coils
are connected to the remaining phase. As shown in fig 2.12, wattmeter W1 is inserted
in phase a and wattmeter W2 in phase c. Their potential coils are connected across a
and b, and c and b respectively. The current coil (c.c) of each wattmeter measures the
line current, while the respective potential coil (p.c) measures the line voltage. The
algebraic sum of the two wattmeter readings equals the total power absorbed by the
loads. Thus
W1 + W2 = Total three phase power (√3 VLILcosφ)
(2.16)
The following are important with regard to two wattmeter method.
1.
For a lagging Power factor,
tanφ = √3 W2 – W1
W2 + W1
2.
(2.17)
For a leading power factor
tanφ = _
√3 (W1 – W2)
(2.18)
W1 + W2
25
a
c.c
W1
ia
p.c
ZA
b
ZB
ZC
p.c
ic
c
c.c
Fig 2.12: schematic diagram of a two wattmeter method
2.9.2 Single- Wattmeter Method
In this method, one wattmeter is used to get two readings which are obtained by two
wattmeter in the two wattmeter method. The total power is 3 times the reading of that
one wattmeter. This method is used only when the load is balanced. Figure 2.13
shows the schematic diagram of a single wattmeter method.
a
W1
c.c
Z
p.c
Z
Z
b
c
Fig_ 2.13: schematic diagram of a single-wattmeter method
26
Week 7
Three Phase Systems
At the end of this week, the students are expected to:
♦
Solve problems on
Line and phase voltages and current in a star and delta connected windings
Power in a three phase circuit
2-wattmeter and single wattmeter methods of measuring 3-phase circuits
2.10 SOLVED PROBLEMS ON LINE AND PHASE VOLTAGES AND
CURRENT IN A STAR AND DELTA CONNECTED WINDINGS,
POWER IN THREE PHASE CIRCUIT AND 2-WATTMETER AND
SINGLE WATTMETER METHOD OF MEASURING 3-PHASE POWER
Example 2.1The impedance per phase of a delta connected load is 10Ω per phase, the line
voltage being 230V. Determine (a) the phase current (b) the line current (c) the apparent
power of the load.
Solution
(a) Iph = Vph/Zph = 230/10 = 23A
(b) IL = √3 Iph = √3 x 23 = 39.8A
(c) S = √3 VL IL = √3 x 230 x 39.8 = 15.9KVA
The circuit arrangement is shown in fig 2.14.
Iph = 39.8A
VL = 230V
Zph = 10Ω
Iph = 23A
Zph = 10Ω
Zph = 10Ω
Fig 2.14 for example 2.1
27
Example 2.2 An unbalanced star-connected load has balanced phase voltage of 120V and
the a.b.c sequence. The line currents are given as Ia = 10∠00A, Ib = 20∠-400A and Ic = 30∠1550A. Find the power consumed by the circuit.
Solution
Ia = 10∠50A, Vpa = 120∠00V
Ib = 20∠400A, Vpb = 120∠-1200V
Ic = 30∠1550A, VPc = 120∠-2400V
Pa = VPIacosφa = 120 x 10cos(0 – 0) = 1200W
Pb = VpIbcosφb = 120 x 20cos{-120 –(-40)} = 416.76W
Pc = VPIccosφc = 120 x 30cos{-240 – (-155)} = 313.76W
P = Pa + Pb + Pc
= 1200 + 416.76 + 313.76 ≅ 1.93KW
Example 2.3: Line voltage and current of a delta connected inductive load is 160V and 40A
respectively. Power factor of load is 0.8lag. Assuming that power is being measured using
two wattmeters, find the readings of the wattmeters.
Solution
VPh = VL = 160V, IL = 40A
Total power, P = √3 VLILcosφ
= √3 x 160 x 40 x 0.8 = 8868W
∴ W1 + W2 = 8868
(2.4.1)
cosφ = 0.8, ⇒ φ = cos-1(0.8) = 36.90
tanφ = 0.75
For a lagging P.f,
tanφ = √3 (W1 - W2)
W1 + W2
⇒ 0.75 = √3 (W1 – W2), ⇒ W1 – W2 = 8868 x 0.75 = 3840
8868
√3
∴ W1 – W2 = 3840
(2.4.2)
Adding eqtn. (2.4.1) and (2.4.2) together gives
2W1 = 12708
⇒ W1 = 12708/2 = 6354W
From eqtn. (2.4.1), W2 = 8868 – 6354 = 2514W
28
Example 2.4: The current coil of a wattmeter is connected to the red phase of a star
connected load, while the potential coil is connected between the line and the neutral. If the
phase voltage and current are respectively 20∠800V and 30∠500A, find the total power
consumed by the load.
Solution
The circuit diagram is shown in fig 2.15
W1
c.c
IR
R
Vph
Z
p.c
Z
Z
Y
B
Fig 2.15: schematic diagram of a single-wattmeter method
P = √3 VLILcosφ
where IR = IL and VL = √3 VPh = √3 x 20∠800 = 34.64∠800V
P = √3 x 34.64 x 30 cos(800 – 500) = 1558.8W
29
Week 8
Time Domain Analysis
At the end of this week, the students are expected to:
♦
Explain the meaning of transient
♦
Derive formulae for current and voltage growth and decay in RC circuits
3.1
TRANSIENT
when a circuit possesses energy storing elements such as inductance and capacitance,
the energy state of the circuit can be disturbed by changing the position of the switch
(fig 3.1) connecting the elements to the source. There is a transitional period during
which the branch current and element voltages change from their former values to
new ones. This period is called the transient.
3.2
DERIVATIONS
OF
FORMULAE
FOR
CURRENT
AND
VOLTAGE GROWTH AND DECAY IN RC CIRCUIT
3.2.1 Current and Voltage Growth in RC Circuit
Let us consider the circuit shown in figure 3.1
R
C
S
V
Fig 3.1: RC series circuit
Let the current in the circuit at any instant t be i, and capacitor voltage be v. When
switch S is closed and Kirchhoff’s law is being applied, this yield
Ri + v = V
(3.1)
Since i = C(dv/dt), equation (3.1) becomes
V = RC (dv/dt) + v
∴ V – v = RC (dv/dt)
so that dt/RC = dv/(V – v)
Integrating both sides, we have
t = -ln (V – v) + A
RC
where A is the constant of integration
when t = 0, v = 0
30
∴ A = ln V
V .
So that t = ln
RC
V–v
∴
and
V = e t/RC
V–v
v = V (1 – e
-t/RC
) volts
(3.2)
Also i = C dv = CV . d (1 – e-t/RC)
dt
dt
∴ i = V/R . e
-t/RC
(3.3)
3.2.2 Current and Voltage Decay in RC Circuit
Let the capacitor C in fig 3.1 be charged initially to V volts and then be discharged
through a resistor R as shown in figure 3.2
C
R
Fig 3.2: Decay of current and voltage in RC circuit
By Kirchhoff’s law
iR + v = 0 or -v = C dv
R
dt
{ i = C dv/dt}
So that dt/RC = -dv/v
Integrating both sides, we have
t = - ln v + A
RC
When t = 0, v = V, so that A = ln V. Hence
t = ln V/v
RC
So that V/v = et/RC
and v = Ve
-t/RC
(3.4)
Also i = -v/R
The negative sign indicates that the direction of the discharge current is the reverse of
that of the charging current
∴ i = -V/R e-t/RC
(3.5)
31
Week 9
Time Domain Analysis
At the end of this week, the students are expected to:
♦
Sketch the growth and decay curves in RC circuits
♦
Define time constant in RC circuit
♦
Solve problems involving transient in RC circuit
3.3
SKETCH FOR GROWTH AND DECAY CURVES OF CURRENT
AND VOLTAGE IN RC CIRCUIT
The sketch for growth of current and voltage in RC circuit is shown in figure 3.3
while the sketch for decay of current and voltage in RC circuit is as shown in fig 3.4
i,v
V
Charging current
Voltage across the capacitor
Voltage across the capacitor
T
t
Discharging current
t
i
Fig 3.3
3.4
Fig 3.4
TIME CONSTANT IN RC CIRCUIT
Consider fig 3.3 above. T is known as the time constant in RC circuit. It is given by
the expression
T = RC, (seconds)
(3.6)
Time constant of an RC circuit is the time required for the voltage across
capacitor to increase from zero to its final value if it continued increasing at its initial
rate.
32
3.5
SOLVED
PROBLEMS
INVOLVING
TRANSIENT
IN
RC
CIRCUIT
Example 3.1: A 20µF capacitor is charged to a p.d of 400V and then discharged through a
100,000Ω resistor. Determine:
(a)
the time constant
(b)
the initial value of discharged current
Solution
(a)
T = RC = 100 000 x
20
= 2s
100 000
(b)
Initial value of discharge current is
V/R = 400/100 000 = 4mA
Example 3.2: An 8µF capacitor is connected in series with a 0.5MΩ resistor across a 200V
d.c supply. Calculate:
(a)
the time taken for the P.d across the capacitor to grow to 160V:
(b)
the current and the P.d across the capacitor 4s after it is connected to the supply
Solution
(a)
-6
T = RC = 0.5 x 106 x 8 x 10 = 4s
From v = V(1 – e-t/RC)
-t/4
⇒e
= 0.2
From mathematical table
t/4 = 1.61
∴ t = 6.44s
(b)
At t = 4s
-4/4
v = 200(1 – e
and
i = v/R e
) = 200( 1 – 0.68) = 126.4V
-t/RC
= 400 e-1 = 147µA
Example 3.3: A 15µF capacitor in series with a 10KΩ resistor is connected across a 300V
d.c supply. The fully charged capacitor is disconnected from the supply and is discharged by
connecting a 11200Ω resistor across it terminals. Determine (i) the initial value of the
charging current, and (ii) the initial value of the discharge current.
33
Solution
(i)
Initial value of charging current, I0 = V/R =
300
= 30mA
10 x 103
(ii)
The initial value of discharging current is given by
i = V/R, where R = 11200Ω
∴ i = 300 = 26.7mA
11200
34
Week
10
Time Domain Analysis
At the end of this week, the students are expected to:
♦
Derive expression for the growth and decay of voltage and current in RL circuit
♦
Sketch curves for growth and decay of current and voltage in RL circuit
♦
Define time constant in RL circuit
3.6
DERIVATIONS
OF
FORMULAE
FOR
CURRENT
AND
VOLTAGE GROWTH ANF DECAY IN RL CIRCUIT
3.6.1 Current and Voltage Growth in RL Circuit
Consider the circuit shown in figure 3.5 and suppose i amperes to be the current t
seconds after the switch is closed, and di amperes to be the increase of current in dt
seconds. Then rate of change of current is
di/dt ampere per second
and induced e.m.f is
L di/dt volts
i
Switch
L
V
R
Fig 3.5: RL series circuit
Applying KVL yield
V = Ri + L di/dt
(3.7)
so that V – Ri = L di/dt
and
V/R – i = L/R . di/dt
But V/R = final value of current = I
∴ R dt = di .
L
I–i
Integrating both sides, we have
Rt/L = -ln (I – i) + A
Where A is the constant of integration
35
When t = 0, i = 0, so that, A = ln I.
∴ Rt/L = -ln(I – i) + ln I
= ln
I .
I–i
Hence I – i = e
I
-Rt/L
∴ i = I(1 – e
-Rt/L
)
(3.8)
The voltage across the inductance,
VL = L di/dt
= L d/dt [ I (1 – e
VL = IR e
-Rt/L
)]
-R/L . t
(3.9)
3.6.2 Current and Voltage decay in RL circuit
Consider the circuit shown in figure 3.6. When the switch is moved from position X
to Y, the battery voltage is short circuited. Hence, since battery is cut off from the
circuit, applying Kirchoff’s law yield.
R
X
L
Switch
Y
V
Fig 3.6
0 = Ri + L di/dt
∴ Ri = -L di/dt
The negative sign indicate a decaying current, hence
(R/L) dt = -1/t dt
Integrating both sides, we have
(R/L) t = -ln i + A
where A is the constant of integration
At the instant of closing switch, t = 0 and i = I, so that
0 = -ln I + A
∴ (R/L) t = ln I – ln i
Hence I/i = e-Rt/L
and i = I e
-Rt/L
(3.10)
Also, VL = L di/dt
36
= L [d/dt {Ie
-Rt/L
}]
VL = -IR e-Rt/L
3.7
(3.11)
SKETCH FOR GROWTH AND DECAY CURVES OF CURRENT
AND VOLTAGE IN RL CIRCUIT
The sketch for growth of current and voltage in RL circuit is shown in figure 3.7
while figure 3.8 depicts the sketch for decay of current and voltage in RL circuit
i,vL
i
Voltage across
the inductance
Decaying current
Growing current in an
inductive circuit
t
Voltage across the
inductor
t
VL
Fig 3.7
3.8
Fig 3.8
TIME CONSTANT IN RL CIRCUIT
The time constant of an RL circuit is define as the time during which current
would have reached its maximum value had it maintained its initial rate of rise.
The expression for the time constant in an RL circuit is given by
T = L/R
(3.12)
37
Week
11
Time Domain Analysis
At the end of this week, the students are expected to:
♦
Explain the need for connecting a resistor in parallel with an inductor
♦
Solve problems involving transient in RL circiuits
3.9
ESSENCE OF CONNECTING A RESISTOR IN PARALLEL
WITH AN INDUCTOR
A
S
R
L
B
E
RD
Fig 3.9: Decay of current in an RL series circuit
Let us consider the circuit shown in figure 3.9. A discharge resistor RD is connected in
parallel with the inductive circuit. When the switch S is connected to A, the current
becomes E/R. When the switch is moved to position B, the circuit is disconnected
from the supply.
When the position of the switch is change from A to B, the circuit in the
inductive circuit is switched off and a voltage is induced in it. Arcing is produced at
the time of charging the position of the switch and there is a possibility of damaging
the insulation because the induced voltage may be many times greater than the
applied voltage. To avoid this arcing, a discharge resistor RD is connected as shown in
figure 3.9
3.10 SOLVED
PROBLEMS
INVOLVING
TRANSIENT
IN
RL
CIRCUIT
Example 3.4: A coil having a resistance of 4Ω and a constant inductance of 2H is switched
across a 20V d.c supply. Calculate:
(a)
the time constant
(b)
the final value of the current
Solution
(a)
T = L/R = 2/4 = 0.5s
38
(b)
I = V/R = 20/5 = 5A
Example 3.5 For the network shown in fig 3.10:
(a)
Determine the mathematical expression for the variation of the current in the inductor
following the closure of the switch at t =0 on to position 1
(b)
The switch is closed on to position 2 when t = 100ms, determine the new expression
for the inductor current
1
2
s
R1 = 10Ω
R = 15Ω
10V
L = 0.1H
Fig 3.10
Solution
(a)
For the switch in position 1, the time constant is
T1 = L/R1 = 0.1/10 = 10ms
∴i1 = I(1 – e-t/T) = 10/10 (1 – e-t/10x10
-t/10x10 -3
= (1 – e
(b)
-3
)
) amperes
For the switch in position 2, the time constant is
T2 =
L = 0.1 = 4ms
R1 + R 10 + 15
2
t/4x10-3
∴i2 = Ie-t/T = 10/10 (e=e
Example 3.6:
-t/4x10
)
-3
amperes
A coil having 10Ω resistance and 14H inductance is connected across a
d.c voltage of 140V. After the current has reached its final value, how much time it would
take for the current to reach a value of 8A after switching off the supply.
39
Solution
i = Ie
-t/T
where i = 8A, and T = L/R = 14/10 = 1.4s
and I = V/R = 140/10 = 14A
∴ 8 = 14 e
-t/1.4
⇒ e-t/1.4 = 8/14
∴ t = 0.783s
40
Week
12
Magnetic Coupling
At the end of this week, the students are expected to:
♦
Describe magnetic coupling
♦
Define mutual inductance
♦
Derive an expression for mutual inductance between two coils
4.1
MAGNETIC COUPLING
R1
i1
R2
I2
φ12
φ2
φ1
Coil 1
Fig 4.1:
Coil 2
Magnetic Coupling
Figure 4.1 shows magnetic coupling between two coils. In this case, a portion of the
magnetic flux established by one coil interlinks the other. In this case, energy is
transferred from one circuit to the other through the medium of magnetic flux that is
common to both.
From fig 4.1, φ1 is the total flux produced by i1, and φ2 is the total flux
produced by i2, and φ12 is the magnetic flux that linked both coils
4.2
MUTUAL INDUCTANCE
The mutual inductance between two coils is defined as the Weber-turns in one coil
due to current through the other coil
41
4.3
DERIVATION FOR MUTUAL INDUCTANCE BETWEEN TWO
COILS
Consider fig 4.1. The induced voltage in coil 2 due to a change in flux can be written
as (by faraday’s law)
e2 = N2dφ12
dt
(4.1)
where N2 is the number of turns in coil 2
Also, e2 is proportional to the rate of change of i1, that is
e2 α di1 or e2 = M di
dt
dt
(4.2)
In equation (4.2), M is the constant of proportionality known as mutual inductance
between two coils. It is measured in Henry (H)
Equating equation (4.1) to (4.2) we have
N2 dφ12 = M di1
dt
dt
⇒ M = N2 dφ12
di1
(4.3)
42
Week
13
Magnetic Coupling
At the end of this week, the students are expected to:
♦
Describe the polarity of coupled coils
♦
Define coefficient of coupling
♦
Derive an expression for the coupling coefficient
4.4
POLARITY OF COUPLED COILS
Polarity of coupled coils could be determined by dot convention. Dot convention is
very useful to determine the nature of mutually induced e.m.f instead of showing the
actual mode of the winding. If the dotted terminals of the winding correspond to each
other, the mutually induced e.m.f is positive otherwise it is negative. Figure 4.2
clearly explains the sign or polarity of mutually induced e.m.f.
M positive
M negative
M negative
M negative
Fig 4.2: Dot Convention
43
4.5
COEFFICIENT OF COUPLING
Coefficient of coupling is defined as the ratio of mutual inductance M to the square
root of the product of inductances of coil 1 and coil 2.
4.6
DERIVATION FOR COEFFICIENT OF COUPLING
In figure 4.1, let the two inductively coupled coils 1 and 2 have the number of turns
N1 and N2 respectively. Their individual coefficients of self induction are,
L2 =
N2
and L2 = N2 .
L/µ0µrA
L/µ0µrA
The flux φ1 produced in coil 1 due to a current I1 ampere is
φ1 =
N1 I1 .
L/µ0µrA
Suppose a fraction K1 of this flux i.e K1φ1 is linked with coil 2. Then
M = K1 φ1 x N2
I1
where K ≤ 1.
Substituting the value of φ1, we have,
M = K1 x
N1 N2
L/µ0µrA
(4.4)
Similarly, the flux φ2 produced in coil 2 due to I2 ampere is
φ2 =
N2 I2 .
L/µ0µrA
Suppose a fraction K2 of this flux i.e K2φ2 is linked with coil 2 then
M = K2 φ2 x N1
I1
= K2 x N1 N2
L/µ0µrA
(4.5)
Multiplying eqtn. (4.4) and (4.5), we get
M2 = K1K2
N12
x
L/µ0µrA
N22 = K1K2L1L2
L/µ0µrA
Putting √(K1K2) = K, we have M = K√(L1L2)
∴K=
M
√(L1L2)
where K is the coefficient of coupling.
(4.6)
44
Week
14
Magnetic Coupling
At the end of this week, the students are expected to:
♦
Define an ideal transformer
♦
Draw the equivalent circuit of an ideal transformer
♦
Explain with the aid of a diagram an equivalent circuit of a practical transformer
4.7
IDEAL TRANSFORMER
An ideal transformer is one with perfect coupling (K = 1).
4.8
EQUIVALENT CIRCUIT OF AN IDEAL TRANSFORMER
I1
P
I2
S
V1
V2
load
Fig 4.5 : Equivalent circuit of an ideal transformer
4.9
EQUIVALENT CIRCUIT OF A PRACTICAL TRANSFORMER
Z1
I1
Z2
X1
R1
I11
I0
I1 Z1
R
V1
P
E1
X2
S
E2
R2
I2
I2 Z2
X
To load
IC
V2
Img
Ideal transformer
Fig 4.6: Equivalent circuit of a practical transformer
Fig 4.6 is an equivalent circuit of a practical transformer. P and S represent the
primary and secondary windings of the ideal transformer, R1 and R2 are resistances
equal to the resistances of the primary and secondary windings of the practical
45
transformer. Similarly, inductive reactances X1 and X2 represent the reactance of the
windings due to leakage flux in the practical transformer.
The inductive reactor X is such that it takes a reactive current equal to the
magnetizing current Img of the practical transformer. The core losses due to hysteresis
and eddy currents are allowed for, by a resistor R of such value that it takes a current
Ic equal to the core loss of the practical transformer. The resultant of Img and Ic is Io,
which is the current the transformer take on no load.
46
Week
15
Magnetic Coupling
At the end of this week, the students are expected to:
♦
Solve problems involving magnetic coupling s
4.10 SOLVED PROBLEMS INVOLVING MAGNETIC COUPLING
Example 4.1: If a coil of 150turns is linked with a flux of 0.01wb when carrying current of
10A, calculate the inductance of the coil.
Solution
L = Nφ = 150 x 0.01 = 0.15H
i
10
Example 4.2: Two coils having 30 and 600turns respectively are wound side by side on a
closed iron circuit. If the mutual inductance between the coils is 0.226H, and the change of
current in the first coil is 20A in 0.02s, find the e.m.f induced in the second coil
Solution
e2 = M di/dt = 0.226 x 20/0.02 = 226V
Example 4.3: The coefficient of coupling between two coils is 0.75. There are 250turns in
coil 1. The total flux linking coil 1 is 0.4mWb, when the current in this coil is 3A. When i1 is
changed from 3A to zero linearly in milliseconds, the voltage induced in coil 2 is 70V.
Calculate: (a) the self inductance of coil 1 (b) the mutual inductance between the two coils.
Solution
(a)
L1 = N1φ1 = 250 x 0.4 x 10-3 = 3.4 x 10-3H
i1
(b)
3
e2 = M di, ∴ M = e2 dt = 70 x 3x 10-3 = 70 x 10-3H
dt
di
3
Example 4.4: Two coils X and Y have self inductance of 12H and 20H respectively, the
mutual inductance between the coils being 8.5H. Find the coupling coefficient between the
two coils
Solution
K=
M
=
8.5 = 0.55
√(LXLY ) √(2 x 20)
47
Example 4.5: The current in the primary winding of a pair of mutually coupled coil is 5A,
and the primary flux of 6mWb links with the secondary winding. If the secondary winding
has 950turns and the primary winding current is reduced to zero in 4ms, determine: (a) the
average value of the e.m.f induced in the secondary winding (b) the mutual inductance
between the two coils
Solution
(a)
e2 = N2 dφ2 = 950 x 6 x 10-3 = 1425V
dt
4 x 10-3
(b)
M = e2 dt, whee dI1 = 5 – 0 = 5A
dI1
∴ M = 1425 x 4 x 10-3 = 1.14H
5
48