Power Distribution Engineering
MSc/PG Diploma/PG Certificate
Pre Programme Preparation
Degree Programme Director: Dr B Zahawi
© Copyright Newcastle University
Pre-Programme Preparation
Section 1
Mathematical Tools for Power Calculations
Introduction
Virtually all of the electricity generated, transmitted, and utilised in homes and in industry
comes in the form of sinusoidal voltages and currents. In this section we will examine
some of the basic mathematical tools used in the analysis of electrical circuit. Our topics of
study will include vectors and vector algebra, phasor representation of sinusoidal
quantities, complex numbers (including the representation of circuit impedances) and
simple matrix algebra. Hopefully, a lot of the material will be simple revision for most of
you.
Learning Outcomes
On completion of this section you will be able to:
ƒ perform simple calculations involving vector addition, vector subtraction and
multiplication of a vector by a scalar
ƒ use phasors to represent sinusoidal, time-varying quantities, e.g. voltages and currents
ƒ use complex numbers and complex number algebra to perform phasor calculations
ƒ express circuit impedances and reactances as complex numbers
ƒ perform simple matrix calculations
Time
You will need about three to four hours to complete this section.
Resources
Calculator and scrap paper
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1.1 Vectors
Scalars and vectors
Physical quantities come in two kinds, scalars and vectors. A scalar is a quantity that is
entirely determined by its magnitude. For example, temperature is described by its number
of degrees, mass by the number of kilograms, length by the number of centimetres, voltage
by the number of volts, speed by its number of units (m/sec), density by its own number of
units (kg/m3) and so on.
A vector, on the other hand, is a quantity that has not only magnitude but also direction.
For example, force is clearly a vector. If you want to open a door, you have to apply the
appropriate amount (magnitude) of force in the correct direction. There is no point pulling
when the sign says Push. Another vector is velocity, which gives the speed (magnitude) as
well as the direction of travel. In our case the vectors we are most concerned with are those
that can be used to represent the sinusoidal signals of voltage and current at a fixed point in
their cycle.
A vector is usually represented by an arrow with a starting point and a terminal point, as
shown in Fig. 1.1 The magnitude of the vector is represented by the length of the arrow
and its direction by the direction of the arrow. A unit vector is a vector of magnitude 1.
A
Fig. 1.1 Vector A
A vector is usually denoted by boldface letters, such as A, B, C, etc. The magnitude of a
vector is usually denoted by ⏐A⏐ or simply A.
-A denotes a vector with magnitude⏐A⏐and a direction opposite to that of A. The zero
vector 0 is a vector with magnitude zero and no direction.
By definition, two vectors are equal if they have the same magnitude and direction,
regardless of the position of their starting or terminal points (Fig. 1.2). For example, a
force of 3 Newtons acting vertically on M1 is equal to another force of 3 Newtons acting
vertically on M2. This means that we can arbitrarily translate a vector (i.e. move it around)
when performing vector calculations. This is an extremely useful property because it
means we can construct phasor diagrams and add our voltage and current sinusoidal
signals vectorially.
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C
A
B
3N
3N
M1
M2
Fig. 1.2 Equal vectors
The Position Vector
Using a Cartesian co-ordinate system (i.e. an x-y-z co-ordinate system), the position vector
r of point A with co-ordinates (x1, y1, z1) is defined as the vector with a starting point at the
origin (0, 0, 0) and a terminal point at A, as shown in Fig. 1.3. The vector r may then be
represented by the three real numbers
r = [x1, y1, z1]
y
y1
A
r
z1
x1
x
z
Fig. 1.3 Position vector r
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Vector Addition
The sum of two vectors r1 = [x1, y1, z1] and r2 = [x2, y2, z2] is obtained simply by adding the
corresponding components
r1 + r2 = [(x1+ x2), (y1+ y2), (z1+ z2)]
Geometrically, this is equivalent to placing the starting point of r2 at the terminal point of
r1. The sum r1 + r2 is then the vector drawn from the starting point of r1 to the terminal
point of r2, as shown in Fig. 1.4.
r2
r1
r
Fig. 1.4 Vector addition
Because we are simply adding numbers when we perform a vector addition, a number of
familiar rules follow:
A+B=B+A
A – B = A + (-B)
(A + B) + C = A + (B + C) = A + B + C
A+0=A
A – A = A + (-A) = 0
Worked Example 1.1
A and B are two vectors given by A = [2, 0, 1], B = [-1, 2, 3], then
A + B = [1, 2, 4], B + A = [1, 2, 4],
A – B = [3, -2, -2], B – A = [-3, 2, 2].
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Multiplication of a vector by a scalar
The product of a vector A = [a1, a2, a3] and a scalar q (i.e. the number q) is obtained by
multiplying each component of A by q.
q A = [qa1, qa2, qa3]
q A is a vector with magnitude q⏐A⏐ and the same direction as A (Fig. 1.5).
Some very familiar properties of scalar multiplication follow from these basic definitions:
q(A + B) = qA + qB
1A=A
p(qA) = (pq)A = pqA
-1 A = -A
(p + q)A = pA + qA
0A = 0
qA
A
Fig. 1.5 Multiplication of a vector by a scalar
Worked Example 1.2
Let P and Q be two vectors given by P = [1, 1, 3], Q = [0, -2, -1]. Then
-P = [-1, -1, -3], -Q = [0, 2, 1], 2(P-Q) = 2[1, 3, 4] = [2, 6, 8] = 2P – 2Q.
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Cartesian unit vectors i, j and k.
Another popular method of representing the vector A is by using i, j, k, the three unit
vectors in the positive directions of the three axes in the Cartesian co-ordinate system (Fig.
1.6).
A = a1 i + a2 j + a3 k = [a1, a2, a3]
where a1, a2 and a3 are the components of A along the x, y and z axes, respectively.
z
z
A
a3 k
k
j
a1 i
i
y
a2 j
x
x
Fig. 1.6 The unit vectors i, j, k and vector representation
Worked Example 1.3
In Example 2 we have:
P = i + j + 3 k, Q = -2 j - k, 2(P + Q) = 2(i - j + 2 k) = 2 i - 2 j + 4 k.
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Magnitude of a vector in terms of its i, j, k components
By definition, the magnitude of a vector A is the distance between its starting point and its
terminal point. Thus:
⏐A⏐= √(a12 + a22 + a32)
Worked Example 1.4
The vector A with starting point at (2, 1, 1) and terminal point at (3, 5, 2). From Fig. 1.7
we can see that r1 + A = r2, or
A = r2 - r1 = [1, 4, 1] = i + 4 j + k
Hence ⏐A⏐= √(12 + 42 + 12) = √(1 + 16 +1) = √(18) = 3√2.
y
r2
A
r1
x
z
Fig. 1.7 Vector components and magnitude
Exercise 1.1
Let A = i + 7 j + 4 k, B = 2 i − 4 j + 3 k, C = 5 i − k. Find:
i) C + B, B + C
ii) A + (B + C), (A + B) + C
iii) 2 A + (B − C)
iv) 3 A – B + 2 C,
v) (A − C) /5
vi) ⏐A⏐, ⏐B⏐, ⏐C⏐
Turn to the end of the booklet for suggested answers.
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1.2 Phasor Representation of Sinusoidal Time Varying Quantities
The study of sinusoidally driven electrical circuits and networks can be greatly simplified
by using the concept of phasors to represent sinusoidal time functions.
Consider the two mathematical time functions
x(t) = X sin ωt, y(t) = Y sin (ωt + θ)
where ω is the angular frequency measured in radians per second, t is the time in seconds
and θ is the phase difference between x and y.
Both functions vary sinusoidally with time and may be used to represent steady-state
currents, voltages, temperatures, velocities or any other sinusoidal variable.
x(t), and y(t) may be graphically represented as the projections of two rotating vectors X
and Y on the vertical axis, as shown in Fig. 1.8.
ω rad/sec
t=0
y(t)
W
Y
x(t)
t
θ
X
θ
Fig. 1.8 Phasor representation of sinusoidal signals
These vectors have the magnitudes X and Y, respectively, and rotate with an angular speed
ω (rad/sec), corresponding to the angular frequency of the two sinusoidal functions.
The sum of the two sinusoidal functions w(t) = x(t) + y(t) is also a sinusoidal function (you
might need to revise your school trigonometry at this point) and may be represented in Fig.
1.8 by a third rotating vector W given by the vector sum of the vectors X and Y.
Sinusoidal time varying quantities can thus be analysed by a geometric vector
representation. If the rotating vectors are frozen at an instant in time, as shown in Fig. 1.8,
the corresponding vector diagram is referred to as a phasor diagram.
A big advantage of using phasors is concerned with the addition and subtraction of
sinusoidal quantities. Suppose that we have two voltages v1(t) = V1 sin(ωt + θ1), v2(t) = V2
sin(ωt + θ2) of the same frequency ω. To calculate the sum v1(t) + v2(t) without using
phasors we would have to make use of trigonometric identities and go through some fairly
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lengthy algebraic manipulations. If we use phasors, however, then the sum of the two
sinusoidal voltages is simply the vector sum of the two corresponding phasors.
Worked Example 1.5
A set of balanced three-phase sinusoidal voltages given by va = V sin(ωt),
vb = V sin(ωt-2π/3) and vc = V sin(ωt-4π/3) may be represented by the phasor diagram
shown in Fig. 1.9.
Vc
2π/3
2π/3
Va
Vb
Fig. 1.9 Phasor diagram representing three-phase sinusoidal voltages
The three vectors Va, Vb and Vc are shown in the x-y plane. For simplicity, Va was taken as
the reference phasor, i.e. the vectors were frozen at the instant when va is zero and thus
coincides with the x-axis.
Note: Phasor currents and voltages are denoted by bold upper case typeface, e.g. I, V, Ia,
Vab, etc.
Exercise 1.2
Sketch the following six sinusoidal signals as phasors.
i) A sin ωt,
iv) A cos (ωt − 18°)
ii) A sin (ωt + 94°)
v) −A cos (ωt + 18°)
iii) −A sin (ωt − 36°),
vi) −A cos (ωt − 54°)
Turn to the end of the booklet for suggested answers.
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Exercise 1.3
In Fig. 1.10 the three sinusoidal currents flowing into the node n are given by:
i1(t) = 4 sin ωt Amps
i2(t) = 3 cos (ωt + 30°) Amps
i3(t) = − 5 cos (ωt − 60°) Amps
Using phasors, calculate the current i(t) flowing out of the node.
i3(t)
i2(t)
i(t)
i1(t)
Fig. 1.10 Calculate i(t)
Turn to the end of the booklet for suggested answers.
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Phasor diagram representations of simple L, C and R circuits
For an inductor, the relationship between v(t) and i(t) is given by the equation
v(t) = L di(t) ⁄ dt.
Assuming sinusoidal current flow such that i(t) = Imax sin ωt, the voltage across the
inductor is given by,
⎛ di ⎞
v(t ) = ⎜ L ⎟ = (L I max ω cos ω t ) = ω L I max cos ω t
⎝ dt ⎠
i.e. the inductor current lags the voltage by 90°, as shown in Fig. 1.11.
v(t)
Vmax
Imax
i(t)
t
Fig. 1.11 Inductor voltage and current waveforms
Giving the following phasor diagram representation,
V
i
v
L
I
Fig. 1.12 Inductor voltage and current phasor diagram
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For a capacitor, the relationship between voltage and current is given by the equation
i(t) = C dv(t) ⁄ dt.
Assuming sinusoidal conditions such that v(t) = Vmax sin ωt, the instantaneous current
through the capacitor is given by,
i(t ) = C
dv
= (C Vmax ω cos ωt ) = ω C Vmax cos ωt
dt
i.e. the capacitor current leads the voltage by 90°, as shown in Fig. 1.13.
v
Vmax
Imax
i
t
Fig. 1.13 Capacitor voltage and current waveforms
Giving the following phasor diagram representation,
i
I
v
V
C
Fig. 1.14 Capacitor voltage and current phasor diagram
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For a resistor, the relationship between the voltage across the resistor R and the current
through R is given by Ohm’s law v(t) = R i(t).
If we assume a sinusoidal voltage supply v(t) = Vmax sin ωt, the current through the resistor
will also be sinusoidal, i(t) = Imax sin ωt, as shown.
Vmax
Imax
v(t)
i(t)
t
Fig. 1.15 Resistor voltage and current waveforms
Giving the following phasor diagram representation,
i
v
R
I
V
Fig. 1.16 Resistor voltage and current phasor diagram
The resistor current and voltage are in phase with each other.
The length of the voltage and current phasors in the above phasor diagrams should of
course correspond to the peak value of the voltage and current sinusoidal waveforms.
However, we as electrical engineers are more interested in the rms values of ac currents
and voltages because these give us a direct relationship between voltages, currents and
power flows in ac circuits. Therefore, we will always express all phasor diagrams in terms
of the rms values of voltage and current.
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1.3 Complex Numbers
In the previous section we saw how the concept of phasors is useful in dealing with
sinusoidal quantities. However, the application of phasors to even simple electrical circuits
tends to become awkward when expressed in terms of the algebra of real numbers. The use
of complex numbers allows us to describe the mathematical operations of multiplication,
division, addition and subtraction of phasors more easily.
Definitions
Consider the equation x2 + 4 = 0, or x2 = - 4
The solution to the above equation is not a real number. In other words, no real number,
positive or negative, will give you -4 when you multiply it by itself.
To deal with such equations, we introduce the imaginary number j defined by the equation
j=
−1 , hence
j2 = − 1
so that we can obtain a solution to the above equation given by x = ± j 2.
A complex number z is usually written in the form,
z = x + jy
x is called the real part of z (x = Re z) and
y the imaginary part of z (y = Im z).
If y = 0, then z = x and it’s a real number. Also, if x = 0, z = jy and is then said to be purely
imaginary.
Written in the above form, z is defined by its two co-ordinates x and y. Fig. 1.17 shows an
example. The point z = x + jy is represented by a point with co-ordinates x along the real xaxis, and y along the imaginary y-axis. The x-y plane in which the complex number is
represented is called the complex plane.
imaginary axis
z=x+jy
y
x
real axis
Fig. 1.17 Graphical representation of a complex number z
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Addition and multiplication of complex numbers
The addition of two complex numbers z1 = x1 + jy1 and z2 = x2 + jy2 is defined by:
z1 + z2 = (x1 + x2) + j (y1 + y2)
Multiplication is carried out simply by multiplying each term by each term in the normal
way:
z1 z2 = (x1 + jy1)( x2 + jy2) = x1 x2 + j x1 y2 + j y1 x2 + j2 y1 y2
= (x1 x2 - y1 y2) + j(x1 y2 + y1 x2)
Subtraction and division of complex numbers
The difference of two complex numbers z1 = x1 + jy1 and z2 = x2 + jy2 is defined by,
z1 - z2 = (x1 - x2) + j (y1 - y2)
z1/z2 is obtained by multiplying both numerator and denominator by x2 - jy2 (to obtain a
real number in the denominator) and simplifying:
z1 z 2 =
( x + jy1 )( x2 − jy 2 ) = x1 x2 + y1 y 2 + x2 y1 − x1 y 2
x1 + jy1
= 1
j
x 2 + jy 2 ( x 2 + jy 2 )( x2 − jy 2 )
x2 2 + y 2 2
x2 2 + y 2 2
Worked Example 1.6
Let z1 = 3 + 2j, z2 = 5 – j. Then, Re z1 = 3, Im z1 = 2, Re z2 = 5,
Im z2 = -1.
z1 + z2 = (3 + 5) + j(2 – 1) = 8 + j, z1 - z2 = (3 - 5) + j(2 + 1) = -2 + 3j.
z1 z2 = (3 + 2j)( 5 – j) = 15 - 3 j + 10 j – 2 j2 = 17 + 7 j.
z1
(
)
3 + 2 j (3 + 2 j )(5 + j ) 15 + 3 j +10 j + 2 j 2 (13+13 j ) 1 1
z2 =
=
=
=
= + j
(5 − j )(5 + j )
5− j
26
2 2
52 − j 2
Check the result by multiplying by z2 to obtain z1.
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Complex conjugates
The complex conjugate z of a complex number z = x + jy is defined as
z = x − jy
Geometrically, it is obtained on the complex plane by reflecting z in the real x-axis (Fig.
1.18).
imaginary axis
z=x+jy
=r∠θ
y
r
θ
−θ
x
real axis
r
−y
z* = x − j y
=r∠−θ
Fig. 1.18 Complex conjugates
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Complex numbers in polar form
Complex numbers are often used in their polar form, i.e. expressed in terms of the polar
co-ordinates (r, θ) instead of the Cartesian co-ordinates (x, y), as shown in Fig. 1.19.
imaginary axis
z=x+jy
=r∠θ
y
r
θ
x
real axis
Fig. 1.19 Complex numbers in polar form
The relationship between the polar co-ordinates (r, θ) and the more familiar (x, y) coordinates can easily be defined by referring to Fig. 1.19.
x = r cos θ
y = r sin θ
Hence, the complex number z = x + jy may be expressed in its polar form as
z = r (cos θ + j sin θ)
r is called the modulus of z and is usually denoted by ⏐z⏐or sometimes simply z.
z = r = x2 + y2
(1.1)
Geometrically, ⏐z⏐is the distance between z and the origin on the complex plane.
θ is the argument of z and is usually denoted by arg z.
θ = arg z = arc tan (y/x)
(1.2)
θ is measured in radians and is positive in the counter clockwise direction.
Note: Two complex numbers z1 and z2 are equal if and only if the real parts are equal and
the imaginary parts are equal. If expressed in polar forms, this is the same as saying that z1
and z2 are equal if and only if
⏐ z1 ⏐=⏐ z2⏐and arg z1 = arg z2 .
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Worked Example 1.7
The polar form of the complex number z = 3 + 2j is
⏐z⏐= √(32 + 22) = √13, arg z = arc tan (2/3) = 33.7°,
z = √13 (cos 33.7° + j sin 33.7°) Or simply z = √13 ∠ 33.7°
Multiplication and division in polar form
Multiplication and division of complex numbers is a lot easier and of more practical
significance when expressed in their polar form.
Let z1 and z1 be two complex numbers given by
z1 = r1 (cos θ1 + j sin θ1), z2 = r2 (cos θ2 + j sin θ2). Then,
z1 z2 = r1 r2 [cos(θ1 + θ2) + j sin(θ1 + θ2)]
In other words,
⏐ z1 z2⏐= ⏐ z1 ⏐x⏐ z2⏐
and
arg (z1 z2) = arg z1 + arg z2
Also,
z1
z1
=
z2
z2
and
arg (z1 ⁄ z2) = arg z1 − arg z2
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Worked Example 1.8
Let z1 = 1 + j and z2 = 2j. Then,
⏐z1⏐= √2, arg z1 = π/4, ⏐z2⏐= 2, arg z2 = π/2
Therefore,
⏐ z1 z2⏐= ⏐ z1 ⏐⏐ z2⏐= 2√2, arg (z1 z2) = arg z1 + arg z2 = π/4 + π/2 = 3π/4
⏐z1 ⁄ z2⏐= ⏐ z1 ⏐ ⁄ ⏐ z2⏐= √2 ⁄ 2, arg (z1 ⁄ z2) = arg z1 − arg z2 = π/4 − π/2 = −π/4
Exercise 1.4
Multiplication by j is geometrically equivalent to a counter clockwise rotation through 90°
in the complex plane. Verify this statement by plotting z and jz for z = 2 − j.
Turn to the end of the booklet for the answer to the exercise.
Exercise 1.5
z1 and z2 are two complex numbers given by z1 = 2 − j 5, z2 = −3 + j 4.
Express z1 and z2 in their polar form.
Calculate 1/z1, 1/z2, z1 + z2, z1 - z2, z1 x z2, z1/z2 and express your results in polar form.
Turn to the end of the booklet for the answer to the exercise.
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Use of complex numbers to represent phasor quantities
Previously, we saw how the concept of phasors is useful in dealing with sinusoidal
quantities. However, the application of phasors to even simple electrical circuits tends to
become awkward when expressed in terms of the algebra of real numbers. The use of
complex numbers allows us to describe the mathematical operations of multiplication,
division, addition and subtraction of phasors more easily.
Comparing a phasor diagram with the complex number representation on the complex
plain, it should be apparent that each phasor can be represented by a corresponding
complex number. For example, in Fig. 1.20, if we choose the phasor X as our reference
phasor, it may be written as X ∠ 0°. The phasors Y may then be written as Y∠θ°.
Im
Y = Y ∠ θ° = Ycosθ + j Ysinθ
θ X = X ∠ 0° = X cos0 + j Xsin0 = X
Re
Fig. 1.20 Complex number representation of phasor quantities
This is a very important step as it allows the use of complex number algebra in the study of
ac electrical circuits.
More specifically, multiplication by j results in a 90° counter clockwise rotation with no
change in magnitude. Similarly, multiplication by − j results in a 90° clockwise rotation
with no change in magnitude. This is an extremely useful property and means that we can
represent the effects of circuit inductance and capacitance and the associated 90° phase
shifts between circuit and current voltages using complex numbers. Using complex
numbers we can thus greatly simplify the steady-state solution of sinusoidally driven ac
electrical circuit replacing the time-domain circuit differential equations by simple
algebraic phasor equations.
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1.4 Reactance and Impedance
The reactance and impedance of an inductor
In an inductor, assuming sinusoidal current flow such that i(t) = Imax sin ωt, the voltage
across the inductor is given by;
⎛ di ⎞
v(t ) = ⎜ L ⎟ = (L I max ω cos ω t ) = ω L I max cos ω t
⎝ dt ⎠
We can now define the reactance of the inductor XL as:
XL = ωL (Ω)
And the relationship between the voltage across an inductor and the current through the
inductor may now be expresses in one simple phasor equation given by,
V = j XL I = j ωL I
(1.3)
where V and I are phasor quantities (i.e. complex numbers) and XL is the reactance of the
inductor given by ωL or 2πfL. Equation (1.3) expresses not only the relationship between
voltage and current magnitudes but also the fact that inductor current lags the voltage by
90°.
The complex impedance, or simply the impedance Z, of an inductor is defined as,
ZL = j XL (Ω)
Hence,
V = ZL I
The reactance and impedance of a capacitor
Similarly, for a capacitor, and assuming sinusoidal conditions such that v(t) = Vmax sin ωt,
the current through the capacitor is given by,
i(t ) = C
dv
= (C Vmax ω cos ωt ) = ω C Vmax cos ωt
dt
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We can now define the reactance of the capacitor XC such that,
XC = 1 ⁄ ωC (Ω)
And the relationship between the voltage across an inductor and the current through the
inductor may now be expresses in one simple phasor equation:
V=
1
jω C
I=−j
1
I = − j XC I
ωC
(1.4)
The negative sign in the above equation indicates that the capacitor current leads the
voltage by 90°. The impedance of the capacitor Z is given by,
ZC = − j XC (Ω)
And,
V = ZC I
The impedance of a resistor
For a resistor, the voltage and current are in phase with each other and we can write,
V=RI
And, even for a resistor, we can still define an impedance Z given by,
ZR = R (Ω)
And,
V = ZR I
In all cases we can now write a phasor version of Ohm’s law given by,
(1.5)
V=ZI
where V and I are phasor quantities and Z is the complex impedance of the circuit.
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The differential equations we started with have now been replaced by algebraic equations.
Note how the indicated differentiation is replaced by jω in equation (1.3) and by (1 ⁄ jω) in
equation (1.4).
Complex impedances are subject to the same rules for circuit simplification as resistances
in dc circuits: rules for combining in series or parallel are applicable to complex
impedances, as are star/delta and delta/star transformations.
For example, the impedance of a simple series RL circuit is given by (R + j XL), the
impedance of a simple series RC circuit is given by (R − j XC) and the impedance of the
RLC circuit shown below is simply (R + j XL − j XC).
Z = R + jω L +
1
1
(Ω)
= R + jω L − j
j ωC
ωC
where ω = 2πf and f is the supply frequency.
i
R
vR
Z = R + jω L +
L
C
vL
1
1
= R + jω L − j
j ωC
ωC
V=ZI
Fig. 1.21 Series connected R-L-C circuit
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Worked Example 1.9
The total impedance of a series connection (Fig. 1.21) of a 10 Ω resistor, an inductor of
1mH and a 1500 μF capacitor operating from a 50 Hz supply is given by
Z = 10 + j (2π×50) (1×10-3) – j [1 ⁄ (2π×50) (1500×10-6)]
= 10 + j (0.3) – j (2.1) = 10 – j 1.8
Worked Example 1.10
If a voltage of 100 Volts rms is now applied to the above circuit, the rms values of the
current in the circuit and the voltage drops across each of the three circuit elements may be
found as follows.
First, we choose the applied voltage V as our reference phasor. In other words we choose
to represent the voltage by a complex number of modulus 100 and zero argument (often
this is written as 100 ∠ 0°). V will then lie on the real axis on the complex plane and will
have no imaginary part.
The current that will flow in the circuit is then given by
I=
100 ∠ 0°
V 100 ∠ 0°
=
=
= 9.84∠ + 10.2° (Amps)
Z 10 − j 1.8 10.16 ∠ − 10.2°
i.e. the current will have a rms value of 9.84 Amps and will be leading the applied voltage
V by 10.2°.
For the voltage drop across the circuit elements we have
VR = R I = 10×(9.84 ∠ 10.2°) = 98.4 ∠ 10.2°,
i.e. a voltage drop of 98.4 volts in phase with the current.
VL = j (ω L) I = j (2π×50) (1×10-3) × (9.84 ∠ 10.2°)
= (0.3 ∠ 90°) × (9.84 ∠ 10.2°) = (3 ∠ 100.2°)
i.e. a voltage drop of 3 volts leading the current waveform by 90°.
VC = − j
1
I = – j [1 ⁄ (2π×50) (1500×10-6)] × (9.84 ∠ 10.2°)
ωC
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Pre-Programme Preparation
= (2.1 ∠ – 90°) × (9.84 ∠ 10.2°) = (20.66 ∠ – 79.8°)
i.e. a voltage drop of 20.66 volts lagging the current waveform by 79.8°.
If we now wish to express the above variables as functions of time, we can write:
v(t) = √2 (100) sin ωt, i(t) = √2 (9.84) sin (ωt + 10.2°),
vR(t) = √2 (98.4) sin (ωt + 10.2°)
vL(t) = √2 (3) sin (ωt + 100.2°) and vC(t) = √2 (20.66) sin (ωt – 79.8°).
Note: Strictly speaking we should convert all our angles to radians before we write an
expression like sin (ωt + 10.2°), since ωt is measured in radians, but sin (ωt + 0.057)
somehow doesn’t look right.
Exercise 1.6
Draw the phasor diagram representation of the current and voltages of the above circuit
(Fig. 1.21).
Turn to the end of the booklet for the answer to the question.
Exercise 1.7
Determine the steady-state solution of the circuit shown in Fig. 1.22 through the use of
complex numbers. The frequency of the supply is 50 Hz.
Turn to the end of the booklet for the answer to the question.
R1 = 10 Ω
C = 2500 μF
190 sin ωt
R2 = 100 Ω
Fig. 1.22 Solve for the steady state currents and the voltage drops across the circuit elements
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Pre-Programme Preparation
Exercise 1.8
An R-L-C series circuit is connected to a sinusoidal voltage supply (Fig. 1.21). At what
frequency will the circuit have minimum impedance?
Turn to the end of the booklet for the answer to the question.
Exercise 1.9
Fig. 1.23 shows a parallel LC arrangement connected to a sinusoidal voltage supply. At
what frequency will the circuit have maximum impedance?
Turn to the end of the booklet for the answer to the question.
XL
XC
Fig. 1.23 Parallel LC circuit.
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Pre-Programme Preparation
1.5 Matrices
Definitions
A matrix is a rectangular arrangement of numbers or mathematical functions enclosed in
brackets. Each number or function is called an element of the matrix. Their main area of
application, as far as we are concerned, is to systems of linear equations, which we often
have to deal with as electrical engineers, for example, when attempting to obtain a solution
for load flow problems or when performing fault current calculations.
The following are all matrices,
⎛1 3⎞
⎜⎜
⎟⎟ , (6 1 3 5) ,
⎝0 4⎠
⎛ a1
⎜
⎜ b1
⎜c
⎝ 1
a2
b2
c2
a3 ⎞
⎛ 2 ⎞
⎟ ⎛ 3x
8
sin x ⎞ ⎜ ⎟
⎟
b3 ⎟ , ⎜⎜
x ⎟ , ⎜ − 1⎟
0
cos
x
e
⎝
⎠ ⎜ ⎟
c3 ⎟⎠
⎝1 2 ⎠
The first is a 2×2 (read 2 by 2) square matrix. It has 2 horizontal lines or rows and two
vertical lines or columns. The second consists of one row only and is termed a row matrix
(or sometimes row vector). The third is a 3×3 square matrix with 3 rows and three
columns. The fourth consists of 2 rows and 3 columns and the last is a column matrix (or
column vector) consisting of one column only.
A matrix is usually referred to as an m×n matrix meaning a matrix with m rows and n
columns.
The elements of a matrix are denoted by the double subscript aij, where the first subscript
denotes row and the second the column in which the element aij stands. For example, a13 is
the element in the first row and third column.
A matrix is usually denoted by capital boldface letters such as A, B, C or by the general
double subscript element between brackets, such as [aij], [bij], [cij] and so on.
Matrix Transposition
The transposition AT of an m×n matrix A is defined as the n×m matrix that has the first
row of A as its first column, the second row of A as its second column, and so on.
⎛1 7⎞
⎟
⎜
Example: If A = ⎜ 0 2 ⎟ , then AT =
⎜5 3⎟
⎠
⎝
⎛ 1 0 5⎞
⎟⎟ .
⎜⎜
⎝ 7 2 3⎠
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Pre-Programme Preparation
Matrix Equality
Two matrices P and Q are said to be equal (P = Q) if and only if they have the same size
and all the corresponding elements are equal, i.e. p11 = q11, p12 = q12, p21 = q21, and so on.
⎛ a11
Example: The matrices A = ⎜⎜
⎝ a 21
a12 ⎞
⎟ and B =
a 22 ⎟⎠
⎛ 4 1⎞
⎟⎟ are equal only if a11 = 4, a12 = 1,
⎜⎜
⎝ 2 0⎠
a21 = 2 and a22 = 0.
⎛ − 1 3⎞
⎛ −1 0⎞
⎟⎟ and ⎜⎜
⎟⎟ are not equal because not all
Similarly, the two matrices ⎜⎜
⎝ 7 0⎠
⎝ 7 3⎠
corresponding elements are equal.
And, in the previous example A and AT are not equal because they are not even of the
same size (i.e. the same number of rows and the same number of columns).
Matrix Addition
You can add matrices only if they are of the same size. The sum is then obtained by adding
the corresponding elements. Matrices of different size cannot be added together.
⎛1 −1 0 ⎞
⎟
⎜
Example: If A = ⎜ 3 − 2 1 2 ⎟ and B =
⎜4 0
5 ⎟⎠
⎝
⎛− 2 0 1 ⎞
⎟
⎜
⎜ 1 4 − 1 ⎟ then A + B is given by the matrix
⎜ 1 0 − 3⎟
⎠
⎝
1 ⎞
⎛−1 −1
⎜
⎟
⎜ 4 2 −1 2⎟ .
⎜5 0
2 ⎟⎠
⎝
Multiplication by a Scalar
The product of any matrix A by any scalar k (i.e. number k) is obtained by multiplying
each element of the matrix by k.
The negative of A written as –A is obtained by multiplying each element of A by (-1).
⎛ 3 1 − 5⎞
⎟
⎜
Example: Let A = ⎜ 1 2 0 2 ⎟ , then
⎜ − 4 4 −1⎟
⎠
⎝
⎛ − 3 −1 5 ⎞
⎟
⎜
- A = ⎜ −1 2 0 − 2⎟ , 2 A =
⎜ 4
− 4 1 ⎟⎠
⎝
⎛ 6 2 − 10 ⎞
⎟
⎜
4 ⎟, 0 A =
⎜ 1 0
⎜− 8 8 − 2 ⎟
⎠
⎝
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⎛0 0 0⎞
⎟
⎜
⎜0 0 0⎟
⎜0 0 0⎟
⎠
⎝
Pre-Programme Preparation
The last matrix denotes a zero matrix (of size m×n) with all elements zero.
Matrix addition and multiplication by a scalar follow all the familiar rules for numbers.
For matrices of the same size m×n we can write:
A+B=B+A
A – B = A + (-B)
(A + B) + C = A + (B + C) = A + B + C
A+0=A
A – A = A + (-A) = 0
And for multiplication by a scalar,
k(A + B) = kA + kB
c(kA) = (ck)A = ckA
(c + k)A = cA + kA
1A=A
Matrix Multiplication
When multiplying a matrix by another matrix we do not multiply corresponding elements.
Instead, the definition of matrix multiplication is fairly complicated and looks very
artificial. The reason matrix multiplication is defined in the way it is has its roots in the
study of linear transformations and linear systems of equations, but this is beyond the
scope of our studies.
In any case, the product C = A B of an m×n matrix A and an r×p matrix B is defined if
and only if the number of columns of the first matrix A equals the number of rows of the
second matrix B, i.e. only if n = r.
If n = r, the product is then defined as the m×p matrix C with elements cij obtained by
multiplying each element in the ith row of A by the corresponding element in the jth column
of B and then adding the n products. The result will be a matrix with the same number of
rows as A and the same number of columns as B.
All this Gibberish will hopefully be clarified by the following examples:
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Pre-Programme Preparation
⎛ 1 4⎞
⎜
⎟
Example: Let A = ⎜ 2 7 ⎟ , B =
⎜ −1 0⎟
⎝
⎠
⎛5 0⎞
⎟⎟
⎜⎜
⎝1 2⎠
⎛ 1× 5 + 4 × 1 1× 0 + 4 × 2 ⎞
⎟
⎜
A B = ⎜ 2 × 5 + 7 ×1 2 × 0 + 7 × 2 ⎟ =
⎜ − 1× 5 + 0 × 1 − 1× 0 + 0 × 2 ⎟
⎠
⎝
⎛ 9 8⎞
⎟
⎜
⎜ 17 14 ⎟
⎜− 5 0 ⎟
⎠
⎝
In this example, A had 2 columns, the same number of rows of B, so that the product AB is
defined as a matrix with the same number of rows as A (3) and the same number of
columns as B (2). The product BA is undefined, i.e. it doesn’t exist.
⎛ − 1 2 ⎞ ⎛ 2 ⎞ ⎛ − 1 × 2 + 2 × 1⎞ ⎛ 0 ⎞
⎟⎟ = ⎜⎜ ⎟⎟ , whereas
⎟⎟ ⎜⎜ ⎟⎟ = ⎜⎜
Example: ⎜⎜
⎝ 4 7 ⎠ ⎝ 1 ⎠ ⎝ 4 × 2 + 7 × 1 ⎠ ⎝15 ⎠
⎛ 2⎞ ⎛ −1 2⎞
⎟⎟ is undefined.
⎜⎜ ⎟⎟ ⎜⎜
⎝1⎠ ⎝ 4 7⎠
⎛1⎞
⎜ ⎟
Example: (1 6 3) ⎜ 2 ⎟ = (1 × 1 + 6 × 2 + 3 × 0) = (13)
⎜0⎟
⎝ ⎠
⎛ 1×1 1× 6 1× 3 ⎞ ⎛ 1 6 3 ⎞
⎛1⎞
⎟
⎟ ⎜
⎜
⎜ ⎟
Example: ⎜ 2 ⎟ (1 6 3) = ⎜ 2 × 1 2 × 6 2 × 3 ⎟ = ⎜ 2 12 6 ⎟
⎜ 0 ×1 0 × 6 0 × 3⎟ ⎜ 0 0 0 ⎟
⎜0⎟
⎠
⎠ ⎝
⎝
⎝ ⎠
Matrix multiplication differs from the normal rules of number multiplication in three
important respects,
Firstly, matrix multiplication is not commutative, i.e. A B does not equal B A, in general.
This applies even for square matrices.
Secondly, A B = 0 does not necessarily imply that A = 0 or B = 0 or B A = 0.
Thirdly, A C = A D does not necessarily imply that C = D.
Other properties of matrix multiplication are similar to that of normal multiplication of
numbers. For example:
(kA)B = k(AB) = A(kB) = kAB, where k is a scalar.
A(BC) = (AB)C
(A + B)C = AC + BC
C(A + B) = CA + CB
Provided of course that the matrix products on the left-hand side are defined.
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Pre-Programme Preparation
Summary
In this section, we’ve considered the basic mathematical tools used in the analysis
of electrical circuits with sinusoidal quantities. In particular, we saw how the use of
phasors and complex algebra could simplify the analysis of electrical circuits operating
under sinusoidal conditions. You should familiarise yourself with the derivation of all the
formulae used in this section and be able to use them.
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Pre-Programme Preparation
Section 2
AC Power and Reactive Power
Introduction
In this section, some of the basic definitions, concepts and ideas of ac power systems will
be outlined. These include the definitions of power, reactive power and complex power as
well as simple ac power calculations as applied to single phase systems. In subsequent
sections we will see how these principles are applied to three-phase systems such as the
grid network.
Learning Outcomes
On completion of this section you will be able to:
ƒ Define the terms power, reactive power, complex power, voltamperes and power factor
ƒ Perform power and reactive power calculations in simple electric circuits.
Time
You will need between 3 and 4 hours to complete this section
Resources needed
Calculator, scrap paper
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Pre-Programme Preparation
2.1 Energy and power
The figure below shows a voltage source v(t) supplying a load with current i(t). The energy
W(t) absorbed by the load from time t0 to t seconds later is given by the equation,
t
W (t ) = ∫ v(t ) i (t ) dt (Joules)
t0
power
+
i
v
Load
Fig. 2.1 Power flowing into the load is positive (sink convention)
Power is the rate at which energy is absorbed and is given by,
P (t ) =
dW
= v(t ) i(t ) (Watts)
dt
The above two equations are general relationships that apply to all two-terminal electrical
networks regardless of the shape of the current and voltage waveforms i.e. not simply sine
waves and of the type of load. They also apply whether energy, and thus power, is being
produced, transmitted, utilised, or dissipated.
The convention for positive power flow is as shown in Figure 2.1. A numerically positive
product v(t)i(t) indicates positive instantaneous power flow from the supply to the load. A
numerically negative product v(t)i(t) indicates an instantaneous power flow into the supply
from the load.
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Pre-Programme Preparation
Exercise 2.1
Figure 2.2 shows a drawing of the voltage across a two-terminal electric circuit and the
current that flows in the circuit. Calculate the energy absorbed by the circuit per cycle and
the average power absorbed by the circuit.
v
100 V
0.1
0.2
0.3
0.4
t (sec)
- 100 V
i
12 A
0.1
0.2
0.3
0.4
t (sec)
Fig. 2.2 Calculate the energy absorbed per cycle
Turn to the end of the booklet for the answers to the exercises.
Next, we will derive expressions for the power absorbed by the three basic circuit
elements, the resistor, the inductor and the capacitor.
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Pre-Programme Preparation
2.1.1 Power absorbed by a resistor
For a purely resistive load, if we assume a sinusoidal voltage supply v(t) = Vmax sin ωt, the
current through the resistor will also be sinusoidal, i(t) = Imax sin ωt.
The instantaneous power P(t) absorbed by the resistor is then given by,
P(t) = v(t) i(t) = Vmax sin ωt × Imax sin ωt = Vmax × Imax sin2 ωt = Vmax Imax [1 - cos 2ωt] ⁄ 2,
or,
P(t) = Vrms Irms [1 - cos 2ωt]
The cos 2ωt term has an average value of zero, so the average power, P, delivered to the
load is given by:
P = Vrms Irms (Watts)
Or, combining this relationship with Ohm’s law,
P = (Irms)2 R = (Vrms)2/R (Watt)
The instantaneous power absorbed by the resistor pulsates at twice the supply frequency
around the average value P, but never goes negative, as shown in Fig. 2.3.
Vmax
Imax
v(t)
i(t)
t
P(t)
average
power
P
t
Fig. 2.3 Instantaneous power flow in a resistor
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Pre-Programme Preparation
2.1.2 Power absorbed by an inductor
Assuming sinusoidal current flow such that i(t) = Imax sin ωt, the voltage across the
⎛ di ⎞
inductor is given by v(t ) = ⎜ L ⎟ = (L I max ω cos ωt ) = ω L I max cos ωt , and the inductor
⎝ dt ⎠
current lags the voltage by 90°, as shown in Fig. 2.4.
Vmax
v(t)
i(t)
Imax
t
reactive
power
Q
P(t)
t
Fig. 2.4 Instantaneous power flow in an inductor
The instantaneous power P(t) absorbed by the inductor is given by,
P(t ) = v(t )i (t ) = (Vmax cos ω t ) (I max sin ω t ) = Vmax I max sin ω t cos ω t
However, sin ωt cos ωt = (sin 2ωt) ⁄ 2. So, the above relationship can be written
as P(t ) =
1
Vmax I max sin 2ωt , or,
2
P(t ) = Vrms I rms sin 2ωt
The above equations show that the average power absorbed by the inductive load is zero.
Power absorbed in one half cycle is returned to the supply during the next half cycle and
no net power is delivered to the load.
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Pre-Programme Preparation
Although the net power absorbed by the inductive load is zero, our source still has to
supply the peak of the pulsating power given by Vrms Irms. Because of the importance of this
quantity, we give it a special name, reactive power, the units of which are measured in volt
amperes reactive, or VAr for short. Reactive power is usually denoted by the letter Q.
Q = Vrms Irms (VAr)
Combining this with Ohm’s law, we obtain,
Q = XL (Irms)2 = (Vrms)2 ⁄ XL (VAr)
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Pre-Programme Preparation
2.1.3 Power absorbed by a capacitor
Assuming sinusoidal conditions such that v(t) = Vmax sin ωt, the instantaneous current
dv
= (C Vmax ω cos ωt ) = ω C (Vmax ) cos ωt , and the
through the capacitor is given by i(t ) = C
dt
capacitor current leads the voltage by 90°, as shown.
v
Vmax
Imax
i
t
reactive
power
Q
P(t)
t
Fig. 2.5 Instantaneous power flow in a capacitor
The power P(t) absorbed by the capacitor is given by,
P (t ) = v(t )i (t ) = (Vmax sin ωt ) (I max cos ωt ) = Vmax I max sin ωt cos ωt =
Vmax I max
sin 2ωt , or,
2
P(t ) = Vrms I rms sin 2ωt (Watt)
Again, the power absorbed by the capacitor is first positive then negative with a mean
value of zero, as shown. Power absorbed by the capacitive load in the first half cycle is
returned to the supply during the second half cycle and no net power is delivered to the
load.
The reactive power, Q, supplied by the source in this case is given by,
Q = Vrms Irms (VAr)
Combining this with Ohm’s law, we obtain,
Q = (Vrms)2 ⁄ XC = XC (Irms)2 (VAr)
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Pre-Programme Preparation
2.2 Real and Reactive Power in a General Two-Terminal Power Network
Let’s consider a simple ac circuit similar to that shown in Fig. 2.1. This time the load is not
purely resistive, inductive or capacitive, but a general load made up of a combination of all
three circuit elements. Load voltage and current waveforms will be out of phase by an
angle φ as shown in Fig. 2.6.
v(t) = Vmax sin ωt, and i(t) = Imax sin(ωt - φ)
v(t)
Vmax
i(t)
Imax
t
φ
Fig. 2.6 Voltage and current waveforms in a general ac circuit
Note that a positive value of the phase angle φ means that the current lags the voltage and
the load is inductive. A negative value of φ means that the current leads the voltage and the
load is capacitive.
The power P(t) supplied to the load is given by,
P(t) = v(t) i(t) = (Vmax sin ωt) (Imax sin(ωt - φ)) = Vmax Imax sin ωt sin(ωt - φ)
Using the trigonometric identity sin ωt sin(ωt - φ) = [cos φ - cos(2ωt- φ)] ⁄ 2, we can write
the above power formula as,
P(t ) =
Vmax I max
[cos φ − cos(2ωt − φ)]
2
Written in terms of the rms values of voltage and current, the above equation becomes,
P(t ) = Vrms I rms [cos φ − cos(2ωt − φ)]
Using trigonometric identities (see appendix), the above equation may be written as,
P(t) = Vrms Irms cos φ (1 – cos 2ωt) − Vrms Irms sin φ sin 2ωt
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Pre-Programme Preparation
v(t)
Vmax
i(t)
Imax
t
φ
P(t)
Q
P
t
Fig. 2.7 Instantaneous power flow in a general ac circuit
Written in this form, it may be seen that the power P(t) is made up of two components,
both pulsating at twice the supply frequency (Fig. 2.7).
The first component, given by Vrms Irms cos φ (1 – cos 2ωt) and shown in red in the above
diagram, pulsates around an average value of Vrms Irms cos φ but never actually goes
negative.
The second component, given by Vrms Irms sin φ sin 2ωt and shown in brown in the above
diagram, has a peak value of Vrms Irms sin φ and a mean of zero.
The real power P is defined as the average value of P(t) given by,
P = Vrms Irms cos φ (Watts)
This is the useful power delivered to the load.
The peak value of the second component that is forever travelling backward and forward
between the load and the supply is the reactive power Q given by,
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Pre-Programme Preparation
Q = Vrms Irms sin φ (VArs)
Note: Reactive power consumed in an inductor (i.e. when the current lags the voltage) is
regarded as positive reactive power. Reactive power consumed in a capacitor (i.e. when the
current leads the voltage) is therefore regarded as negative reactive power.
The reactive power input to the two terminals is the algebraic sum of the reactive power of
each branch or element. What we mean by algebraic sum is that we must attach the
appropriate sign to the reactive power of each element, positive for lagging reactive power
(inductive loads) and negative for leading reactive power (capacitive loads).
Principle of conservation of power and reactive power
Note that in any linear two-terminal network with sinusoidal voltages and currents, the
total power and reactive power inputs to the two terminals is equal to the sum of the
powers and reactive powers dissipated in each branch or element within the network.
Exercise 2.2
You arrive home after a long day. You switch on four 60 W light bulbs and two 100 W
light bulbs. You feel a bit cold so you also turn on your 1 kW electric heater. What are the
average and the maximum values of the power you are drawing from the supply?
Hint: remember that the above load is purely resistive.
Turn to the end of the booklet for the answers to the exercises.
Exercise 2.3
A farmer uses a 240 V single-phase diesel generator which delivers 12.5 kW from its
terminals at a current of 55 A. Calculate the reactive power.
(Note: there are two possible answers to this exercise)
Turn to the end of the booklet for the answers to the exercises.
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Pre-Programme Preparation
Now consider the simple RL circuit shown below. Here, the current lags the voltage by an
angle φ between 90° and 0°, as shown.
i
V
IP
R
v
φ
L
IQ
I
Fig. 2.8 In-phase and quadrature components of current
The phasor diagram representation above shows that the current I may be resolved into two
components. The first component is in phase with the voltage phasor and the second
component is 90° out of phase with the voltage. Power and reactive power may now be
expressed as,
P = V I cosφ = V IP
Q = V I sinφ = V IQ
where V and I are the rms values of voltage and current, respectively. IP is the rms value of
the in-phase current component, and IQ is the rms value of the quadrature current
component, i.e. the current component 90° out of phase with the current.
Power is given by the product of the voltage and the in phase current component, while the
reactive power is the product of voltage and the quadrature out of phase component of the
current waveform.
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Pre-Programme Preparation
2.3 Complex Power
The voltage and current phasors of a simple series RL circuit may be expressed in complex
form as,
V=V∠0
I=I∠-φ
The complex power S is defined by the equation
S = V I*
where I* is the complex conjugate of I given by,
I* = I ∠ + φ
Substituting into our equation for S, we obtain
S = V I* = V I ∠ φ = V I cosφ + j V I sinφ
Or,
S=P+jQ
The real part of S is thus the real power P and the imaginary part of S is the reactive power
Q.
The magnitude of S is usually referred to as the apparent power (or voltamperes) and is
given by,
S = V I = P 2 + Q 2 (VA)
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Pre-Programme Preparation
2.4 Power Factor
The power factor PF is an important parameter of ac electrical systems. It is defined as the
ratio of the real power (Watts) to the apparent power (VA) flowing through the system, i.e.
PF =
V I cos φ
Power
=
= cos φ
ApparentPo wer
V I
The terms lagging and leading are always included when specifying the power factor. A
lagging power factor means that the current lags the voltage and indicates the presence of
an inductive load. A leading power factor means that the current leads the voltage and
indicates the presence of a capacitive load.
Operation at a low power factor means that the required VA rating of the distribution plant
is much higher than the average power it delivers. Most electric circuits draw lagging VArs
from the supply, i.e. they will operate at a lagging power factor because of the presence of
transformers, etc.
A common method of power factor correction is to place a capacitive load (usually a bank
of capacitors) across the line. The capacitors will draw leading VArs thus reducing the
total lagging reactive power drawn from the supply and bringing its power factor nearer
unity.
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Pre-Programme Preparation
Worked Example 2.1
Calculate the power, reactive power and power factor of the simple electrical circuit shown
in Figure 2.9. The supply voltage V has an rms value of 150 V.
2Ω
V
1Ω
2Ω
2Ω
5Ω
Fig. 2.9 Calculate the power, reactive power and power factor
Solution
The impedance of the RL branch is given by (2 + j 5) Ω. The impedance of the RC branch
is given by (1 – j 2) Ω. Hence the impedance of the parallel combination Zpar is given by
1
Z par
=
1
1
+
(2 + j 5) (1 − j 2)
or
Z par =
(2 +
(12 + j ) = (12 + j ) = (12 + j ) (1 − j )
j 5) (1 − j 2)
=
(1 − j 2) + (2 + j 5) (3 + 3 j ) 3(1 + j ) 3(1 + j ) (1 − j )
Z par =
13 − j 11
= (2.17 − j 1.83) Ω
6
The total circuit impedance Ztot is thus given by
Ztot = 2 + (2.17 – j 1.83) = (4.17 – j 1.83) Ω
The current I is thus given by
I=
150 〈 0°
V
150
=
=
= 32.97 〈 23.7° (Amps)
Z tot 4.17 − j 1.83 4.55 〈 − 23.7°
Thus, the current has an rms value of 32.97 Amps and leads the voltage by 23.7°.
The power factor is given by
PF = cos (23.7°) = 0.9157 leading
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Pre-Programme Preparation
To calculate the power and reactive power, we need to calculate the complex power S.
S = V I = 150 (32.97 ∠ − 23.7°) = 4945.5 ∠ − 23.7° = 4528.4 − j 1987.8
Thus the power consumed by the circuit is approximately 4.5 kW and the reactive power
consumption is (−2 kVAr), or 2 kVAr leading.
Note that the convention is to regard the lagging reactive power consumed by a circuit as
positive reactive power. Consequently, a circuit consuming leading VArs is said to be
consuming negative reactive power.
Alternatively, we can calculate the power and reactive power consumption by calculating
IP and IQ, the in phase and quadrature components of I.
P = V IP = V I cos φ = (150) (32.97) cos -23.7° = 4528.4 Watts
Q = V IQ = V I sin φ = (150) (32.97) sin -23.7° = -1987.8 VArs or 1987.8 VArs leading
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Pre-Programme Preparation
Exercise 2.4
Figure 2.10 shows a general two-terminal network. If v = 120 sin ωt volts and i = 10 sin
(ωt − 60°), calculate the average power, the reactive power and the voltamperes.
i
v
linear
network
Fig. 2.10 Two-terminal network
Turn to the end of the booklet for the answers to the exercises.
Exercise 2.5
For a two-terminal network like that shown in Figure 2.10, calculate the power, reactive
power, voltamperes and power factor if the voltage and current are given by
(i)
V = 120 ∠ 0° volts, I = 25 ∠ 20° amps
(ii)
V = 240 ∠ 0° volts, I = 80 ∠ – 30° amps
(iii)
V = 280 ∠ 30° volts, I = 90 ∠ 10° amps
(iv)
V = 100 volts, I = 15 – j 2 amps
Turn to the end of the booklet for the answers to the exercises.
Exercise 2.6
A capacitor C is connected in parallel with an RL series combination whose resistance is 5
Ω and whose reactance is 8 Ω. What should the reactance of the capacitor be to produce a
lagging power factor of 0.98 for the combination? If you are operating from a 240V, 50 Hz
supply, determine the value of C.
Calculate the rms current of the supply before and after the connection of the capacitor
bank.
Turn to the end of the booklet for the answers to the exercises.
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Pre-Programme Preparation
Worked Example 2.2
A load draws 3000 VA from a 50 Hz, 240 volt single-phase line at a power factor of 0.8
lagging.
a) Calculate the number of voltamperes required of a capacitor bank to be connected in
parallel with the load for an overall power factor of
(i)
0.96 lagging
(ii)
0.96 leading
Solution
Without the capacitor bank, cos φ = 0.8 lagging, φ = 36.87°,
P = 2400 W, Q = 1800 VAr lagging.
For operation with a power factor of 0.96, the amount of VArs drawn from the supply is
given by the equation
PF = 0.96 =
P
P2 + Q2
=
2400
(2400)2 + Q 2
Thus Q = 700 VAr.
The load consumes 1800 VAr lagging. Thus:
a) (i) For operation at a lagging PF of 0.96, we need to draw 700 VAr from the supply. The
capacitor bank must then supply the remaining 1100 VArs.
a) (ii) For operation at a leading PF of 0.96, we need to inject 700 VAr into the supply).
The capacitor bank must then supply 2500 VArs.
Exercise 2.7
A single phase induction motor is connected to a 240 V, 50 Hz supply. At full load the
motor consumes 3.5 kW at a power factor of 0.84 lagging. Calculate the capacitance which
when connected across the motor terminals will bring the supply power factor to 0.98
lagging.
Turn to the end of the booklet for the answers to the exercises.
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Pre-Programme Preparation
Summary
It is perhaps appropriate at this stage to summarise some of the concepts we’ve come
across in Section 2 because of their fundamental importance.
ƒ The real power P is defined as the average value of the instantaneous power flow and is
therefore the useful power transmitted from the supply to the load.
ƒ The reactive power Q is defined as the peak value of the power component that travels
backwards and forwards between the supply and the load. It has an average value of
zero.
ƒ We can associate P and Q with two different components of supply current, the in-phase
component IP (i.e. the component in-phase with the supply sinusoidal voltage) and the
quadrature or out of phase component IQ (the component 90° out of phase with the
supply voltage).
ƒ The reactive power required by an electric load is positive for an inductive load (lagging
power factor) and negative for a capacitive load (leading power factor).
ƒ The complex power S is defined as the complex number
ƒ S = P + j Q. The magnitude of S has the units volt-amperes and is usually referred to as
the apparent power.
ƒ The power factor PF is defined as the ratio of the real power to the apparent power and
for sinusoidal conditions is given by cos φ, where φ is the phase shift between supply
voltage and current waveforms. It is lagging for inductive loads and leading for
capacitive loads.
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Pre-Programme Preparation
Section 3
Three-Phase Systems
Introduction
The material presented in the previous section dealt with single-phase ac power and
reactive power relationships, i.e. power and reactive power calculations in electrical
systems with a single sinusoidal voltage source. However, most ac power is generated,
transmitted and distributed as three-phase power using a set of three sinusoidal voltages
that are 120° out of phase with each other.
In this section we will look at the nature of three-phase systems and the relationship
between line and phase voltages and currents.
Learning Outcomes
On completion of this section you will be able to:
ƒ Recognise the characteristics of star and delta connected three-phase voltage supplies.
ƒ Recognise the characteristics of star and delta connected three-phase loads.
Time
You will need 2 to 3 hours to complete this section
Resources needed
Calculator and scrap paper
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Pre-Programme Preparation
3.1 Three-Phase Supplies
The main reason for using a three-phase power supply is that it makes more efficient use of
the copper and iron of the conductors, transformers and other power equipment. The
weight of the copper and iron used in a three-phase system is considerably less than in a
single-phase system delivering the same amount of power. Another reason for preferring a
three-phase supply is that it is capable of delivering a constant flow of power, unlike a
single-phase system in which the power pulsates at twice the supply frequency. Also, a
three-phase supply is capable of providing starting torque for industrial three-phase
induction motors, unlike a single-phase induction motor that needs a starting capacitor to
develop torque at standstill.
A balanced three-phase system is one that has three sinusoidal voltages of equal amplitude
and frequency displaced by 120° from each other, as shown in Fig. 3.1. The voltage in the
a phase reaches its peak 120° before the b phase which reaches its peak 120° before the c
phase, and so on. The phase sequence of a three-phase system is simply the order in which
the three phase voltages reach their maximum values. Thus, the phase sequence in the
above example is a-b-c.
vc
va
vb
vc
va
ωt
Fig. 3.1 Three-phase voltages
Each of the three voltages is 120° out of phase with the others, so we may write
va = √2 Vp sin ωt
vb = √2 Vp sin (ωt − 120°)
vc = √2 Vp sin (ωt − 240°)
where Vp is the rms value of each of the three phase voltages.
Observe that for a balanced set of three-phase voltages, the sum of the instantaneous
voltages is always equal to zero.
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Pre-Programme Preparation
Note: Although the total three-phase power in a balanced three-phase system is constant,
we should not be tempted to think that the corresponding system reactive power is zero.
The reactive power is very much a reality in each of the three phases. This means that the
rms currents in each line will be higher that what is strictly needed to deliver the load
power, and all cables, transformers, machines, etc. will have to be rated for the full
voltamperes in each phase. In electrical engineering lingo we would say that the total
reactive power drawn from the supply is three times the per phase value.
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Pre-Programme Preparation
3.2 Phase and line voltages
Now let’s consider a balanced three-phase circuit in which the three source voltages va, vb
and vc are connected as shown in Fig. 3.2. The three voltage sources are connected together
at the neutral or star point usually denoted by n.
In phasor notation, the three voltages may be written as,
Va = Vp ∠ 0°
Vb = Vp ∠ −120°
(3.2)
Vc = Vp ∠ −240° = Vp ∠ +120°
a
va
vc
vab
vca
n
b
c
vb
vbc
Fig. 3.2 Star-connection of three-phase voltages
This type of three-phase connection is called a star or wye connection. Two sets of
voltages exist in the system
The phase, or line-to-neutral voltages va, vb, vc described by the above equations.
The line-to-line voltages vab, vbc, vca, where vab is the voltage of line a with respect to line b
given by (va − vb), vbc is the voltage of line b with respect to line c given by (vb − vc) and vca
is the voltage of line c with respect to line a given by (vc − va).
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Pre-Programme Preparation
Fig. 3.3 shows the phasor diagram corresponding to the above voltages. The line-to-line
voltages may be calculated relative to the phase voltages as follows,
Vab = Va − Vb = Vp ∠ 0° − Vp ∠ −120° = √3 Vp ∠ 30°
Vbc = Vb − Vc = Vp ∠ −120° − Vp ∠ −240° = √3 Vp ∠ −90°
Vca = Vc − Va = Vp ∠ −240° − Vp ∠ 0° = √3 Vp ∠ 150°
-Va
Vab
Vc
Vca
-Vb
30°
30°
Va
30°
Vb
-Vc
Vbc
Fig. 3.3 Phasor diagram for star connection
It can thus be seen, from the above calculations or from the phasor diagram representation,
that the magnitude of the line-to-line voltages is √3 times the magnitude of the phase
voltages. The line voltages also lead the corresponding phase voltages by 30°, i.e. Vab leads
Va by 30°, Vbc leads Vb by 30° and Vca leads Vc by 30°.
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Pre-Programme Preparation
3.3 Three-Phase Loads
Three-phase loads may also be connected in star or in delta.
Balanced star connected three-phase loads
Fig. 3.4 shows a balanced three-phase circuit in which a three-phase load is connected
using the star configuration. In a star connection, the phase currents ia, ib, ic are equal to the
currents in the lines. If the three-phase load is also a balanced load, i.e. made up of
identical load elements Z = Z ∠ φ, then the currents will also form a balanced three-phase
set given by:
Ia = Va / Z =
Ib = Vb / Z =
Ic = Vc / Z =
V p ∠ 0°
Z ∠φ
= Ip ∠ − φ
V p ∠ − 120°
Z ∠φ
V p ∠ − 240°
Z ∠φ
= Ip ∠ − φ −120°
= Ip ∠ − φ −240°
where Ip is the rms value of each of the three phase currents. Fig. 3.5 shows a phasor
diagram representation of the above voltages and currents.
The sum of the three phase currents is zero at any instant in time. This is an important
feature of a balanced three-phase system and means that a fourth wire (the neutral
connection) is not required, since the neutral current is zero at all times.
If the three load impedances are not equal to each other, the neutral current will not be zero
and a neutral conductor, usually of a small cross sectional area, will have to be used. Such
a four-wire system is used in 240 V low voltage distribution systems in which users are
supplied with a single phase supply taken from one of the three lines and the neutral.
va
n
a
ia
Z
vb
b
ib
Z
vc
c
Z
ic
neutral connection
Fig. 3.4 Balanced three-phase star-connected load
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n
Pre-Programme Preparation
-Va
Vab
Vc
Vca
-Vb
Ic
30°
φ
30°
φ
Ib
φ
30°
Va
Ia
Vb
-Vc
Vbc
Fig. 3.5 Phasor diagram for balanced three-phase star-connected load
Exercise 3.1
A 50 Hz, 415 V (line voltage), three-phase supply is loaded by three equal impedances Z =
1.8 + j 0.5 Ω connected between each phase and ground. Calculate the rms line currents,
the power and reactive power delivered to the load.
Turn to the end of the book for suggested answers to the exercise
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Pre-Programme Preparation
Balanced delta-connected three-phase loads
It is also possible to connect a balanced load in the delta configuration as shown in Fig.
3.6. The voltage that appears across each phase or load element Z is now the line-to-line
voltage and not the phase voltage.
va
a
Z
vb
n
ia
b
vc
ib
c
Z
ic
Z
Fig. 3.6 Balanced three-phase delta-connected load
The three phase currents Iab, Ibc, Ica are given by
I ab =
Vab
V
V
, I bc = bc , I ca = ca
Z
Z
Z
The three line currents Ia, Ib, Ic are given by
Ia = Iab – Ica , Ib = Ibc – Iab , Ic = Ica – Ibc
For a balanced load, the currents in each phase will also be balanced. The magnitude of
line currents is √3 times the phase currents, as shown in the phasor diagram of Fig. 3.7.
Ic
-Ibc
Vc
Ica
30°
φ
Iab
Va
φ
φ
30°
-Ica
30°
Ibc
Ib
-Iab
Ia
Vb
Fig. 3.7 Phasor diagram for balanced three-phase delta connected load
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Pre-Programme Preparation
Exercise 3.2
Re-connect the three impedances in Exercise 3.1 into a delta load and calculate the rms line
and phase currents as well as the power and reactive power delivered to the load.
Turn to the end of the book for suggested answers to the exercise
Star-delta equivalence
For any balanced delta connected load with impedances Zdelta, there must be an equivalent
star connected load Zstar that would draw the same current from the supply. The two line
currents will be the same if:
Z star =
Z delta
3
(3.4)
Ia
Ia
Zstar
Zdelta
Zdelta
Zstar
Ic
Zdelta
Ib
Ic
Ib
Fig. 3.8 Star-delta transformation
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Zstar
Pre-Programme Preparation
Worked Example 3.1
Consider the system shown in Figure 3.9a, in which two loads, one star-connected and the
other delta-connected, are supplied from a 415 V delta connected source. Our task is to
calculate the supply and load power factors, the power and reactive power supplied by the
three-phase source, the power delivered to the load and the magnitude of all the currents in
the system.
Z1
415 V
Z2
Z1
Z2
415 V
415 V
Z1
Z2
balanced
three-phase
source
Z3 Z3 Z 3
Fig. 3.9a Three-phase system for example 3.1 where z1 =0.1 + j0.5, z2 = 3 + j6, z3 = 2 – j3
Under balanced conditions, the delta-connected supply may be replaced by an equivalent
star connection with a hypothetical neutral connection, thus allowing a simple per-phase
solution to be obtained. The phase voltage of the equivalent star connection is of course
given by Vline/√3 yielding the circuit shown in Figure 3.9b.
Z1
240 V
Z2
Z1
Z2
240 V
Z1
Z2
240 V
balanced
three-phase
source
Z3 Z3 Z3
Fig. 3.9b Balanced star-connected supply.
The next step is to simplify the circuit by converting the delta-connected load to an
equivalent star, giving the circuit of Figure 3.9c. The impedance of each phase of the
equivalent star connection is Z4 = Z2/3 = (3 + j 6)/3 = (1 + j2) Ω.
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Pre-Programme Preparation
Z1
Z4
Z1
Z4
Z1
Z4
Z3 Z3 Z3
Fig. 3.9c Star-equivalent load.
The two star loads (the actual star load and the equivalent star load) are now in parallel.
The equivalent combination of the two parallel loads is given by
Z5 = Z3 Z4 ⁄ (Z3 + Z4) = (2 – j3)(1 + j2) ⁄ (3 – j)
= (2.3 + j 1.1) Ω = 2.55 ∠ 25.56° Ω,
as shown in Figure 3.9d.
Z1
Z5
Z1
Z5
Z1
Z5
Fig. 3.9d Z3 and Z4 replaced by Z5.
Now we can draw the per-phase circuit representing the actual three-phase system as
shown in Figure 3.9e. The phase-to-neutral voltage is 415 ⁄ √3 = 240 V. As we said earlier,
the fact that the actual three-phase source has no neutral terminal does not matter, since the
system is balanced.
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Pre-Programme Preparation
Z1
240 V
Z5
Fig. 3.9e Per-phase circuit representation.
From the single-phase diagram shown in Figure 3.9e, we can see that the impedance
presented to the source is given by
Z1 + Z5 = (0.1 + j 0.5) + (2.3 + j 1.1) = (2.4 + j 1.6) Ω
= 2.9 ∠ 33.7° Ω.
The line current is thus given by
Ia = 240 ∠ 0° ⁄ 2.9 ∠ 33.7° = 82.8 ∠ −33.7°.
In other words, the line current has an rms value of 82.8 A and lags the voltage by 33.7°.
The supply power factor is given by PF = cos φ = cos (33.7°) = 0.832, the power supplied
by the three-phase source is
P = 3 (240) (82.8) cos (33.7°) = 49.6 kW and the reactive power supplied is given by
Q = 3 (240) (82.8) sin (33.7°) = 33.1 kVAr.
The rms value of the current flowing in each of the actual delta-connected supply phases is
82.8 ⁄ √3 = 47.8 A.
The magnitude and phase angle of the equivalent load impedance Z5 have already been
calculated as 2.55 Ω and 25.56°, respectively. Thus the rms phase voltage across the
combined load is given by
(82.8) (2.55) = 211.1 V.
This means that the rms value of the line-to-line voltage across the actual delta connected
load is √3 (211.1) = 365.7 V. The power factor of the load is given by cos (25.56°) =
0.902. The power taken by the load is thus given by 3 (211.1) (82.8) cos(25.56°) = 47.3
kW.
The power delivered to the load may also be calculated by computing the I2R product per
phase and multiplying by three, where
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Pre-Programme Preparation
I is the rms line current and R is the real part of Z5 the equivalent load impedance. Thus
load power = 3 (82.8)2 (2.3) = 47.3 kW.
We now need to calculate the rms values of the currents flowing in the star and delta loads
to complete our circuit analysis. To do so, we need to consider the impedances of the star
and delta load elements and the rms value of the voltages imposed across them.
The magnitude of the star-connected load elements is √(22+32) = 3.6 Ω and across this we
have the phase-to-neutral voltage of 211.1 V, giving an rms current of 58.5 A. The
magnitude of the impedance of each of the delta elements is √(32+62) = 6.7 Ω. Thus, the
current drawn by each phase of the delta load is 365.7 ⁄ 6.7 = 54.6 A. The line current
drawn by the delta load is √3 (54.6) = 94.5 A.
Figure 3.9f gives a summary of all the values of voltages and currents calculated in this
example.
82.8 A
415 V
94.5 A
365.7 V
47.8 A
54.6 A
balanced
three-phase
source
58.5 A
Fig. 3.9f Circuit voltages and currents
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Pre-Programme Preparation
3.4 Single-Line Representation of a Balanced System
In a balanced 3-phase system the phase and line voltages and currents have the same
magnitude, but are displaced by 120o. Therefore only one phase needs to be considered and
the system may be treated as a single-phase system using phase voltages.
The total power and reactive power consumption of the three-phase system are then 3
times the single-phase values.
Fig. 3.10 shows an example of a single-line diagram representation of a three phase loop
distribution network.
33kV/11 kV
transformer
33kV/11 kV
transformer
n/o point
Fig. 3.10 Single-line diagram of three-phase loop network
Summary
Three-phase power systems offer significant advantages over single-phase systems and are
used to generate, transmit, and distribute almost all of the electric power consumed all over
the world. In this section we examined the characteristics of three-phase voltage supplies
and three-phase loads.
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Pre-Programme Preparation
Solutions
Section 1
1.1
(i)
7i–4j+2k
(iii) - i + 10 j + 12 k
(ii) 8 i + 3 j + 6 k
(iv) 11 i + 25 j + 7 k
(v) – 0.8 i + 1.4 j + k (vi) √66, √29, √26
1.2
Using A sin ωt as our reference phasor,
A sin (ωt + 94°)
A cos (ωt − 18°)
− A sin (ωt − 36°)
A sin ωt
− A cos (ωt − 54°)
− A cos (ωt + 18°)
Note that cos ωt leads sin ωt by 90°.
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Pre-Programme Preparation
1.3
Using i1(t) as our reference phasor,
i2
30°
30°
i1
i3
We can now add the three phasors geometrically to obtain i(t),
i2
i3
i
i1
i(t) = 1.87 sin(ωt + 176.5°)
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Pre-Programme Preparation
1.4
imaginary axis
2
jz
real axis
1
2
-1
z
z = 2 – j, Re(z) = 2, Im(z) = -1
j z = j (2 – j) = 1 + j 2, Re(jz) = 1, Im(jz) = 2
1.5
In polar form
z1 = 5.385 ∠ − 68.2°
z2 = 5 ∠ 126.9°
1 / z1 = 0.186 ∠ 68.2°
1 / z2 = 0.2 ∠ − 126.9°
z1 + z2 = √2 ∠ − 135°
z1 − z2 = 10.3 ∠ − 60.9°
z1 z2 = 26.925 ∠ 58.7°
z1 / z2 = 1.077 ∠ 164.9°
1.6
VL
VR
VC
I
10.2°
V
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Pre-Programme Preparation
1.7
The impedance of the parallel RC branch ZRC is given by the equation
1
1
1
= +
Z RC R − j X C
where R = 100 Ω,
XC = 1 / ωC = 1 / 2 fC = 1 / 2  (2500×10-6) = 1.273 Ω.
ZRC = (0.0162 – j 1.273) Ω or (1.2731 ∠ − 89.27°)
The total circuit impedance Z is then given by
Z = (10.0162 – j 1.273) Ω or (10.097 ∠ − 7.24°)
Using rms values, the supply voltage may be represented by the phasor V = (190/√2)
∠ 0°.
The supply current I is then given by
I=
134.35 ∠ 0°
V
=
= 13.31 ∠ 7.24°
Z 10.097 ∠ − 7.24°
i.e. I has an rms value of 13.31 Amps and leads the supply voltage by 7.24°.
In order to calculate the curents flowing in the 100 resistor and in the 2500 μF
capacitor, we first need to calculate VR, the voltage drop across the 10 resistor.
VR = (133.1 ∠ 7.24°) Volts
VRC, the voltage across the parallel RC branch is then given by
VRC = V – VR = (16.93 ∠ - 82.16°) Volts
Hence,
IR = (0.1693 ∠ - 82.16°) Amps
IC = (13.3 ∠ 7.84°) Amps
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Pre-Programme Preparation
1.8 Total circuit impedance Z is given by
1
ωC
Z = R + jωL − j
The circuit will have minimum impedance when j ω L − j
i.e. when f =
1
2π LC
1
= 0 , or ω =
ωC
.
1.9 Total circuit impedance Z is given by
1
1
1
1
1
=
+
=
+
Z j XL − j XC
j ω L 1 j ωC
1
Thus, Z =
j ωC − j
1
ωL
Z will be maximum when j ω C − j
i.e. when f =
1
= 0 , or ω =
ωL
1
LC
1
2π LC
Section 2
2.1
Current only flows in the half cycle from t = 0 to t = 0.1 sec.
i = 120 t,
0 < t ≤ 0.1
0.1 < t ≤ 0.2
0,
0.1
0.1
0
0
Energy absorbed per cycle W = ∫ v i dt = ∫ (100 )(120 t ) dt = 60 (J)
Average power = 60 ⁄ 0.2 = 300 (W)
2.2
Light bulbs and heating elements are resistive circuit elements.
Average power = 4 (60) + 2 (100) + 1000 = 1440 W
Peak power = 2 (1440) = 2880 W
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1
LC
Pre-Programme Preparation
2.3
P = V I cos φ
cos φ = 12500 ⁄ (240)(55) = 0.9470
φ = 18.74° or − 18.74°
Q = V I sin φ = 4.24 kVAr lagging, or 4.24 kVAr leading
2.4
Using the supply voltage as our reference phasor
V=
I=
120
2
10
2
∠ 0° = 84.85 ∠ 0° Volts
∠ − 60° = 7.07 ∠ − 60° Amps
S = V I = 600 ∠ 60° = 300 + j 519.6 (VA)
P = 300 W, Q = 519.6 VAr, voltamperes = ⏐S⏐= 600 VA.
2.5
S = V I , PF = cos φ
i.
S = 3000 ∠ − 20° = 2819 – j 1026, P = 2819 W, Q = 1026 Var leading, voltamperes
= 3 kVA, PF = 0.9397 leading
ii.
S = 19200 ∠ 30° = 16628 + j 9600, P = 16628 W, Q = 9600 Var lagging,
voltampers = 19.2 kVA, PF = 0.866 lagging
iii. S = 25200 ∠ 20° = 23680 + j 8619, P = 23.68 kW, Q = 8.62 kVar lagging,
voltamperes = 25.2 kVA, PF = 0.9397 lagging
iv. S = 1513 ∠ 7.6° = 1500 + j 200, P = 1.5 kW, Q = 0.2 kVar lagging, voltamperes =
1.5 kVA, PF = 0.991 lagging
2.6
For a power factor of 0.98 lagging, the circuit current I must lag the supply voltage
V by an angle φ such that cosφ = 0.98, or φ = 11.48°.
The total circuit impedance Z is given by
1
1
1
=
+
Z (5 + j 8) (− j X C )
or
Z=
8 XC − j 5 XC
5 + j (8 − X C )
arg (Z) = arctan (−5 ⁄ 8) − arctan [(8 − XC) ⁄ 5]
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Pre-Programme Preparation
For a power factor of 0.98 lagging, arg (Z) = 11.48°.
Thus: 11.48° = − 32° − arctan [(8 − XC) ⁄ 5]
XC = 12.74 Ω, C = 250 μF.
Without the capacitor I = 25.45 ∠ − 58° (Amps)
With the capacitor
I = 13.75 ∠ − 11.48° (Amps)
2.7
PF = 0.84 =
P
P2 + Q2
Thus
3500
(3500)2 + Q 2
= 0.84
Q = 2260 VAr.
Using the above equations, we can also calculate the amount of VArs drawn from the
supply for operation at a PF of 0.98 lagging.
Qsupply = 710 VAr
Thus, the amount of VArs required from the capacitor bank is (2260 – 710) = 1550
VAr.
XC = 37.16 Ω, C = 85.6 μF.
Section 3
3.1 Taking the supply voltage as our reference phasor
I=
240 ∠ 0°
V
240
=
=
= 128.13 ∠ − 15.5° (A )
Z 1.8 + j 0.5 1.87 ∠15.5°
S = V I = (240 ∠ 0°)(128.13 ∠ + 15.5°) = 30.7 ∠15.5° = (29.58 + j 8.2) kVA
I = 128.13 A, P = 29.58 kW, Q = 8.2 kVAr lagging
Total power delivered to the load = 3 × P = 88.74 kW
Total reactive power delivered to the load = 3 × Q = 24.61 kVAr lagging
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Pre-Programme Preparation
3.2 Taking the supply voltage as our reference phasor
I=
415 ∠ 0°
V
415
=
=
= 221.9 ∠ − 15.5° ( A )
Z 1.8 + j 0.5 1.87 ∠15.5°
S = V I = (415 ∠ 0°)(221.9 ∠ + 15.5°) = 92.09 ∠15.5° = (88.74 + j 24.61) kVA
I = 221.9 A, P = 88.74 kW, Q = 24.61 kVAr lagging
Total power delivered to the load = 3 × P = 266.22 kW
Total reactive power delivered to the load = 3 × Q = 73.83 kVAr lagging
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Pre-Programme Preparation
Appendix A – Calculus
Standard Derivatives
y
dy/dx
xn
nx(n−1)
sin x
cos x
cos x
− sin x
tan x
sec2 x
ln x
1/x
eax
a eax
ax
ax ln a
Standard Integrals (constant of integration omitted)
function
integral
xn
x(n+1)/(n+1), n ≠ − 1
sin x
− cos x
cos x
sin x
1/x
ln x
eax
eax/a
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Pre-Programme Preparation
Appendix B – Trigonometry
sin (− θ) ≡ − sin θ
cos (− θ) ≡ cos θ
tan (− θ) ≡ − tan θ
sin2 θ + cos2 θ ≡ 1
tan2 θ + 1 ≡ sec2 θ
Addition Formulas
sin(A+B) ≡ sin A cos B + cos A sin B
sin(A−B) ≡ sin A cos B − cos A sin B
cos(A+B) ≡ cos A cos B − sin A sin B
cos(A−B) ≡ cos A cos B + sin A sin B
Double Angle Formulas
sin (2θ) ≡ 2 sin θ cos θ
cos (2θ) ≡ cos2 θ − sin2 θ ≡ 2 cos2 θ − 1 ≡ 1 − 2 sin2 θ
tan (2θ) ≡ 2 tan θ / (1 − tan2 θ)
Also,
⎛ A+ B ⎞
⎛ A− B ⎞
sin A + sin B ≡ 2 sin ⎜
⎟ cos ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
⎛ A+ B ⎞
⎛ A− B ⎞
sin A − sin B ≡ 2 cos ⎜
⎟ sin ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
⎛ A+ B ⎞
⎛ A− B ⎞
cos A + cos B ≡ 2 cos ⎜
⎟ cos ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
⎛ A+ B ⎞
⎛ A− B ⎞
cos A − cos B ≡ − 2 sin ⎜
⎟ sin ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
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