6.1 GENERAL
It is necessary here, while dealing
with fundamentals, to look again at two important mathematical concepts which
many will remember from their schooldays.
Both are widely used in calculations associated with electrical
engineering, especially when dealing with alternating currents. The two concepts are Pythagoras’ Theorem and,
associated with it, the basic trigonometrical functions. At the end of this chapter are some examples
showing how these functions are used.
6.2 PYTHAGORAS’ THEOREM
If ABC is a right-angled triangle
with the right angle at C, the side AB opposite the right angle is called the
‘hypotenuse’ (Figure 6.1).
FIGURE 6.1
PYTHAGORAS’ THEOREM
Pythagoras’ Theorem states that ‘the
square on the hypotenuse is equal to the sum of the squares on the other two
sides’. Put into mathematical form:
AB2 = AC2
+ BC2
(i)
For example, if AC = 4 inches and BC
= 3 inches, then:
AB2
= 42
+ 32
= 16 + 9
= 25
\ AB = √ 25
=
5 inches
Thus, if the ‘other two’ sides are
both known, the length of the hypotenuse can be calculated.
Expression (i) can also be written:
AC2
= AB2
- BC2 ….(ii)
or
BC2 = AB2
- AC2 ….(iii)
For example, if AB = 13 inches and BC = 5
inches:
AC2
= 132
- 52 from equation (ii)
=
169 - 25
=
144
\ AC = √ 144
= 12 inches
Similarly, if AB and AC are known,
BC can be calculated from equation (iii).
To sum up: in a right-angled
triangle, if the lengths of any two of the sides are known, that of the third
side can be calculated by using one of the three forms (equations (i), (ii) or
(iii)) of Pythagoras’ Theorem:
AB2 = AC2 + BC2
6.3 TRIGONOMETRICAL FUNCTIONS
The word ‘trigonometry’ is derived from the Greek, meaning ‘three-cornered’, and, as may be expected, it deals with triangles and the relationships between their sides and angles. For the sake of this discussion it will be limited to right-angled triangles and so stays within the realm of Pythagoras.
FIGURE 6.2
TRIGONOMETRICAL FUNCTIONS
Figure 6.2 shows a right-angled
triangle with the right angle at C and with AB as its hypotenuse. The angle ABC will be called ‘j’; the side BC adjoining j is called the
‘adjacent side’ (symbol ‘A’), and the side AC opposite j is called the ‘opposite side’ (symbol ‘O’). Keeping to the symbol notation, the
hypotenuse AB will be called ‘H’. Then, by Pythagoras:
AB2 = AC2
+ BC2
or, using the
symbol notation,
H2 = A2
+ O2
This is the Pythagoras relationship
between the hypotenuse and the other two sides.
So far only the relationship between
the triangle’s three sides have been discussed.
Look now at the angle j and consider three more definitions:
Each of these three is the ratio of
a pair of sides, the ratio of different pairs having different names, but all
being functions of the angle j.
Tables are available of sines, cosines and tangents from which, if the
ratio is known, the angle j can be read off directly. So, if any two sides of a right-angled
triangle are known, not only can the third side be calculated by Pythagoras,
but either of the two acute angles can be determined by use of the appropriate
table. (Nowadays ‘scientific’ pocket
calculators are used instead of tables.)
Relations (a), (b) and (c) above can
also be written:
It is in these forms that the
relationships are most useful.
Example
In case of the triangle of Figure
6.2, suppose the opposite side ‘O’ had length 10cm and the adjacent side ‘A’
had length 15cm. Then by Pythagoras:
H2
= 102
+ 152
=
100 - 225
=
325
\ H = √ 325
= 18.03 cm.
Also:
From the tangent tables or pocket
calculator:
(Note that this can be calculated
without working out H first.)
But if H is known:
From the sine
tables From the cosine
tables
or pocket
calculator: or
pocket calculator:
It will be seen that, whichever of
the three functions is used, the calculation leads to the same answer for 
When it is stated that 
, it could equally well be said that j is the angle whose sine is 
. This may be written
mathematically as:
, it could equally well be said that j is the angle whose sine is . This may be written
mathematically as:
The expression ‘sin-1’ or
‘arcsin’ means simply ‘the angle whose sine is ...’, and one or other of these
expressions will be found on most scientific pocket calculators. It is important to distinguish between ‘sin’
(the sine of the keyed-in angle) and ‘sin-1 (the angle whose sine is
the keyed-in figure).
Thus: sin 30° = 0.5, but sin-1 0.5 = 30°
6.3.1 Use of Trigonometric Functions
In many a.c. electrical power
problems much use is made of Pythagoras and of the trigonometrical functions
sine, cosine and tangent. Chapters 8 to
10 show that currents in purely resistive circuits are in phase with the
applied voltage, but currents in purely reactive circuits lag or lead 90° on the applied voltage. We
therefore have all the elements of a right-angled triangle to which Pythagoras
and the simple trigonometrical functions can be applied.
The principal uses of these methods
in electrical circuits are for COMBINING and for RESOLVING. These two terms are explained below.
6.3.2 Combining
Anticipating for a moment the use of
vector notation described more fully in Chapter 7, Figure 6.3 shows an applied
voltage V supplying a purely
resistive circuit R in parallel with
a purely reactive (inductive) circuit X. The resistive circuit draws a current IR which is in phase with the
voltage, whereas the reactive circuit draws a current Ix which lags 90° on the voltage. These two
currents are so shown in the figure on the right.
The diagram shows that two separate
currents are flowing in the two limbs, although of course only one current can
come out of the voltage source, and that must clearly be the combination of the
two separate currents. It is therefore
required to combine the two currents
IR and Ix to give a single
equivalent or ‘resultant’ current I.
FIGURE 6.3
COMBINED CURRENTS
In the diagram on the right of
Figure 6.3, OA represents the current IR
in phase with the voltage V, and OB
the current Ix lagging 90° on V. Complete the rectangle OACB and draw the
diagonal OC. Then OC represents the
resultant or equivalent current I to
the same scale as IR and Ix
By Pythagoras:
That is to say, the resultant
current is the square root of the sum of the squares of the individual
resistive and reactive currents. Putting
numbers to these, if IR =
40A and Ix = 30A, then:
I2 = 402 + 302
= 1 600 + 900
= 2 500
\ I =
50A
Thus the resultant current equivalent
to the 40A and 30A with this arrangement is not 70A but 50A. This is an example of combining. (Compare Chapter 7, Figure 7.5.)
This combining of currents can also
be used to determine the phase angle (j) of the
resultant current with respect to the applied voltage.
For, as already defined,
Expressed mathematically:
That is to say, the phase angle j is the angle whose tangent is the ratio of the reactive current (lx) to the resistive current
(lR). It will be shown later that this is the same
as the ratio of the actual reactance (X)
to the resistance (R); these two
terms are explained in Chapters 8 to 10.
6.3.3 Resolving
When the actual current is known and
also its angle of lag (or lead) j, as shown in
Figure 6.4, it is often required to break down, or ‘resolve’, that current into
components which are in phase with the applied voltage and at right angles to
that voltage - that is the reverse process to ‘combining’ which has just been
described.
FIGURE 6.4
RESOLVED CURRENTS
All that it is necessary to do is to
apply the trigonometrical functions cos j and sin j to the current. In the figure, V is the applied voltage and I
the total current which flows. The
angle between I and V is j. If perpendiculars are
dropped from C to the horizontal (CB) and to the vertical (CA), then OA is the
in-phase component IP, of
the current I (that is, in the same
phase as the voltage), and OB is the right-angled (or ‘quadrature’) component lQ of the current I (that is, at right angles to the
voltage).
Remembering equations (v) and (iv),
But OC is the total current I, and OA and OB are the in-phase and
quadrature components lP
and IQ respectively.
Thus, if the total current and phase
angle are known, the in-phase and quadrature components can be immediately
calculated. (Usually the actual angle j is not given, but the power factor, cos j, is. IP then makes direct use of it, but to find sin j it is necessary to use the cosine tables to find j, then the sine tables to find sin j to evaluate IQ. A good scientific calculator can do this
transition in one operation.) In finding
IP, and IQ from I in this manner, the current I is said to be resolved into in-phase and quadrature
components.
Example
A circuit has a current of 100A at a
lagging power factor of 0.8 on the applied voltage. Find the in-phase and quadrature components
of this current.
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Power factor is 0.8
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=
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cos j
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From the cosine tables (or
calculator) j
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=
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36°52’
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From the sine tables (or
calculator) sin j
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=
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0.60
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\
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In-phase current
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=
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100 cos j = 80A
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Quadrature current
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=
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100 sin j = 60A
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(Note that, if these were recombined
by Pythagoras,
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(Total current)2
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=
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802 + 602
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=
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6 400 + 3 600
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=
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10 000
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 |
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\
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