Vectors and phasors
Vectors and Phasors
A supplement for students taking BTEC National, Unit 5, Electrical
and Electronic Principles
Owen Bishop
Copyrught © 2007, Owen Bishop
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Electronics — Circuits and Systems
Vectors and phasors
Physical quantities are of two kinds:
•
Scalars - specified by their magnitude
(= size). Examples are mass, electric
charge, temperature, speed.
•
Vectors - specified by their magnitude
and their direction. Examples are weight
(its direction is ‘down’), electric field,
velocity, acceleration.
Scalars of the same kind can be added together,
using the ordinary rules of arithmetic.
Example: Distance is a scalar; if I walk 3 km to
the shopping centre and then walk 2 km to the
swimming pool, the total distance I walk is 3 +
2 = 5 km. Note that I may change direction
several time on my walk.
Vectors of the same kind can be added by
techniques that include both magnitude and
direction. Two ways of adding vectors are by:
•
making an accurate scale drawing.
•
using trig equations.
As the boat steers north-east across the
river (vector AB)), it is carried downstream
by the current (vector AC). Its resultant
velocity relative to the banks is AD,
The drawing has sides AB and AC at the
correct angles. The lengths of the sides are
proportional to the magnitudes of the
velocities.
The drawings below shows the stages in
solving the problem.
Adding vectors by drawing
Problem::
A boat is headng north-east at 1.4 m/s across a
river that flows from west to east at 2 m/s. Find
the boat’s resultant velocity, the sum of these
two velocities.
Solution:
The drawing above shows the addition of the
two velocities: AB, the velocity of the boat
relative to the water and AC the velocity of the
water relative to the banks.
Their sum, or resultant, is a vector AD,
obtained by completing a parallelogram with
sides AB and AC, and then drawing the
diagonal.
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Vectors and phasors
To solve this by drawing, first decide on a
scale, for example let 25 m = 1 m/s.
We do not need to draw the whole
parallelogram because side CD is the same
length as AB and at the same angle. Instead,
draw a triangle ACD as shown. To scale, AC is
200 mm and CD is 140 mm The angle between
AC and the extension of AB is 45°.
Summing up: Draw the vectors start to end, to
scale and at the given angles. The resultant is
from the start of the first vector to the end of
the second (or last - Yes, you can add three or
more vectors by this method).
Important note: All vectors must be of the same
kind. In the example we added velocities and
the resultant is a velocity. You can not, for
example, add a velocity to a weight.
Adding vectors by trigonometry
The drawing method is easy to understand but
its accuracy depends on how good you are at
drawing and measuring. For absolute accuracy
we rely on trig(onometical) relationshops. A
few essention ones are listed in the box below.
Sketch for the trig solution, with known
values written on it.
In the triangle ADE:
DE = 0.99
AE = AC + CE = 2 + 0.99 = 2.99
–1
–1
angle A = tan DE/AE = tan 0.99/2.99
= 18.3°
This example shows how use the equations
below to calculate the size and direction of the
resultant. There are more examples in the
following pages.
Trig facts
For the trig technique, sketch a diagram of the
vectors. There is no need to draw to scale or
plot angles accurately..
You need to complete a right-angled triangle,
obtained by dropping a perpendicular DE from
D down to the extension of AC. Mark on it the
actual velocities (no scales in this method), as
shown at top right.
In the triangle CED:
In a right-angled triangle:
sin C = DE/DC
DE = DC × sin C = 1.4 × sin 45 = 0.99
sin B = b/a
In the triangle CED:
cos C = CE/DC
CE = DE × cos C = 1.4 × cos 45 = 0.99
In this example CE = DE because the angle is
45. This is not so when C has some other value.
cos B = c/a
–1
tan b/c = B
2
2
b +c =a
2
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Electronics — Circuits and Systems
Phasors
We are used to representing an alternating
voltage by plotting a graph showing how the
voltage varies with time (see pp. 48 and the
diagram below). Another way of representing
the voltage is to draw a phasor diagram.
In a phasor diagram, the voltage is represented
by a point (A) moving in a circular path, as on
the right at the bottom of this page. At any
instant, the vertical distance of A above the
horizontal axis is the instantaneous value of
the voltage.
The angle marked in the diagram is the phase
angle. This increases with time at a constant
rate as A moves round the circle. The radius of
the circle equals the peak voltage, the peaks
occuring when A is directly above and below
the centre of the circle. In other words, the
radius of the circle equals the amplitude V of
the signal.
The vertical distance of A from the horizontal
axis is alternately positive and negative — the
voltage is alternating. Using the trig equation
for the sine of an angle, the vertical distance of
A at any instant is:
v = V sin φ
Where v is the instantaneous voltage, V is the
amplitude and φ is the phase angle.
The graph on the left, which shows v plotted
against t, is therefore a sine curve. The signal is
a sinusoid.
In the equation above we usually know V and
want to find v. We need to know φ, which
increases steadily with time t. The rate of
change of the phase angle is called the angular
velocity, symbol ω. This is related to the
frequency of the signal by the equation:
ω. = 2πƒ
Where ω is the angular velocity in radian per
second and ƒ is the frequency in cycles per
second. The constant value 2π refers to the fact
that 2π radian equals one complete cycle of
360°.
If the phase angle is zero at zero seconds, its
value at t s is:
ωt = 2πƒt
Entering this into the equation for v, we find
that:
v = V sin ωt = V sin 2πƒt
If the signal leads or lags by a given angle, ϕ,
the value of v is calculated from:
v = V sin (ωt + ϕ) = V sin (2πƒt + ϕ)
Remember that all angles must be expressed in
radian.
An alternating voltage signal (left) can be represented by a phasor diagram (right). Think of
point A moving anti-clockwise round the circle. At zero time it is at the extreme right of the
dagram, and moes once around the circle in one period of the signal
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Vectors and phasors
Adding phasors
The discussion on the opposite page shows that
an alternating voltage can be represented as a
phasor and that a phasor has two properties:
•
•
Magnitude: depends on the amplitude V
of the signal.
Direction: depends on time t.
Vectors have the same two properties, so a
phasor can be treated in the same way as a
vector. In particular, two or more phasors can
be added together using the same techniques
(drawing or trig) as we use with vectors. This
provides a way of analysing circuits in which
there are two or more alternating signals at the
same time. We can add them and obtain their
resultant. The only requirement is that all the
signals have the same frequency.
Problem
In this problem the frequency of the signals is
represented by the value 500. Referring to the
equation on the opposte page:
2πƒ = 500
ƒ = 500/2π = 80 Hz
Inductive impedance
This is an example of the kind of electrical
problem that can be solved by using phasors.
Problem
Using a phasor diagram, find the sum of two
alternating voltages, represented by the
equations:
v1 = 60 sin 500t
v2 = 40 sin (500t + π/5)
Solution
An alternating voltage with the equation v =
100 sin 1600πt, is applied to a coil with
inductance 200 mH and resistance 500 Ω.
Calculate the impedance of the coil and the
current that passes through it.
Solution
Both voltages have the same frequency so they
can be summed by vector addition of their
phasors (see diagram). The phasor for v1 has no
phase lead or lag, so it is drawn as a horizontal
vector pointing right. Its length is scaled to
represent a peak amplitude of 60 V. The phasor
for v2 leads v1 by π/5 rad (= 36°), so this is
drawn as shown, rotated by π/5 rad
anticlockwise of v1, and with a length scaled to
represent an amplitude of 40 V .
Completng the parallelogram (or drawing a
triangle as on p. 2), the length of the resultant is
equivalent to 95 V. The phase angle of the
resultant is 13.75° (= 0.24 rad).
The equation for the resultant phasor is:
v = 95 sin (500t + 0.24)
Note that all the signals have the same
frequency.
The standard equation for a sinusoidal
alternating voltage is:
v = V sin (2πƒt + φ)
where:
v = the instantaneous voltage
V = the peak voltage
2πƒ = ω = angular velocity in rad/s
φ = phase lead or lag, in radian
Comparing the data supplied in the problem
with this equation, the peak voltage is 100 V
and the frequency is 1600π/2π = 800 Hz.
The phasor diagram is drawn overleaf. At
800 Hz, the impedance of the coil is:
–3
XL = 2πƒL = 2π × 800 × 200 × 10 = 1005 Ω
In the diagram, the resultant reperesents the
impedance of the whole circuit, the vector sum
of the resistive and inductive phasors.
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Electronics — Circuits and Systems
Solution
As frequency increases, the voltage waveform
across the capacitor has a phase lag that
approaches –90°. We assume that the circuit is
operating in this region.
The rms current is the same throughout the
circuit, so the rms voltage drop across each
component is proportional to its impedance/
resistance.
Given that the same current flows through the
impedance and resistance of the coil as well as
through the circuit as a whole, the sizes of the
voltage phasors is proportional to their
impedances.
The magnitude of the resultant, the total
impedance, is found by using the Pythagoras
equation (p. 2):
2
Draw a triangle to represent the impedance of
the capacitor and the resistance of the resistor
(below). The triangle is right-angled because of
the phase lag between the resistance and the
capacitance. We know the resistance and can
calculate the impedance of the whole circuit:
Z = supply volts/current = 50/0.8 = 62.5 Ω
Now we find XC by using the Pythagoras
equation (p. 2):
2
2
XC = √(62.2 – 56 ) = 27.07 Ω
2
Z = √(500 + 1005 ) = 1123 Ω
From this we obtain the current flowing in the
circuit:
I = V/Z = 100/1123 = 89 mA
and the phase angle:
–1
φ = tan XL(Z/R) = 1123/500 = 1.15 rad
The current lags VZ by this angle.
We can now calculate the frequency:
Capacitative impedance
Phasors can also be used when solving
problems about impedance.
Problem
A capacitor 47 µF and resistor 56 Ω are wired
in series across an alternating supply of rms
voltage 50 V, and draw an rms current of 0.8 A.
What is th e frequency of the supply, and what
are the voltages across the capacitor and
resistor? State the equations for the total
voltage and current.
ƒ = 1/(2πCXC) = 125 Hz
Given the frequency, the angular velocity is:
ω = 2πƒ = 785 rad/s
To plot the phasor diagram we calculate the
magnitudes of the vectors, using the Ohm’s
Law equation:
VR = IR = 0.8 × 56 = 44.8 V
VC = IXC = 0.8 × 27.07 = 21.7 V
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Vectors and phasors
The phasor diagram below shows VC lagging
90° behind VR. The phase angle of the resultant
is found by calculating:
–1
tan (VC/VR) = 21.7/44.8 = 25.8° or 0.45 rad
This equals the expression in brackets
785t – 0.45 = 0.5235
t = (0.5235 + 0.45) /785 = 1.24 ms
v first reaches half-peak after 1.24 ms
Phase angle in an RLC series
resonant circuit
This is an example in which three phasors are
added.
Problem
The problem quotes the supply voltage as 50 V
rms, but the value of the resultant must be
based on the instantaneous voltage. The peak
instanteous voltage is 50 ×√2 = 70.7 V.
We now have all the values needed for writing
the equations for the total voltage and current.
For the current, the instantaneous peak value
is:
0.8 × √2 = 1.13 A
The current equation is:
i = 1.13 sin 785t A
For the voltage:
v = 70.7 sin (785t – 0.45) V
A 10 mH inductor with coil resistance 25 Ω is
connected in series with a 4.7 µF capacitor. As
in the first problem, the resistance of the coil
can be taken to be in series with its inductance.
An allternating signal frequency 1.2 kHz is
applied to the circuit. What is the impedance
of the circuit at t his frequency? Calculate the
current and voltage phase angle.
Solution
Calculate the inductances
frequency:
XL = 2πƒL = 75.4 Ω
XC = 1/2πƒC = 28.2 Ω
at
the
given
These two inductances act in opposite
directions (inductive leads, capacitive lags) as
shown in the left-hand drawing below.
These equations can be used to calculate the
current or voltage at any given instant. Times
are taken from t = 0 when i or v = 0. For
instance, when t = 5 µs,
–6
v = 70.7 sin (785 × 5 × 10 – 0.45) = –30.5 V
Another example is to find when the voltage
first reaches half the peak, that is, 35.35 V.
This occurs when:
70.7 sin (785t – 0.45) = 35.35
Rearranging terms:
sin (785t – 0.45) = 35.35/70.7 = 0.5
Find the angle which has 0.5 as its sine:
–1
sin 0.5 = 0.5235
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Electronics — Circuits and Systems
If two phasors have exactly opposite directions
the magnitude of their resultant is found by
ordinary subtraction. In this case, we replace
the two phasors with a single phasor XLC which
has magnitude 47.2 Ω and acts in the direction
of XL. This is shown in the right-hand drawing.
Now there are only two phasors to add
together and the drawing shows the resuktant
Z of all three original phasors.
The magnitude of Z is found by using the
Pythagors equation. Having found this we use
it to calculate the current:
I = V/Z = 100/53.4 = 1.87 A
The phase angle of the resultant is:
–1
φ = tan (47.2/25) = 1.08 rad
The voltage across the circuit leads the current
by 1.08 rad.
Questions
There are problems on phasors on this
Companion Site. Click on ‘Questions’
and then on ‘Phasors’.
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