Download Test preparation in complex numbers

1
Test preparation in complex numbers
Point 1
If you see a problem stating;
“Simplify the following: (2 + j3)(5 + j4)” then that means that you have to
multiply the two brackets together the way you have done in the previous
lessons; therefore:
(2 + j3)(5 + j4) =
10 + j15
+ j 8 + j2 12
10 + j23 + j2 12
(But j2 is -1 by definition and therefore)
10 + j23 – 12 =
10 –12 + j23 =
-2 + j23
Now you do this one:
(6 + j)(3 + j8) =
Point 2
The same goes if the expression is “Simplify the following: (2 + j7)/(4 + j5)”
then that means that you have to divide the two expressions the way you
have done in the previous lessons.
You must remember that one cannot divide a complex number by another
complex number. Therefore the denominator needs to be changed into a real
number. This is done by multiplying both the numerator and denominator by
the conjugate number of the denominator (The conjugate number of the
denominator is in this example (4 – j5). So:
2
2 + j7
4 + j5
= (2 + j7)(4 – j5)
(4 + j5)(4 – j5)
= 8 + j(28 – 10) -j2 35 = 43 + j18 =
16 + j(20 – 20) – j2 25
16 + 25
43 + j18 = 1.05 + j0.44
41
Now you do this one:
3 – j8
2+j
=
Point 3
Those of you who do electrical engineering are familiar with the expression
“The impedance of a circuit”. This is resistance of the circuit to the flow of
a.c. current and is measured in Ohm. As you may realise that in such circuits
the current may lag the voltage and vice versa and it is therefore convenient
to express voltage and current as complex numbers.
If you are given the task of drawing an Argand diagram of a circuit
impedance given as a complex number, then you do it the way you have done
it in session 3 of complex numbers. Just remember that the real part of the
complex number is plotted along the horisontal axis and the imaginary part is
plotted along the vertical axis. So if the impedance is Z = (3 + j2), then this
is represented as:
Imaginary
axis
3
2
(3 + j2)
1
1 2 3 4
Real axis
3
If you are asked to derive an expression for the impedance Z = (3 + j2) in
polar form then you need to calculate R and φ where R is the distanse from
origin to the point and φ is the angle this distance makes with the real axis
in the Argand diagram.
First you find R:
R = √ 32 + 22 = √ 9 + 4 + √ 13 = 3.6
You find φ from the relation:
φ = Tan –1 2/3 = 33.69°
So Z in polar form is:
3.6
33.69°
Now here is one for you to do:
The impedance of a circuit is given by the complex number Z = (4 + j3)
Construct an Argand diagram for Z = (4 + j3)
From the Argand diagram derive an expression for the impedance in
polar form


Point 4
You may come across problems where you are asked to simplify expressions
like:
7
44° x 9
22°
Remember that here you multiply the distances and add the angles so that:
7
44° x 9
22° = 63
66°
4
The same goes for:
7
44° / 9
22°
Remembering that in situations like this you divide the first distance by the
second and you subtract the second angle from the first so that:
7
44° / 9
22° = 0.777...
22°
Now you do these:
Simplify:
11
34° x 2
12°
36
28° / 9
15°
Point 5
The last point you need to deal with is when you have both current and
voltage given as complex numbers. An example will clarify this:
An e.m.f. of voltage V = (2 + j3) is applied to an electric circuit. If the
current drawn is I = (5 + j4), determine:
 The phase difference between V and I
 The power, given that the formula for power is: VI Cos φ
First convert both numbers into polar form:
Vv = √ 22 + 32 = √13 = 3.6
φv = Tan –1 3/2 = 56.3°
II = √ 52 + 42 = √ 41 = 6.4
φI = Tan –1 4/5 = 38.7°
5
The phase difference between V and I is: 56.3° - 38.7° = 17.6°
The power is:
Power = 3.6 x 6.4 x cos 17.6° = 3.6 x 6.4 x 0.9532 = 21.96 [W]
Now you do this one:
An e.m.f. of voltage V = (3 + j5) is applied to an electric circuit. If the
current drawn is I = (2 + j3), determine:


The phase difference between V and I
The power, given that the power is VI Cos φ