Engineering Science (Electrical) Assignment Solution & Explanation

This document presents a comprehensive solution to an Electrical Engineering Science assignment, covering various topics within the field. It includes spectrum diagrams for given voltage equations, plots of voltage waveforms, and analysis of circuit transformations and trigonometry phases. The solution details calculations for capacitive reactance, impedance, current, and phase angles in a circuit, along with voltage drops across resistors and capacitors. Furthermore, it addresses active and reactive power calculations, power factor rectification, and transformer voltage and current relationships. The document also includes calculations related to current transformers and concludes with a bibliography of relevant textbooks. Desklib offers a platform for students to access similar solved assignments and study resources.

Engineering Science (Electrical) 1
ENGINEERING SCIENCE (ELECTRICAL)
By Name
Course
Instructor
Institution
Location
Date

Engineering Science (Electrical) 2
QUESTION ONE
a. Spectrum diagram
For V1
V1= 8.1 sin (314.2 t) – 0.9 sin (942.6 t) + 0.324 sin (1571t)
For V2
V2= 2.5-1.592 sin ( 314) – 0.796 sin ( 628.4t) – 0.5305 sin ( 942.6t)

Engineering Science (Electrical) 3
For V3
V3= 20+25.46 sin (314.2 t ) + 8.48 sin( 942.6 t)+ 5.092 sin( 1571t)

Engineering Science (Electrical) 4
b. Plotting
For V1= 8.1 sin (314.2 t) – 0.9 sin (942.6 t) + 0.324 sin (1571t)
For V2 = 2.5-1.592 sin ( 314) – 0.796 sin ( 628.4t) – 0.5305 sin ( 942.6t)

Engineering Science (Electrical) 5
For V3= 20+25.46 sin (314.2 t ) + 8.48 sin( 942.6 t)+ 5.092 sin( 1571t)
c)
V1= 8.1 sin (314.2 t) – 0.9 sin (942.6 t) + 0.324 sin (1571t)
Is a sine wave
V2= 2.5-1.592 sin ( 314) – 0.796 sin ( 628.4t) – 0.5305 sin ( 942.6t)
Is a transformation: scaling function
V3= 20+25.46 sin (314.2 t ) + 8.48 sin( 942.6 t)+ 5.092 sin( 1571t)

Engineering Science (Electrical) 6
Is a trigonometry: Phase
QUESTION TWO
a) From
Xc= 1
2 πfC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
Xc= 1
2× 3.142×1100 × 1×10−5
Xc= 1
0.069124
Xc= 144.4667
And R= 10Ω
Z=√ R2 + Xc2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2
Z=√ 102 +14.46672
Z=√309.287
Z= 17.586
b) From V=IZ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
V=10
Z= 17.586
I= V
Z
I= 10
17.586
I= 0.5686 A

Engineering Science (Electrical) 7
c) Phase angle
From cos ɵ = R
Z
cos ɵ = 10
17.586
cos ɵ = 0.5686
ɵ = 5.340
d) Voltage drop across each components
i) Voltage drop across resistor
From V=IR
V= 10 × 0.5686
V= 5.686
ii) Voltage drop across capacitor
Since the current flows to the resistor before the capacitor, therefore there will be voltage
drop in the resistor before capacitor hence the voltage drop in the capacitor is given as
below;
10- 5.686
V=4.314 volts
e)
Voltage Phasor diagram

Engineering Science (Electrical) 8
QUESTION THREE
Cos ɵ = 0.3
ɵ = 72.540
Tan ɵ = Inductive current
Resistive currrent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
Tan ɵ = 2.1
Resistive current = 3.1798
Resistive current = 0.6604 A
Active power therefore is given as below
Active power = 0.6604 × 230
Active power = 151. 892 watts
b)
Reactive power
Tan ɵ= Reactive power
Active power
Tan ɵ =3.1798
Active power = 151.892 watts

Engineering Science (Electrical) 9
Reactive power = 151.892 × 3.1798
Reactive power = 483 Var
c)
Power factor rectification
1- 0.3 = 0.7
Cos ɵ = 0.7
ɵ = 45.570
Tan ɵ = New reactive power
active power = 1.0202
New reactive power = 151.892 × 1.0202
New reactive power= 154.94 Varl
Reactive power improvement
New reactive power = V 2
Xc
154.96 = 2302
Xc
Xc= 341.378
From Xc= 1
2 πfC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
Xc= 1
2× 3.142×50 × C = 341.378
107260.967C = 1
C= 9.323μF
d)
P= I2Xc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

Engineering Science (Electrical) 10
341.37 I2
341.378 = 154.96
341.378
I2= 0.4559
I = 0.6737 A
QUESTION FOUR
Voltage at secondary is given by the following expression
240
Vs = 11
6
Vs= 130.91 Volts
And from V= IR
130
24 = 24 Is
24
Is= 5.45 A
Current at the source
5.45
Ip = 11
6

Engineering Science (Electrical) 11
Ip= 2.97A
QUESTION FIVE
From 240
Vs = 11
6
Vs= 130.91 Volts
And from V=IR
130
5 = 5 Is
5
Is= 26.182 A
And from
26.182
Ip = 11
6
Ip= 14.28 A
QUESTION SIX

Engineering Science (Electrical) 12
From cos ɵ = Active power
Apparent power
0.9 = 30
Apparent power
Apparent power = 33.3VA
Vs= 1100/100
Vs= 110 Volts
33.3
110 = 110 I
110
I= 0.3027 A
CT = 100: 1
Ip= 100×0.3027
Ip= 30.27 A