ENS - 4 - TMA (v2.1): AC Circuits Assignment Solution

This document presents a comprehensive solution to an AC circuits assignment, addressing key concepts in electrical engineering. The solution includes detailed calculations for power factor correction, reactive power, and impedance analysis in a circuit with a given voltage, resistance, and power factor. It also covers the analysis of a circuit with a given voltage, current, frequency, and phase, determining the inductance and capacitance. The assignment further explores transformer design, calculating secondary turns, currents, and maximum flux. Finally, it delves into harmonic analysis, providing calculations and graphical representations of voltage waveforms with harmonic components, including the determination of voltage at a specific time and the percentage error due to harmonics. The assignment is based on the Teesside University curriculum.

1. 120V
Z R=10Ω
Given power factor Pf =0.72
1rw power = 375v
1> Apparent power-
Apparent power = prue power/ power factor
S = P/PF
= 375/0.72 = 520.8333 VA
2> Reactive power = √apparent power2-(true power)2
√s2-p2
√ ( 520.8333 ) 2−(375)2
=361.44477627
3>Magnetic current
Power = voltage xcurrent x power fa
P = V x I x PF
375 = 120 x I x 0.72
I = 375/120x0.72
I = 4.34A
4>Total ohms=Voltage/current
=120/4.34
=27.6498 ohms(minus 10)
=17.65ohms
The Power factor is lagging within the circuit therefore the circuit is Inductive
120

2. Given f= 60Hz
For 60% p.f:-
Θ1= Cos-1(0.6)
=53.130
φ1= 50000 tan(53.13)=66.667KVAR
For 90%pf
φ2=Cos-1(0.9)
=25.840
φ2= 50000 tan (25.84) = 24.22 KVAR
φcapacitor =φ2-φ1
=24.22-66.667
=-42.45
Xc= VRMS/φ c
= (104)2/-42.45x103
= -2356Ω
[ here we get –ve value the minus sign reflects the fact that it is capacitive reactance
W = 2πf
W = 2x3.14x6 =377
C=1/w.Xc= 1/377x2356
=1.126μf

3. Given Votage = 5V I = 11.9Ma
Frequency = 2.5 KHZ
Φ =70
R = 5/11.9x10-3
=5x103/11.9
=50000/119 => 420.20Ω
Φ =WL/R
W=2πf
W = 2πx2.5x103
=15700
Φ = WL/R = 15700X2/420.16
L = 420.20X70/15700 =>1.871
LC=1/W2
C = 1/W2XL => 1/(15700)2x1.871
=2.16nf
4. Given-> N1 – Primary turns =1000
V2 – Voltage=100
V1- -2000
a> Secondary turns –N2
V2/V1=N2/N1
N2=V2xN1/V1
=100x1000/2000
=50 turns
b> The Rated primary and secondary current
I1=120000/2000
=60Ω
I2= 60/.05=6000/5

=1200A
c> Maximum flux:-
Vp =4.44xfxN1xФmax
Фmax=V1/4.44x60x1000
=200/444x60
= 0.00751wb
Фmax=7.51mwb
d>Bm = 0.25T
Φm = Bm*A Or , A = Φm/Bm
= 0.0075/0.25 m2
= (0.0075/0.25)*10000 cm2= 300 cm2
5. Given Vrms=100V
F=120 Hz
i.V=Vrms*√2
=100*1.414=1.414V at 120 Hz
3rd Harmonic=20% of 141.4
=20/100*141.4
=28.28V at 360 Hz
5th Harmonic=10% of 141.4
=10/100*141.4
=14.14V at 600 Hz
V=(141.4sin(240πt))+(28.3sin(720πt))+(14.1sin(1200πt-1.2))
ii. Sketch w/f for harmonic component for fundamental = 141.4sin(240πt)
3rd harmonic = 28.3 sin(720πt)
5th harmonic = 14.1 sin(120πt-1.2)
Time Fundamental 3rd harmonic 5th harmonic

1 0 0 0 -13.17903
2 0.001 96.7949 21.8055 7.6504
3 0.002 141.1209 -27.7987 0.8004
4 0.003 108.9905 13.6336 -8.9455
5 0.004 17.7221 10.4179 13.673719
6 0.005 -83.112834 -26.9148 -13.1790
7 0.006 -138.8954 23.8944 7.65040
8 0.007 -119.3879 -3.54693 0.80041
9 0.008 -35.1647 -19.37268 -8.9455
10 0.009 68.1199 28.24415 13.67371
11 0.01 134.4793 -16.63432 -13.17903
iii. Determine voltage at 20ms
V = (141.4sin(240πx0.02))+(28.3sin (720πx0.02))+(14.1 sin(1200πx 0.02-1.2))
V = (141.4xsin(15.072)+(28.36 in (45.216)+(14.1 sin(74.16))
= 83.11 + 26.91 -13.14
= 96.88v
iv. V = 10 Vrms what is % error at 20mg
Actual instantaneous value at 20mg
V = 96.88 From 13 calculation ideal instantaneous value at 20 mg
= 141.42 sin (240πx0.02) = 83.11
Error = [actual instan value- ideal istan value/ideal instan value]x100
= 96.88-83.11/83.11x100
= 13.77/83.11x100 => 16.5%