University Electrical Engineering: DC and AC Circuit Analysis Report

This assignment report presents a comprehensive analysis of DC and AC electrical circuits. The solution begins with an analysis of a DC circuit, applying Kirchhoff's and Ohm's Laws to determine voltage drops and currents. It then explores Thevenin's equivalent circuit. The report continues with an examination of various circuit components, including low-pass filters and resonant circuits, calculating impedance, current, voltage drops, and gain. The assignment also covers the principles of electromagnetic induction, explaining its applications. All calculations are presented with detailed working steps and Harvard referencing as required, ensuring a logical and coherent structure suitable for engineering analysis.

SOLUTION REPORT
FOR
ENGINEERING MATHEMATICS QUESTION

Part: 1
RP = R 1 R 2
R 1+R 2 = 100× 120
100+120 = 12000
220 = 54.54Ω
RS = R4 +R3 =150 +50= 200Ω

By Kirchoff’s Voltage law: (Mukhituly 2016)
12=54.54 I1 +1000 I3
9= 200 I2 + 1000I3
I3=I1+I2 (By Kirchoff’s Current Law)
12 = 54.54 I1+1000 I1 +1000I2
12=1054.54 I1 +1000 I2 ……Equation (1)
9= 1200 I2 + 1000 I1 ………Equation (2)
By resolving these equations:
I2= -0.46 A
I1 = 1.782 A
I3 = 1.782 + (-0.46)
I3 = 1.322 A
By using Ohm’s Law:
VP = 54.54 .I1
VP = 54.54 × 1.782
VP = 97.19V
VS= 200. I2

VS= 200 × -0.46 A (- sign shows the opposite direction of current so by neglecting it)
VS= 92 V
VL =1000.I3
VL =1000× 1.322
VL = 1322 V

Part: 2
(a)
R1 +R2 = 2+1 = 3KΩ
R4 +R3 = 2+1 = 3KΩ
IL = VTH / (RTH + RL )

Applying KVL:
3000 I + 3000 I +4.5 – 9 = 0
6000I – 4.5 =0
6000I = 4.5
I = 4.5 / 6000
I = 0.00075 A
VTH = 9 – 3000 × 0.00075 = 6.75 V
VTH = 3000 × 0.00075 + 4.5 = 6.75 V
By putting the value of internal resistance as 1 kΩ in place of power supplies:
RTH = 4× 4 /(4 +4) = 2 kΩ
IL = VTH / (RTH +RL) = 6.75 / (2000 +2000)

IL = 6.75 / 4000 = 0.0016 A
So the Thevnin’s equivalent Circuit is:
(b)
The Thevnin’s equivalent circuit:

For RL = 1KΩ,
IL = VTH / (RTH + RL) = 6.75 / (2000 +1000) = 0.0022 A
PL = IL2. RL
PL = (0.0022) 2. (1000) = 0.0050 W.
For RL = 1.5 KΩ,
IL = VTH / (RTH + RL) = 6.75 / (2000 +1500) = 0.0019A
PL = IL2. RL
PL = (0.0019) 2. (1500) = 0.0054 W.
For RL = 2 KΩ,
IL = VTH / (RTH + RL) = 6.75 / (2000 +2000) = 0.0016A
PL = IL2. RL
PL = (0.0016) 2. (2000) = 0.0051 W.
For RL = 2.5 KΩ,
IL = VTH / (RTH + RL) = 6.75 / (2000 +2500) = 0.0015A
PL = IL2. RL
PL = (0.0015) 2. (2500) = 0.0056 W.
(c)
According to Maximum Power Transfer Theorem:
For RL = 1 KΩ,
PMAX = (6.75)2 / (4 × 1000) = 0.0113 W
For RL = 1.5 KΩ,
PMAX = (6.75)2 / (4 × 1500) = 0.0075 W
For RL = 2 KΩ,
PMAX = (6.75)2 / (4 × 2000) = 0.0056 W
For RL = 2.5 KΩ,

PMAX = (6.75)2 / (4 × 2500) = 0.0045 W
So, RL = 1 KΩ that provides Maximum Power Transfer.

Part: 3
(a)
R1 +R2 = 4+8 = 12KΩ
R4 +R3 = 2+4 = 6KΩ
Calculation of IN:

I1 + I2 = IN
6= 12000 I1
I1 = 0.0005A
4 = 6000 I2
I2 = 0.0006 A
I1 + I2 = IN
0.0005 + 0.0006 = IN
IN = 0.0011 A
By putting the value of internal resistance as 4 KΩ in place of power supplies :
Calculation of RN:
RN = (16000 × 10000) / (16000 +10000)

RN = 6153.84 Ω
RN = 6.153KΩ
IL = RN / (RN + RL)
IL = 6153.84/ (6153.84 + 2000)
IL = 0.75 A
(b)
The power dissipation in the load:
For RL = 1KΩ,
IL = RN / (RN + RL)
IL = 6153.84/ (6153.84 + 1000)