Electrical and Electronic Principle 1 - Solved Assignments and Essays

Electrical and electronic principle 1
ELECTRICAL AND ELECTRONIC PRINCIPLE
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Electrical and electronic principle 2
ASSIGNMENT ONE
QUESTION ONE
Given that
Using Thevenin’s theorem
Open 50<450 load branch.
Find out the voltage across 50<450 branch
To find Vth

Electrical and electronic principle 3
Apply KVL
586<00 + I j4 + j 6 + 586 <900 = 0
I j10 = 586<00 – 586< 900
I = - 58.6- j 58.6
I = 82.87< -1350
Apply KVL
-586<00 + IJ4 + Vth = 0
Vth = 586<00- (82.87<-135)
Vth = 351.6+j234.3
Vth 422.57 < 33.680
To find Zin
V1 and V2 are short circuited

Electrical and electronic principle 4
Zeq = j 4||j6
= j 4 × j 6
j 4+ j6
Zeq= j2.4
Zeq=2.4<900 ῼ
Thevenin’s equivalent circuit
I = Vth
Zth+ Zeq
I = 422.57<33.68
2.4< 900 +50< 450
I = 7.95-j1.86
I = 8.16<13.2 A
2. By applying superimposition

Electrical and electronic principle 5
Vi is acting in V2 is short circuited
Zeq =j4 + [ j6||50<450 ]
Zeq = 0.42+j9.49
Zeq = 9.50<87.4
IT = 5.86 <00
9.50<87.4
IT = 61.6 <87.4 A
By using current division rule
I’= j 6
j 6+50< 450 × 61.6 <-87.4
I’ = 6.79<46.8
V2 is acting; V1 is short circuited

Electrical and electronic principle 6
Zeq= [ j4||50<450] +j6
Zeq = 9.77<88.8
IT= 59.97 <1.2
By using current division rule
I’’ = j4
j 4 × 50<450 × 59.97 <1.2
I’’ = 4.53<42.12A
Finally
I’=I’+I’’
=6.79<-46.8+4.58<43.13
I = 8.16 <-13.2 A
3. By transforming voltage sources into current sources
Similarly

Electrical and electronic principle 7
In parallel the current source are algebraically added based on their directions.
Here IN1 and IN2 are in some direction so they are added or we can use Kcl method also apply Kcl at
node V
-146.9<-90 + V
j 4 + V
50<450 + V
j 6 – 97.66<00 =0
V [ V
j 4 + V
50<450 + V
j 6 ] = 146.5<-900+ 97.06<00
V [ 0.43<-88.11] = 176 .06<-56.31
V =409.45<31.79
I= V
50<450 = 409.45<31.79
50< 450
I= 8.18< -13.21 A

Electrical and electronic principle 8
QUESTION TWO