Electrical Engineering Assignment: Circuit Analysis and Mechanics

This document provides a detailed solution to an electrical engineering assignment. It includes two questions, the first involving circuit analysis to determine the resultant combination of two voltage sources, calculating the magnitude and phase angle of the combined voltage. The second question focuses on mechanics of materials, calculating the stress at the point of failure, fracture strain, and modulus of elasticity of a sample given its dimensions and the applied force. The solution provides step-by-step calculations and explanations for each part of the problem. Desklib is a platform where students can find similar solved assignments and study resources.

Question 1
Part A
V 1= 30sin ( 4000 πt ) V
V 2= 140sin (4000 πt + π
2 ) V
Phasor angle for V 1 = 0
Phasor angle for V 2 = π
2 radians
Convering to degrees: π
2 * 180
π = 90 degrees
The resultant combination, V 3 of V 1 and V 2 calculated from:
Horizontal component
30cos 0 + 140cos 90 = 30
Vertical component
30sin 0 + 140 sin 90 = 140
Magnitude of V 3:

V 3=√302 +1402 = 143.18
To get the phase angle of V 3:
∅ = tan−1 vertical component
horizontal component
∅ = tan−1 140
30 = 77.9
Converting to radians we multiply the angle in degrees by π
180
77.9 * π
180 = 1.36 rad
Hence the expression of V 3 is:
V 3= 143.18sin ( 4000 πt+1.36 ) V
Part B
Angular velocity is expressed as:
ω=2 πf
Making f (frequency) the subject:

f = ω
2 π
Substituting the values:
f = 4000 π
2 π = 2000
f = 2000 Hz

Question 2
Parameters of the sample:
Thickness: 6mm
Width: 15mm
Original length: 65mm
New length: 72mm
Force: 27kN
Part A
Stress at point of failure, σ :
σ = Force
Cross−sectional area
σ = 27 k
6 mm∗15 mm = 27 k
9∗10−5 = 300 * 106 Pa
σ = 300.00 MPa
Part B

Fracture strain, :
ε = change∈length
originallength
ε = ∆ l
lo
ε = 72−65
65 = 7
65 = 0.1076
ε=0.11
Part C
Modulus of elasticity of the sample, E:
E= σ
ε
E=300 M Pa
0.1076
E=2.7857∗109
E= 2.79 GPa