Analytical & Computational Methods

This document provides solutions for tasks related to Analytical & Computational Methods. It includes equations, phasor diagrams, and signals. The tasks cover topics such as current waveforms, power signals, RLC circuits, and cross products.

Analytical & Computational
Methods

Table of Contents
TASK 1............................................................................................................................................1
TASK 2............................................................................................................................................2
TASK 3............................................................................................................................................3
PART 1........................................................................................................................................3
PART 2........................................................................................................................................3

TASK 1
a) Given equation of current waveform as –
is = 13 cos (2πft – π/4) [A]
where, f = 1 Hz and t represents time
Transforming the given equation into t as –
t = cos-1 (is/13) + π/4
2πf
when is = +10A then, t will be
t = cos-1 (10/13) + π/4
2π. 1
= 0.69 + 45
2 x 180
= 0.13 sec
b) The instantaneous value of a power signal is
12 ∟(3π/8) [W]
here, W is units
From given equation, it has been evaluated that 12 is magnitude and 3π/8 represents argument
then, horizontal component of the signal = cos 3π/8
= -0.04
while, vertical component of the signal = sin 3π/8
= -0.99
1

TASK 2
(a) For a RLC circuit, phasor diagram where RL leads RC by 90 is drawn as –⁰
VL
V
90⁰
VR
Therefore, Resultant voltage across whole RL can be calculated by using pythogoras theorem
V2 = V2R + VL2
= 302 + 402
= 900 + 1600
= 2500
or, V = 50 volts
(b) Given,
current as
I = 2𝑖 + 3𝑗 – 4k
and,
magnetic field as
B = 3𝑖 – 2𝑗 + 6k
then, cross product of current and magnetic field is –
I X B = (2𝑖 + 3𝑗 – 4k) x (3𝑖 – 2𝑗 + 6k)
i j k
2 3 -4
3 -2 6
= i (18 – 8) – j (12 + 12) + k(-4 – 9)
= 10i – 24j – 13k
2

TASK 3
PART 1
Given signals -
𝑣1 = 40 sin (4𝑡)
𝑣2 = A cos (4𝑡)
third signal –
𝑣0 = 50 sin (4𝑡 +α)
on expanding third signal as –
𝑣0 = 50 (sin 4𝑡 cos α + cos 4𝑡 sin α)
𝑣1 + 𝑣2 = 50 (sin 4𝑡 cos α + cos 4𝑡 sin α)
40 sin (4𝑡) + A cos (4𝑡) = 50 (sin 4𝑡 cos α + cos 4𝑡 sin α)
on comparing, α = 36.87 approx.⁰
then,
A cos (4𝑡) = 50 cos 4𝑡 sin α
A = sin α = 50 sin 36.87⁰
= 30 approx.
PART 2
Given,
Third harmonic of a sound wave as –
4 cos (3𝜃) – 6 sin (3𝜃)
𝑅 sin (3𝜃 +𝛽) = R (sin 3𝜃 cos 𝛽 + cos 3𝜃 sin 𝛽)
4 cos (3𝜃) – 6 sin (3𝜃) = R (sin 3𝜃 cos 𝛽 + cos 3𝜃 sin 𝛽)
on comparing –
R sin 3𝜃 cos 𝛽 = -6 sin (3𝜃)
R cos 𝛽 = -6 …(i)
R cos 3𝜃 sin 𝛽 = 4 cos (3𝜃)
R sin 𝛽 = 4 …(ii)
3

adding and squaring both equations –
R2 cos2 𝛽 + R2 sin2 𝛽 = (-6)2 + (4)2
R2 (cos2 𝛽 + sin2 𝛽) = 36 + 16
R2 = 52
or, R = √52 = 7.2
while, on dividing both equation
tan 𝛽 = 4/6
𝛽 = tan-1 (4/6) = 0.58 approx.
so,
4 cos (3𝜃) – 6 sin (3𝜃) = 7.2 sin (3𝜃 + 0.58)
4