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This lecture is part of lecture series on Electrical Circuit Analysis course. It was delivered by Prof. Mursleen Sayed at Bengal Engineering and Science University. It includes: Sinusoidal, Source, Function, Response, Transient, Component, Steady, State, Phasor, Complex, Number
Typology: Slides
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v = V
cos(m
ωt +
φ)
^
^
2
) ( cos 1
2 2 0 0
m
Tt m t
rms
V dt t
V T
V^
^
^
where
θ^
= tan
-1^ (ω
L/R) (depends on
source freq and component vals) ^
First term = transient component of thecurrent. As t increases it becomessmaller, approaching 0 in the limit. ^
Second term = steady-statecomponent of the current. ^
Note the form similarity of the 2
nd
term with source function!
(^22)
2
(^22)
2
m
Rt L
m
^
An RL circuit connected to a sinusoidal voltage sourcev = V
cos(m^
ωt +
φ)
.
^
V^ m
cos(
ωt +
φ) = Ldi/dt + iR
Invented to permit extraction of sq. root of –ve num. ^
Roots of ^
Rect. form
n = a+jb (a – real, b – imaginary)
^
Polar form
^
For ease polar form
^
Conversion from rect
polar form.
^
In which quadrant does
θ^
lie?
^
Graphical representation
of complex number. x-real
axis, y-imaginary axis. ^
Conjugate
n*
reflects n around real axis.
^
Addition (subtraction) ^
Multiplication (division)
2
j
Roots of a complex number:
^
2 2 2 (^22)
k jk k
k j
k
k
) 4 ( 1 3
) 2 ( 1 2
1 1
) 4 ( ) 2 (
k j k
k j k
j k k
j
j
j
j
k
docsity.com
jωt
jωt
^
jωt
jωt
^
jβ
Example 1 (how phasor helps!) If^
y^1
= 20cos(
ωt – 30
0 )^ and
y^2
= 40cos(
ωt + 60
Express
y = y
+ y 1
2
(a)^
Solve by using trigonometric identities (b)^
Solve by using the phasor concept (a) y
1
= 20cos
ωtcos
0 + 20sin
ωtsin
0
y^2
= 40cos
ωtcos
0 – 40sin
ωtsin
0
y^
= (20cos
0 + 40cos
0 )cos
ωt + (20sin
0 – 40sin
0 )sin
ωt
= 37.32cos
ωt – 24.64sin
ωt
= 44.72((37.32/44.72)cos
ωt – (24.64/44.72)sin
ωt)
= 44.72(cos33.
0 cos
ωt – sin33.
0 sin
ωt)
= 44.72cos(
ωt + 33.
(b) Y
=^ Y
+ Y 1
2
= =^
(17.32 –j10) + (20 +j34.64) =^
37.32 + j24. = =^
44.72cos(
ωt + 33.
0
60 40 30 20
