AC Circuits 1004121
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Transcript of AC Circuits 1004121
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AC CircuitsLecture Notes
EEE 165
Basic Electrical Technology
Course Teacher: Mr. Hafiz Imtiaz
BANGLADESH UNIVERSITY OF ENGINEERING AND TECHNOLOGY
Submitted by:
Saad Abd Ar Rafie
1004121
L-1/T-2, Section B
Department of Civil Engineering
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LECTURE12 Date: 08-10-2011
Alternating Current (AC)
A current that reverses at regular time intervals and has alternately positive and
negative values is an alternating current. An alternating current is usually referred
to as sinusoidal current.
A sinusoid is a signal that has the form of the sine or cosine function. Thus, circuits
driven by sinusoidal current or voltage sources are calledac circuits.
Let us consider the alternating current ,
i (t) = Imsin (t)
where,
Im = the amplitude of the current
= the angular frequency in radians/s
t= the argument of the sinusoid
The magnitude of current will be 0 (zero) when its amplitude is also zero.
Alternating current will be transformed into direct current when= 0.
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It is evident that the sinusoid repeats itself every Tseconds ; thus Tis called
theperiodof the sinusoid. From the figure included above, we observe that
T = 2
T = i(t) repeats itself every Tseconds is shown by replacing t by t+T . So, we get
i (t+T) = Imsin (t+T) =Imsin (t + )
= Imsin (t + 2) = Imsin (t) = i (t)
ihas the same value at (t+T)as it does at tand so i(t)is said to beperiodic.
Cycle : The portion of a waveform contained in one period of time.
Frequency : The number of cycles that occur in 1 s. The frequency of the
waveform of Fig. 13.5(a) is 1 cycle per second, and for Fig. 13.5(b), 12
cycles per second. If a waveform of similar shape had a period of 0.5 s [Fig.
13.5(c)], the frequency would be 2 cycles per second.
Thus we get,
f =
.
So, finally we get the relation = 2f, which is the angular frequency.
A more general expression for sinusoid,
i (t) = Imsin (t + )
where,
(t + )is the argument and is thephase.
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LECTURE13 Date: 19-10-2011
Sinusoids are easily expressed in terms of phasors, which are more convenient to
work with than sine or cosine functions.
A phasor is a complex number that represents the amplitude and phase of a
sinusoid. A complex number, can be written in the rectangular form as := +
Where, z=, is the real part and is the imaginary part.Polar form: z= r
Exponential form: z= Given and, we can get r and as
r = + = t n
On the other hand we can obtain
and
as
= rs = r sin
So, z can be written as
=+= r= r (s + sin Eulers Identity:
= ssi n
Where,
s=s i n =
Im axis
Re axis
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Given a sinusoid for alternating voltage,
= s + = = =
where,
= = For alternating current,
= s + =
=
where,
= = = = s + = {} *
= {}
Thus, time domain representation may be transformed into phasor domain
representation.
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LECTURE13 Date: 19-10-2011
Instantaneous expression for reactance:
= s +
Part of this, phasor expression: = Instantaneous Phasor
s + s + Impedance: Inductance
= = sin= s+
= = =
Inductive resistance, = Note: Generally, the instantaneous expression is converted into the cosine
function to obtain the phasor domain representation.
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Impedance: Capacitance
= s =
= = * sin= s+ =
=For AC circuits,
=
= =2 2
=
Capacitative reactance,
=
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LECTURE14 Date: 22-10-2011
The RL Branch:
= + =
Impedance, = + The RC Branch:
= s +
= 2 Impedance, =
=
= 2 =
=
2
=
2
= 2 + = +
= s
= s +
=
2 =
=
=
sin=
s
=
2
= =
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The RLC Branch:
= s = s+ =
s
= 2 + = So,
Impedance,
= +
= 2 = 2 =
2
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LECTURE15 Date: 24-10-2011
Calculation of Power:
= + So, || = + 2
= tn
Instantaneous Power,
=
= ss= *s2+s
The angle difference between voltage and current is the angle of impedance
=
= =
= ||
= s = s || =
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To get the average, well divide the whole thing by 2 instead of T and integrate:
*s2+s
= s + s2
Here,
s2 =
s
= ,sin-
= sin+sin =
Instantaneous Power, = *s2+s
Average Power = s
= |||| sSo, the value for average power can be obtained by the multiplication of the
phasor amplitude by the modulus of voltage, current and cos.
Let, 2 = 2 = 0 2 2
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LECTURE16 Date: 26-10-2011
Stored Power for an Inductor:
= ||
= 2 sin2
= ,s2-
=
s s
= 2=
=
=
+
=
=
=
s
s
+
= 2 { sin2}
= sin= s +
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Stored Power for a Capacitor:
= =
ss+ =
2sin2
= || =
2 sin2
= ,s2- = 2
=
=
=
= s
= * sin=
s +
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LECTURE17 Date: 16-11-2011
Phasor Diagram:
In a series circuit the angle of the current is zero. So, it is taken as the reference
quantity. All the angles are considered with respect to the reference quantity.
VR=IR, so angle ofVRwill be zero
VL=I XL XL=XL= XL So, the angle of VLwill be +90with respect to I.
In the whole circuit
VL+ VR= VS
VL
VRI
VLVR+ + --
VS
= += tn
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Phasor Diagram:
At first a reference point has to be selected, which is parallel to the X axis. All the
angles are then calculated with respect to the X axis. Here, the current is selected
as the reference quantity and set in a direction long the X axis.
VS
R
L
VR= Voltage across the resistor
VL= voltage across the inductor
IXL
I VR
VL
=
+
= tn
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=
=
tn
=So, is at an angle of -90with respect to IR.
VR+ -
VC
+
-
R I
VC
VS
VR
Phasor diagram
VS
VC
VL+ - + -
VR1
VR2
R1
R2
XL
+-XC
-
+
-
VL
VC
VR2
VR1
Phasor Diagram
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Series Resonance:
|| = +
= . /In a series circuit the magnitude of Iis maximum at a certain frequency. This
frequency is called the resonance frequency.
Here, Ishould be differentiated with respect to frequency.
=
(
+ . /)=
= = = At resonance point,
XL= XC
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LECTURE18 Date: 19-11-2011
Wye-Delta Connection:
Balanced 3-phased Circuits:
Wye
Z1
Z2Z3
ZBC
a
ZA
a
b bcc
n
Van
Vbn
Vcn
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The voltage between a to b, b to c or c to a is called line to line voltage. The
voltage from the phase to neutral is said to be the line to neutral voltage.
Delta
Line to line voltage =
Zbc
aa
b bcc
Van
Vbn
Vcn
Vbc
Vca
Vab
b 2
c 24
Reference quantity