AC Circuit Analysis

Wednesday, 20 February 2019

5:17 PM

<<ELEC2134_Circuits2_ACCircuitAnalysis.pdf>>

Machine generated alternative text:
ELEC2134: Circuits and Signals
AC Circuit Analysis
PhasorRepresentations and Phasordiagrams
V-I relationship for resistors/capacitors/ Inductors
Impedances and reactances
Analysis of parallel RC/ RL circuits
Analysis of RLC circuits
Professor Eliathamby Ambikairajah

Machine generated alternative text:
PhasorRepresentation
A phasor is a line drawn from an origin that has both
magnitude and angle
Phasorsare used in AC circuits to indicate time
varying quantities of current and voltage
2
Prof E Ambikairajah, UNSW, Australia
Waveform
Vector representation
Phase angle = 0
o
Phase angle = 0
o
Phase angle = 0
o
Amplitude
V
m1
V
m2
V
m1
V
m2
V
m3
V
m3
Length
v(t)
v(
v(
v(
0
o
0
o
0
o

Machine generated alternative text:
PhasorRepresentation
A phasoris a rotating vector
3
Prof E Ambikairajah, UNSW, Australia
+
Notation:
-instantaneous voltage and
is the
corresponding phasor voltage
att=0
att=t
0
t
0
)
Reference
Rotationat
rad/sec
Phasor
representation
0
o
V
Phasor
representation
V

Machine generated alternative text:
4
Prof E Ambikairajah, UNSW, Australia
Phasor representation
Consider the following waveforms:
|OA| = |OB| = |OC|
Peak values of all three
waveforms are equal
0
o
Waveform
A
is chosen as the reference and is
shown to begin at 0
o
.
Waveform
B
leads waveform
A
by 45 degrees
Waveform
C
lags waveform
A
by 60 degrees

Machine generated alternative text:
Phasors
A phasoris a complex number that represents the amplitude
and phase angle of a sinusoid
Transforming a sinusoid to and from the time domain to the
phasordomain
5
Prof E Ambikairajah, UNSW, Australia
If OP is a vector of magnitude
and the
phase angle is
degrees, then P can be
represented in one of the following three
forms:
Rectangular:
cos(
) +
sin(
)
Polar:
Exponential:
cos(
) +
sin(
)
{Polar to rectangular}
domain
phasor
domain
time
)
cos(
)
(
m
m
V
t
V
t
v
V

Machine generated alternative text:
Example 2.1
6
Prof E Ambikairajah, UNSW, Australia
Determine the phasorscorresponding to the following sinusoids:
(a)
Therefore
10
o
(b)
v
(
t
) = 7 sin(2
t
+ 40
o
);
v
(
t
) = 7 cos(2
t
-90
o
+ 40
o
) [we know that sin
t
= cos (
t
-90
o
)]
Therefore
0
o
Find the sinusoids represented by the following phasors:
(a)
V
Since
j
0
o
,
0
o
=(
0
o
)(
0
o
)=
o
)
Converting this phasorto the time domain gives
V
(b)
M
= 5 and
= 126.87
o
o
.
Transforming this into time domain gives
i
A
Example 2.2

Machine generated alternative text:
Example 2.3
Two sine wave sources are shown in series across a load.
Find the total voltage
V
S
=
V
A
+
V
B
=
+
= 60
+ 30
=[60 cos60
o
+
j
60 sin60
o
] + [30 cos30
o
+
j
30 sin30
o
]
= 56 +
j
67
7
Prof E Ambikairajah, UNSW, Australia
R
L
ohms
V
B
=30V
30
o
V
A
=60V
60
o
V
S
o
0
o
V
A
=60V
V
B
=60V
Reference
Phasordiagram
cos(
) +
sin(
)
Convert to
polar form
Magnitude =
= 87.3 V
=
= 50.2
o
= 87.3V

Machine generated alternative text:
Alternatively,
can be calculated using the phasordiagram
for voltages.
=
+
= 60
+ 30
8
Prof E Ambikairajah, UNSW, Australia
R
L
ohms
V
B
=30V
30
o
V
A
=60V
60
o
V
S
=
+
-2 V
A
V
B
cos(150
o
)
Use cosine rule:
= 87.3V
can be calculated as 50.2
o
Exercise 2.1:
Given
i
1
(
t
)
=
4 cos(
+30
)A
and i
2
(
t
)
=
5 sin(
20
)A
,
find their sum
i
(
t
)
= i
1
(
t
)
+
i
2
(
t
)
.
Ans:
3.218
-57
A

Machine generated alternative text:
Phase difference
Consider two phasors given below and draw the phasor
diagram.
9
Prof E Ambikairajah, UNSW, Australia
Phase difference between the
current and the voltage is:
(
-(-
)) = (
+
)

Machine generated alternative text:
Inductor
10
Prof E Ambikairajah, UNSW, Australia
Capacitor
i
(
t
)
instantaneous current
q
(
t
) -instantaneous
charge on the capacitor
v
c
(
t
)
instantaneous voltage
across the capacitor
If
i
(
t
) is the current through the coil, the voltage
across the inductor
v
L
(
t
) is given by
The greater the rate of change of current through the
coil, the greater will be the induced voltage.
Induced voltage
Assume that the
initial charge in
the capacitor is
zero.
L
+v
L
(t)-
i(t)
i(t)
C
v
c
(t)
+
-

Machine generated alternative text:
Impedance of Circuit Elements
Resistor:
I
=
I
m
0
o
V
=
RI
m
0
o
V
=
RI
m
V
=
R
I
Inductor:
I
=
I
m
0
o
V
L
=
LI
m
90
o
V
L
=
j
LI
m
V
L
=
j
L
I
Capacitor:
V
C
=
V
m
0
o
I
=
CV
m
90
o
I
=
j
CV
m
I
The impedance (
Z
) of a
circuit element of an
AC circuit is defined to
be the ratio of the
voltage phasor to the
current phasor.
= R
=
j
L
=
The impedance (
Z
)of a
resistor is
R
, the
impedance of an
inductor is
j
L
and the
impedance of a
capacitor is 1/
j
C
Reference
R
C
Phasor relationships
for circuit elements

Machine generated alternative text:
15
Inductance and Capacitance
L
j
Z
C
j
Z
1
Z
Z
;
0
;
0
0
;
;
0
Z
Z

Machine generated alternative text:
Reactance
The imaginary part of the impedance is called reactance
Reactance is a magnitude (always positive)
As frequency decreases
the inductive reactance
decreases
As frequency increases
the capacitive
reactance decreases
= -
jX
C
=
jX
L
L
+v
L
(t)-
i(t)

Machine generated alternative text:
Impedance in an RC circuit
Total impedance in an RC circuit is composed of
resistance and reactance.
V
S
R
Z
=
=
R
jX
c
=
=
=
Impedance Triangle
R
X
c
A capacitive reactance of
k
is in
series with a resistance of 1.2 k
Calculate the impedance.
Example 2.4

Machine generated alternative text:
= 1.57 k
Impedance
= 2.2 +
j
1.57
Impedance in an RL circuit
Total impedance in an RL circuit is composed of
resistance and reactance.
=
R
+
jX
L
=
=
=
Impedance Triangle
For the circuit shown below, calculate the value of the
impedance.
R
X
L
Example 2.5

Machine generated alternative text:
A sine wave source of 10V
0
o
is across a 0.22
F capacitor. If the frequency
of the source is 2kHz, compute the current.
C
X
c
I
c
V
S
= 10V
0
o
X
c
=
=
Z
c
=
=
-90
o
=
=27.6 mA
90
o
I
c
leads the input voltage V
s
by 90
o
Example 2.6

Machine generated alternative text:
The circuit below has a phase angle of 36
o
. What frequency would cause
the phase angle to double to 72
o
?
tan
X
c
=
R
tan
= 3900(tan72
o
)= 12k
Compute the frequency that will produce 12.0
reactance.
f
=
=
= 1.326 kHz
V
S
R=3.9
C=0.01
I
V
R
=
I
R
I
Reference
V
S
Vc=
I
Example 2.8

Machine generated alternative text:
Analysis of parallel RC circuits
V
S
R
C
I
T
+
-
I
T
I
R
I
C
PhasorDiagram
In parallel circuits the supply voltage is taken as
the reference.
In series circuits the current is taken as the
reference.
From the phasordiagram, we can write the following equations:
|
tan
=

Machine generated alternative text:
Compute the total current for the following circuit
Method 1:
Z
= 2.45k
×
½
ã
ã
= 10.2 mA
90
o
×
Î
ã
ã
= 22.7 mA
0
o
×
Ì
ã
×

½
õ ×

Î
ã
10.2 mA
90
o
+ 22.7 mA
0
o
×
Ì
ã îîòé õ ¶ ïðòî ³ß
×
Ì
ã îìòç ³ß
Method 2:
Total impedance of the circuit is given
by
Z
T
=
R
||(1/
j
C
) or
R
||-
jX
C
=
=
=1.004
-24.2
o
×

Ì
ã
=
îìòç ³ß
C
I
T
+
-
I
T
I
R
I
C
V
s
=25V
0
o
R=1.1k
X
c
=2.45k
Z
T
Exercise 2.2:
For the circuit shown,
compute the value of the resistance (
R
).
I
T
+
-
I
T
= 280mA
I
R
I
C
V
s
=10V
0
o
R
X
c
=50
Ans: 51
Example 2.9: Parallel R-C circuit

Machine generated alternative text:
Analysis of parallel RL circuits
Using the PhasorDiagram
We can write the following equations:
tan
=
I
R
V
s
Reference
I
L
I
T
Total impedance of the circuit
is given by
Z
T
=
R
||(
jX
L
)
= |
×

Ì
ã
OR
×
Ô
ã
=
×
Î
ã
Using complex numbers
Exercise 2.3:
For the parallel
RL
circuit, calculate the total current.
I
T
+
-
I
T
I
R
I
L
V
s
=30V
0
o
R=
1200
X
L
=2400
Ans: 28mA

Machine generated alternative text:
Current and Voltage Division
Ans: 28mA
Ideal current
source
Current Division
Voltage Division
Ideal voltage
source

Machine generated alternative text:
Example 2.10
Determine the current
and
across the inductor.
Alternatively, by
voltage division:
L
L
L

Machine generated alternative text:
Example 2.11: RLC circuit [4]
Find the steady-state current for the circuit shown below. Find the phasorvoltage
across each element and construct a phasordiagram.
I
+
-
I

V
i
=100V
30
o
=500rad/sec
+
+
+
-
-
-
R=100
V
R
V
L
V
C
L=0.3H
C=40
F
Ê
ã
Z
L
=
j
L
=
j
(500) (0.3) =
j
150
Z
C
= (1/
j
C
) =
= -
j
50
The total impedance:
Z
T
=
R
+
Z
L
+
Z
C
=100+
j
150
ó
j
50 = 100 +
j
100
Z
T
=
The Current
I
×
ã
ã
= 0.707
-15
o
Ê
Î
ã
(0.707
-15
o
) = 70.7
-15
o
Ê
Ô
ã
= (150
90
o
) (0.707
-15
o
)=106.1
75
o
Ê
Ý
ã
= (-
j
/
C
)
= (50
-90
o
) (0.707
-15
o
)
= 35.4
-105
o
I
V
R
Reference
(0
O
)
|V
R
|= 70.7
V
C
|V
C
|= 35.4
V
L
|V
L
|= 106.1
V
S
|V
S
|= 100
-15
o
30
o
75
o
PhasorDiagram

Machine generated alternative text:
Example 2.12: Series and Parallel combinations of complex impedances [4]
Q2:
For the circuit shown below, Find the current through each element and construct
a phasordiagram showing the currents and the source voltage.
Ê
·
ã
and
= 1000 rad/sec
Z
L
=
j
L
=
j
(1000) (0.1) =
j
100
Z
C
= (1/
j
C
) =
= -
j
100
Z
RC
=
=
= 70.71
-45
o
Z
RC
=
Ê
Ý
ã
= 10
-180
o
Now we compute the current in each element
×
ã
ã
= 0.1414
-135
o
×
Î
ã
=
= 0.1
-180
o
×
Ý
ã
=
ã
0.1
-90
o
I
R
Reference
I
C
-135
o
I
V
L
PhasorDiagram
I
+
-
I
R
I
C
v
i
(
÷ ã ïð

øïððð
ó çð
±
÷
C=10
F
+
-
V
C
+
-
R=100
V
R
L=0.1H
+
-
V
L
V
i
V
i
=10V
-90
o
X
L
= j
L
I
+
-
+
-
V
C
+
-
V
L
Z
RC

Machine generated alternative text:
Exercise 2.4
Q1:
A circuit element is subjected to the current
Derive an expression for the voltage
waveform as well as its phasorif the element is (i) a 2k
resistance. (ii) a 0.5
F capacitance and (iii) a 10mH inductance.
Ans: (i) 20
120
o
, (ii) 30.9
30
o
, (iii) 0.0647
-150
o
Q2:
Draw the phasordiagram for the circuit shown in Figure 1.
Compute the following for the above circuit
:
(a) X
C
and Z
T
(total
impedance) (b)
I
R
(
c
)
I
C
(d)
V
C
Figure 1
Q3:
Draw the phasordiagram for the circuit shown in Figure 2.
Compute the following for the above circuit: (a) Z
T
(b)
I
R
(c )
I
C
(d)
I
T
Figure 2
Ans: (a) 397.89
-90
o
, 1076.25
-21.7
o
(b) 18.6
21.7
o
mA (c) 18.6
21.7
o
mA (d) 7.4
-68.3
o
V
Ans: (a) 3.69
-68.3
o
k
(b) 1.5
0
o
mA (c) 3.77
90
o
mA (d) 4.06
68.3
o
mA
V
in
=15V
0
o
f=400Hz
R=10k
I
R
C=0.1
µ
F
I
C
I
T
Z
T
V
in
=20V
0
o
f=400Hz
R=1k
I
R
C=1
µ
F
I
C
+
V
C
-

Machine generated alternative text:
Exercise 2.4
Q4:
For the
RL
circuit shown in Figure 3, compute the output
voltage and the angle of lag.
Q5:
For the
RL
circuit shown in Figure 4, compute the output
voltage and the angle of lead.
Figure 4
Figure 3
Ans: 13.29 V, 27.64
o
lag
Ans: 14.53 V, 54.5
o
lead
V
s
=25V
0
o
+
V
out
-
X
L
=200
R=280
V
in
=15V
0
o
f=2kHz
R=600
V
out
L=25mH

Machine generated alternative text:
Exercise 2.4
Q6:
For the circuit shown in Figure 5, compute the following
values (a)
X
L
and (b)
L
.
Figure 5
Q7:
The following two circuits are equivalent. Derive an expression
for each
R
P
and
C
P
in terms of
R
s
,
C
s
&
.
Ans: (a) 0.6 k
(b) 0.1 H
V
in
=18V
0
o
f=955Hz
R=1k
I
T
=35mA
L

Machine generated alternative text:
Exercise 2.5
Q1:
Find
Z
eq
for the circuit shown below:
Q2:
Consider the circuit given below. Let
V
i
= 120 V operating at
60 Hz. Find:
(a)
V
0
when
R
is maximum
(b)
V
0
when
R
is minimum
(c) the value of
R
that will produce a phase shift of 45
0
in
V
0
Ans: 1 +
j
0.5
Ans: (a) 53.89
63.31
V
(b) 100
33.55
V
(c) 25.4
Ê
±
50
200mH
õ
ó
0<R<100
Ê
·
õ
ó
ïó ¶

ïõ¶î

ïõ¶í

¶ë

Æ»¯

Machine generated alternative text:
[1] Alexander, C. K., & Sadiku
th
edition, McGraw Hill.
th
edition, Wiley & sons.
School of Electrical engineering and Telecommunications, UNSW,
Australia.
[4] Hambley
Publishing.
References

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