Electrical Circuits II: AC Circuits
Alternating current (AC) circuits are stable linear
circuits operating in the steady state with sinusoidal
excitation.
AC circuit analysis involves finding the steady state
response forced by a sinusoidal excitation after the
natural or transient response has elapsed.
Some Major Difference between AC and DC
Characteristic
DC
Electron flow The
electrons
flow steadily in
a
single
direction,
or
“forward”.
Amount of
Energy it can
carry
Each battery is
designed
to
produce
only
one voltage, and
that voltage of
DC cannot travel
very far until it
begins to lose
energy.
Motor
configuration
Requires carbon
brushes
or
commutators to
contact
the
shaft
as
it
rotates.
AC
The
electrons
keep switching
directions,
sometime going
“forward” and
then
going
“backward”.
AC voltage from
a generator in a
power plant can
be bumped up
or down in
strength
by
another
mechanism,
called
a
transformer.
Does
not
require carbon
brushes
or
commutators to
contact
the
shaft
as
it
rotates. This is
an
advantage
because
the
contact creates
both heat and
sparks,
especially
at
high
speeds,
which
could
cause problems
if the motor is
situated in an
environment
near flammable
or
explosive
vapors
Cause of the
direction of
flow of
electrons
Frequency
Current
Passive
Parameters
Types
Power Factor
Steady
magnetism
along the wire
Rotating magnet
along the wire
The frequency The frequency
of direct current of
alternating
is zero.
current is 50 Hz
or
60
Hz
depending upon
the country
Constant
Time – varying
Resistance
Impedance
Pure
pulsating
and Sinusoidal,
Trapezoidal,
Square
It is always 1
Lies between 0
and 1
No
Yes
Useful for all
purpose
GENERATION OF AC VOLTAGE
Electromagnetic – by means of rotating machines
(AC generator or alternator)
a. rotation of a coil within a stationary magnetic
field
b. rotation of a magnetic field within a stationary
coil
𝑒 = 𝐸𝑚 sin 2𝜋𝑓𝑡 = 2𝐵𝑁𝑙𝑣 sin 𝜃 = 2𝜋𝑓𝑁𝐵𝐴
where:
Em – the maximum voltage
B – flux density
l – length
v – speed
f – frequency
A – cross – sectional area
Electronic – by means of oscillator circuits in signal
generators whereby alternating current is produced
from a DC source (inverters)
An oscillator is a circuit that produces an ac
waveform as output when powered by a dc input.
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
GfApolinario
1
Electrical Circuits II: AC Circuits
For rotating machines, the generated or induced
voltage across the coil depends on the following:
1. the number of turns of the coil
2. the speed of rotation of the coil or the magnetic
field
3. the strength of the magnetic field
An alternating current or voltage is one the circuit
direction of which reverses at regularly recurring
intervals.
Nowadays 95% of the total energy is produced,
transmitted and distributed in AC supply.
The reasons are the following:
a. More voltage can be generated (up to 33 kV)
than DC (650 V only).
b. AC voltage can be increased and decreased with
the help of a static machine called the
‘transformer’.
c. AC transmission and distribution is more
economical as line material (say copper) can be
saved by transmitting power at high voltage.
d. AC motors for the same horse power as of DC
motors are cheaper, lighter in weight, requires
less space and require lesser attention in
operation and maintenance.
e. AC can be converted to DC easily, when and
where required but DC cannot be converted to
AC so easily and it will not be economical.
However, DC entails the following merits and hence
finds wide applications.
a. DC series motors are most suitable for traction
purposes in tramway, railways, cranes and lifts.
b. For
electroplating,
electrolytic
and
electrochemical processes (battery charging
etc.), DC is required.
c. Arc lamps for search lights and cinema
projectors work on DC.
d. Arc welding is better than on AC.
e. Relay and operating time switches, etc., and
circuit breakers, DC works more efficiently.
f.
In rolling mills, paper mills, colliery winding,
etc., where find speed control in both directions
is required, DC motors are used.
A sinusoid is a signal that has the form of the sine or
cosine function.
𝑣(𝑡) = 𝑉𝑚 sin 𝜔𝑡
where:
Vm – the amplitude of the sinusoid
𝜔 – the angular frequency in
radians
𝜔𝑡 – the argument of the sinusoid
Cycle is a complete change in value and direction of
an alternating quantity.
1 cycle = 360 electrical degrees = 2π radians
Periodic time (T) is the time taken to complete one
cycle.
2𝜋
𝑇 = 1⁄𝑓 =
𝑠𝑒𝑐𝑜𝑛𝑑𝑠
𝜔
A periodic function is one that satisfies f(t) = f(t+nT),
for all t and for all integers n.
Frequency (f) is the number of cycles per second. It
has a unit of Hertz (Hz).
1
𝑓=
𝑇
𝜔 = 2𝜋𝑓 𝑟𝑎𝑑/𝑠
Phase Difference is the angular difference or
angular
displacement
between
alternating
quantities. It is also called “phase angle”.
If two alternating quantities attains their maximum
and minimum values at the same time it is called in
phase and out of phase if it does not.
Instantaneous value is the value of alternating
quantity at any instant.
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
2
Electrical Circuits II: AC Circuits
Maximum value is the maximum value attained
during positive or negative half cycle.
Average value is the average of all instantaneous
values of half cycle.
𝐴𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑡ℎ𝑒 𝑐𝑢𝑟𝑣𝑒
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑎𝑙𝑢𝑒 = √
𝐵𝑎𝑠𝑒
2
= 𝑀𝑎𝑥 𝑣𝑎𝑙𝑢𝑒
𝜋
= 0.637 𝑀𝑎𝑥 𝑣𝑎𝑙𝑢𝑒 (𝑠𝑖𝑛𝑒 𝑤𝑎𝑣𝑒)
Effective value is that which when applied to a given
circuit for a given time produces the same
expenditure of energy as when dc is applied to the
same circuit for the same interval time. It is also
called “Root Mean Square” (RMS) value.
𝐴𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑡ℎ𝑒 𝑠𝑞𝑢𝑎𝑟𝑒 𝑐𝑢𝑟𝑣𝑒
𝑅𝑀𝑆 𝑣𝑎𝑙𝑢𝑒 = √
𝐵𝑎𝑠𝑒
1 𝑇
1
= √ ∫ 𝑥 2 𝑑𝑡 =
𝑀𝑎𝑥 𝑣𝑎𝑙𝑢𝑒
𝑇 0
√2
= 0.707 𝑀𝑎𝑥 𝑣𝑎𝑙𝑢𝑒 (𝑠𝑖𝑛𝑒 𝑤𝑎𝑣𝑒)
The effective value of a periodic current is the dc
current that delivers the same average power to a
resistor as the periodic current.
The effective value of a periodic signal is its roots
mean square (rms) value.
Form Factor
𝑅𝑀𝑆 𝑉𝑎𝑙𝑢𝑒
𝐹𝑜𝑟𝑚 𝐹𝑎𝑐𝑡𝑜𝑟 =
= 1.11 (𝑠𝑖𝑛𝑒 𝑤𝑎𝑣𝑒)
𝐴𝑣𝑒 𝑉𝑎𝑙𝑢𝑒
Peak Factor
𝑀𝑎𝑥 𝑉𝑎𝑙𝑢𝑒
𝑃𝑒𝑎𝑘 𝐹𝑎𝑐𝑡𝑜𝑟 =
= 1.414 (𝑠𝑖𝑛𝑒 𝑤𝑎𝑣𝑒)
𝑅𝑀𝑆 𝑉𝑎𝑙𝑢𝑒
Half – wave (HW) rectified alternating current is one
whose one half – cycle has been suppressed i.e. one
which flows for half the time during one cycle.
1 𝑇 𝑥2
1
𝑅𝑀𝑆 𝑉𝑎𝑙𝑢𝑒 = √ ∫
𝑑𝑡 = 𝑀𝑎𝑥 𝑣𝑎𝑙𝑢𝑒
𝑇 0 4
2
Half Wave
Rectified
Sine Wave
Full Wave
Rectified
Sine Wave
Rectangular
Wave
Triangular
Wave
0.5𝐼𝑚𝑎𝑥
= 1.57
0.318𝐼𝑚𝑎𝑥
𝐼𝑚𝑎𝑥
=2
0.5𝐼𝑚𝑎𝑥
0.707𝐼𝑚𝑎𝑥
= 1.11
0.637𝐼𝑚𝑎𝑥
𝐼𝑚𝑎𝑥
= 1.41
0.707𝐼𝑚𝑎𝑥
1
1
0.578𝐼𝑚𝑎𝑥
= 1.16
0.5𝐼𝑚𝑎𝑥
𝐼𝑚𝑎𝑥
= 1.73
0.578𝐼𝑚𝑎𝑥
Reasons for using alternating current (or voltage) of
sinusoidal form:
An alternating current (or voltage) of sinusoidal
form is normally used because of the following
reasons:
1. Mathematically, it is quite simple.
2. Its integrals and differentials both are
sinusoidal.
3. It lends itself to vector representation.
4. A complex waveform can be analyzed into a
series of sine waves of various frequencies, and
each such component can be dealt with
separately.
5. This waveform is desirable for power
generation, transmission and utilization.
Complex waves are those which depart from the
ideal sinusoidal form. All alternating complex
waves, which are periodic and have equal positive
and negative half cycles can be shown to be made
up of a number of pure sine waves, having different
frequencies but all these frequencies are integral
multiples of that of the lowest alternating wave,
called the fundamental (or first harmonic). These
waves of higher frequencies are called harmonics. If
the fundamental frequency is 50 Hz, then the
frequency of the second harmonic is 100 Hz and of
the third is 150 Hz and so on. The complex wave
may be composed of the fundamental wave (or first
harmonic) and any number of other harmonics.
𝐻𝑎𝑙𝑓 𝑎𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑡ℎ𝑒 𝑐𝑢𝑟𝑣𝑒
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑎𝑙𝑢𝑒 = √
𝐵𝑎𝑠𝑒
1
= 𝑀𝑎𝑥 𝑣𝑎𝑙𝑢𝑒
𝜋
Waveform
Form Factor
Peak Factor
Sine Wave
0.707𝐼𝑚𝑎𝑥
= 1.11
0.637𝐼𝑚𝑎𝑥
𝐼𝑚𝑎𝑥
= 1.41
0.707𝐼𝑚𝑎𝑥
The first figure shows a complex wave which is
made up of a fundamental sine wave of frequency
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
3
Electrical Circuits II: AC Circuits
of 50 Hz and third harmonic of frequency of 150 Hz.
It is seen that:
a. The two halves of the complex wave are
identical in shape. In other words, there is no
distortion. This is always the case when only
odd harmonic (3rd, 5th, 7th, 9th etc.) are
present.
b. Frequency of the complex wave is 50 Hz i.e. the
same as that of the fundamental sine wave.
The second figure shows a complex wave which is a
combination of fundamental sine wave of frequency
50 Hz and 2nd harmonic of frequency 100 Hz and
3rd harmonic of frequency 150 Hz. It is seen that
although the frequency of the complex wave even
now remains 50 Hz, yet:
a. The two halves of the complex wave are not
identical. It is always so when even harmonics
(2nd, 4th, 6th etc.) are present.
b. There is distortion and greater departure of the
wave shape from the purely sinusoidal shape.
Sometimes, a combination of an alternating and
direct current flows simultaneously through a
circuit. The figure shows a complex wave
(containing fundamental and third harmonic)
combined with a direct current of value ID. It is seen
that the resultant wave remains undistorted in
shape but is raised above the axis by an amount ID.
It is worth noting that with reference to the original
axis, the two halves of the combined wave are not
equal in area.
V and I Relationships in a Purely Resistive AC Circuit
In a purely resistive circuit, the voltage across a
resistor is in phase with the current through it. Both
the voltage and current pass through the zero
points and reach the maximum points of the same
polarity at the same time.
V and I Relationships in a Purely Inductive AC Circuit
In a sine – wave ac circuit environment, voltage
across an inductor leads the current through the
inductor by 90° (in a pure inductance).
RL Circuits in AC
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
4
Electrical Circuits II: AC Circuits
V and I Relationships in a Purely Capacitive AC
Circuit
In a sine – wave ac circuit environment, current
across a capacitor leads the voltage through the
capacitor by 90° (in a pure capacitance).
RC Circuits in AC
A phasor is a complex number that represents the
amplitude and phase of a sinusoid.
𝑣(𝑡) = 𝑉𝑚 cos(𝜔𝑡 + 𝜙) ⇔ 𝑉 = 𝑉𝑚 ∠𝜙
(Time Domain)
(Phasor Domain)
Sinusoid – phasor transformation
Time – domain
Phasor – domain
𝑉𝑚 cos(𝜔𝑡 + 𝜙)
𝑉𝑚 ∠𝜙
𝑉𝑚 sin(𝜔𝑡 + 𝜙)
𝑉𝑚 ∠𝜙 − 90
𝐼𝑚 cos(𝜔𝑡 + 𝜙)
𝐼𝑚 ∠𝜙
𝐼𝑚 sin(𝜔𝑡 + 𝜙)
𝐼𝑚 ∠𝜙 − 90
Graphical representation of phasors is known as a
phasor diagram.
Phasor Relationships for Circuit Elements
Element
Time domain
Frequency
domain
R
𝑣 = 𝑅𝑖
𝑉 = 𝑅𝐼
𝑑𝑖
L
𝑉 = 𝑗𝜔𝐿𝐼
𝑣=𝐿
𝑑𝑡
𝐼
𝑑𝑣
C
𝑉=
𝑣=𝐶
𝑗𝜔𝐶
𝑑𝑡
Impedance (Z) of a circuit is the ratio of the phasor
voltage V to the phasor current I measured in ohms
(Ω).
𝑉
𝑍=
𝑜𝑟 𝑉 = 𝑍𝐼
𝐼
𝑍 = 𝑅 + 𝑗𝑋
Admittance (Y) is the reciprocal of impedance,
measured in siemens (S) or mhos (℧).
1 𝐼
𝑌= =
𝑍 𝑉
𝑌 = 𝐺 + 𝑗𝐵
Element
Impedance
Admittance
1
R
𝑍=𝑅
𝑌=
𝑅
1
L
𝑍 = 𝑗𝜔𝐿
𝑌=
𝑗𝜔𝐿
1
C
𝑌 = 𝑗𝜔𝐶
𝑍=
𝑗𝜔𝐶
Kirchoff’s Law in the Frequency Domain
Basic circuit laws (Ohm’s and Kirchhoff’s) apply to
AC circuits in the same manner as they do for DC
circuits; that is,
𝑉 = 𝑍𝐼
∑ 𝐼𝑘 = 0 (𝐾𝐶𝐿)
∑ 𝑉𝑘 = 0 (𝐾𝑉𝐿)
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
5
Electrical Circuits II: AC Circuits
NOTE: Convert first all parameters such as Voltage,
Current & Impedances in time domain to frequency
domain before applying the following laws.
Impedance Combinations
The equivalent impedance of any number of
impedances connected in series is the sum of the
individual impedances.
𝑴
𝒁𝒆𝒒 = 𝒁𝟏 + 𝒁𝟐 + ⋯ + 𝒁𝑵 = ∑ 𝒁𝒏
𝒏=𝟏
In terms of admittance, the equivalent admittance
of any number of impedances in series is
𝟏
𝟏
𝟏
𝟏
=
+
+ ⋯+
𝒀𝒆𝒒
𝒀𝟏 𝒀𝟐
𝒀𝑵
The main characteristics of a series circuit are:
1. Same current flows through all parts of the
circuit.
2. Different impedances have their individual
voltage drops.
3. Voltage drops are additive.
4. Applied voltage equals the sum of different
voltage drops.
5. Impedances are additive.
6. Powers are additive.
The equivalent impedance of any number of
impedances connected in parallel is
𝟏
𝟏
𝟏
𝟏
=
+
+⋯+
𝒁𝒆𝒒
𝒁𝟏 𝒁𝟐
𝒁𝑵
The equivalent admittance of any number of
impedances connected in parallel is the sum of the
individual admittance.
𝑴
𝒀𝒆𝒒 = 𝒀𝟏 + 𝒀𝟐 + ⋯ + 𝒀𝑵 = ∑ 𝒀𝒏
𝒏=𝟏
The main characteristics of a parallel circuit are:
1. Same voltage acts across all parts of the
circuit
2. Different impedances have their individual
current.
3. Branch currents are additive.
4. Admittance are additive.
5. Powers are additive.
Voltage Divider Rule
𝒁𝒏
𝑽𝒏 = 𝑽
𝒁𝟏 + 𝒁𝟐 + ⋯ + 𝒁𝑵
Current Divider Rule
𝒁𝟐
𝒁𝟏
𝑰𝟏 =
𝑰 𝑰𝟐 =
𝑰
𝒁𝟏 + 𝒁𝟐
𝒁𝟏 + 𝒁𝟐
Delta to Wye Conversion
1
3
Zc
1
2
3
Zc
Zb
Za
2
4
4
NOTE: Each impedance in the Wye Network is the
product of the impedances in the two adjacent
delta branches, divided by the sum of the 3 delta
impedances.
𝒁𝒃 𝒁𝒄
𝒁𝟏 =
𝒁𝒂 + 𝒁𝒃 + 𝒁𝒄
𝒁𝒂 𝒁𝒄
𝒁𝟐 =
𝒁𝒂 + 𝒁𝒃 + 𝒁𝒄
𝒁𝒃 𝒁𝒂
𝒁𝟑 =
𝒁𝒂 + 𝒁𝒃 + 𝒁𝒄
Wye to Delta Conversion
a
b
a
n
n
Z1
Z2
b
Z3
Z3
c
c
NOTE: Each impedance in the Delta Network is the
sum of all possible products of wye impedances
taken two at a time, divided by the opposite wye
impedance.
𝒁𝟏 𝒁𝟐 + 𝒁𝟐 𝒁𝟑 + 𝒁𝟑 𝒁𝟏
𝒁𝒂 =
𝒁𝟏
𝒁𝟏 𝒁𝟐 + 𝒁𝟐 𝒁𝟑 + 𝒁𝟑 𝒁𝟏
𝒁𝒃 =
𝒁𝟐
𝒁𝟏 𝒁𝟐 + 𝒁𝟐 𝒁𝟑 + 𝒁𝟑 𝒁𝟏
𝒁𝒄 =
𝒁𝟑
The Wye and Delta networks are said to be
balanced when
𝒁𝟏 = 𝒁𝟐 = 𝒁𝟑 = 𝒁𝒀
𝒁𝒂 = 𝒁𝒃 = 𝒁𝒄 = 𝒁∆
thus,
𝒁∆
𝒁𝒀 =
𝒐𝒓 𝒁∆ = 𝟑𝒁𝒀
𝟑
EXAMPLES:
1. A square coil of 10 cm side and 100 turns is
rotated at a uniform speed of 1000 revolutions
per minute, about an axis at right angles to a
uniform magnetic field of 0.5 Wb/m2. Calculate
the instantaneous value of the induced
electromotive force, when the plane of the coil
is
a. at right angles to the field
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
6
Electrical Circuits II: AC Circuits
b. in the plane of the field
2. In a linear circuit, the voltage source is
𝑣𝑠 = 12 sin(103 𝑡 + 24°) 𝑉
a. What is the angular frequency of the
voltage?
b. What is the frequency of the source?
c. Find the period of the voltage.
d. Express vs in cosine form.
e. Determine vs at t = 2.5 ms.
3. A current source in a linear circuit has
𝑖𝑠 = 8 cos(500𝜋𝑡 − 25°) 𝐴
a. What is the amplitude of the current?
b. What is the angular frequency?
c. Find the frequency of the current.
d. Calculate is at t = 2 ms.
4. Given 𝑣1 = 20 sin(𝜔𝑡 + 60°) and 𝑣2 =
60 cos(𝜔𝑡 − 10°), determine the phase angle
between the two sinusoids and which one lags
the other.
5. Transform the following sinusoids to phasors:
a. −10 cos(4𝑡 + 75°)
b. 5 sin(20𝑡 − 10°)
c. 4 cos 2𝑡 + 3 sin 2𝑡
6. Using phasors, find:
a. 3 cos(20𝑡 + 10°) − 5 cos(20𝑡 − 30°)
b. 40 sin 50𝑡 + 30 cos(50𝑡 − 45°)
c. 20 sin 400𝑡 + 10 cos(400𝑡 + 60°) −
5 sin(400𝑡 − 20°)
7. Find a single sinusoid corresponding to each of
these phasors:
a. 𝑉 = 40∠ − 60° V
b. 𝑉 = −30∠10° + 50∠60° V
c. 𝐼 = 𝑗6𝑒 −𝑗10° A
2
d. 𝐼 = + 10∠ − 45° A
𝑗
8. A parallel RLC circuit has the node equation
𝑑𝑣
+ 50𝑣 + 100 ∫ 𝑣 𝑑𝑡 = 110 cos(377𝑡 − 10°)
𝑑𝑡
Determine v(t) using the phasor method. You
may assume that the value of the integral at t =
-∞ is zero.
9. Determine the current that flows through an 8
Ω resistor connected to voltage source 𝑣𝑠 =
110 cos 377𝑡 V.
10. The voltage across a 4 mH inductor is 𝑣 =
60 cos(500𝑡 − 65°) V. Find the instantaneous
current through it.
11. A current source 𝑖(𝑡) = 10 sin(377𝑡 + 30°) 𝐴
is applied to a single – element load. The
resulting voltage across the element is 𝑣(𝑡) =
−65 cos(377𝑡 + 120°) 𝑉. What type of
element is this? Calculate its value.
12. Find ix when is = 2 sin 5t A is supplied to the
circuit.
ix
2Ω
is
1H
0.2F
13. A series RL circuit is connected to a 110 V ac
source. If the voltage across the resistor is 85 V,
find the voltage across the inductor.
14. In the circuit below, find io when:
a. ω = 1 rad/s
1H
b. ω = 5 rad/s
c. ω = 10 rad/s
4 cos ωt V
2Ω
0.05 F
15. Find current Io in the network below.
j4Ω
IO
2Ω
5˂0° A
-j2Ω
-j2Ω
2Ω
16. Find the input impedance for the circuit below
given that ω = 10 rad/s.
Zin
2Ω
3F
3H
1.5 F
17. A capacitor in series with a 66 Ω resistor is
connected to a 120 V, 60 Hz source. If the
impedance of the circuit is 116 Ω, determine
the size of the capacitor.
18. A coil with impedance 8+j6 Ω is connected in
series with a capacitive reactance X. The series
combination is connected in parallel with a
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
7
Electrical Circuits II: AC Circuits
resistor R. Given that the equivalent impedance
of the resulting circuit is 5∠0° Ω, find the value
of X and R.
19. Calculate Zeq for the circuit below.
6Ω
2Ω
1Ω
j4 Ω
Zeq
20 μF
b.
40 Ω
Yin
30 Ω
10 mH
24. Determine Yeq for the circuit below.
-j2 Ω
5
Yeq
3
-j4
20. Find Zeq in the circuit below.
-j2
j1
Zeq
1 - j1 Ω
1+j3 Ω
j5 Ω
25. Determine the equivalent impedance of the
circuit below.
-j4
1+j2 Ω
-j6
21. For the circuit below, find the value of ZT.
8Ω
-j12 Ω
-j16 Ω
20 Ω
10 Ω
ZT
a
10 Ω
j15 Ω
10 Ω
22. Determine I and ZT for the circuit below.
-j6 Ω
I
4Ω
j4 Ω
2Ω
3Ω
120˂10° V
ZT
23. At ω = 103 rad/s, find the input admittance of
each of the circuits.
a.
60 Ω
Yin
60 Ω
20 mH
12.5 μF
j6
2
j8
j8
4
j12
b
26. A transmission line has a series impedance of
𝑍 = 100∠75° Ω and a shunt admittance of 𝑌 =
450∠48° 𝜇𝑆. Find:
a. The characteristic impedance 𝑍𝑜 = √𝑍⁄𝑌
b. The propogation constant 𝛾 = √𝑍𝑌
Steps to Analyze AC Circuits:
1. Transform the circuit to the phasor or
frequency domain.
2. Solve the problem using circuit techniques
(nodal analysis, mesh analysis, superposition,
etc.).
3. Transform the resulting phasor to the time
domain.
Methods of Circuit Analysis
a. MESH ANALYSIS or MAXWELL’S LOOP
CURRENT METHOD – provides another general
procedure for analyzing circuits using mesh
currents as the circuit variables. It is a method
that is used to solve planar circuits for the
voltage and currents at any place in the circuit.
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
8
Electrical Circuits II: AC Circuits
Mesh analysis uses Kirchhoff’s voltage law to
solve these planar circuits. The advantage of
using mesh analysis is that it creates a
systematic approach to solving planar circuits
and reduces the number of equations needed
to solve the circuit for all of the voltages and
currents.
Planar circuits are circuits that can be drawn on a
plane with no wires overlapping each other.
A mesh is a loop which does not contain any other
loops within it.
Steps:
1. Assign mesh currents to n meshes.
2. Apply KVL to each of the n meshes. Use Ohm’s
Law to express the voltages in terms of the
mesh currents.
3. Solve the resulting n simultaneous equations to
get the mesh currents.
MESH ANALYSIS WITH CURRENT SOURCES
 When a current source exists only in one mesh
we set the equation of the current equal to the
current source and the other in the usual way.
 When a current source exists between two
meshes, we create a supermesh by excluding
the current source and any elements connected
in series with it and apply KVL and KCL to it.
A supermesh results when two meshes have a
(dependent or independent) current source in
common.
EXAMPLES:
1. Solve for io using mesh analysis.
3. By using mesh analysis, find I1 and I2 in the
circuit depicted.
j4 Ω
3Ω
I1
10 cos 2tV
- j6 Ω
30˂20° V
4. Compute VO in the circuit below using mesh
analysis.
j4 Ω
- j13 Ω
+
4˂90° A
2Ω
VO12˂0° V
-
2Ω
2Ω
2˂0° A
5. Find I1, I2, I3 and IX in the circuit below.
20 Ω
- j15 Ω
iO
12˂64° V
I3
j16 Ω
IX
20 Ω
I1
I2
8Ω
0.25 F 6 sin 2t V
2. Using mesh analysis, find I1 and I2 in the circuit
below.
j10 Ω
40 Ω
I1
40˂30° V
j2 Ω
j1 Ω
2H
4Ω
2 Ω I2
3Ω
- j20 Ω 50˂0° V
I2
b. NODAL ANALYSIS or BRANCH CURRENT
METHOD – provides a general procedure for
analysing circuits using node voltages as circuit
variables. The node-equation method is based
directly on Kirchhoff’s current law unlike loopcurrent method which is based on Kirchhoff’s
voltage law. However, like loop current method,
nodal method also has the advantage that a
minimum number of equations need be written
to determine the unknown quantities.
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
9
Electrical Circuits II: AC Circuits
Steps:
1. Select a node as the reference node. Assign
voltages to the remaining n – 1 nodes. The
voltages are referenced with respect to the
reference node.
2. Apply KCL to each of the n – 1 non – reference
nodes. Use Ohm’s Law to express the branch
currents in terms of node voltages.
3. Solve the resulting simultaneous equations to
obtain the unknown node voltages.
NOTE: Current flows from a higher potential to a
lower potential in a resistor.
𝑣ℎ𝑖𝑔ℎ𝑒𝑟 − 𝑣𝑙𝑜𝑤𝑒𝑟
𝑖=
𝑅
NODAL ANALYSIS WITH VOLTAGE SOURCES
 If a voltage source is connected between the
reference node and a non – reference node, we
simply set the voltage at the non – reference
node equal to the voltage of the voltage source.
 If the voltage source (dependent or
independent) is connected between two non –
reference nodes, the two non – reference
nodes form a generalized node or super node;
we apply both KCL and KVL to determine the
node voltages.
A super node is formed by enclosing a (dependent
or independent) voltage source connected between
two non – reference nodes and any elements
connected in parallel with it.
Properties of super node:
1. The voltage source inside the super node
provides a constraint equation needed to solve
for the node voltages.
2. A super node has no voltage if its own.
3. A super node requires the application of KCL
and KVL.
EXAMPLES:
1. Find Vo in the circuit below.
2. Determine Vx below.
j10 Ω
20 Ω
+
-
4VX
3˂0° A
+
VX
-
20 Ω
3. Use nodal analysis to find V in the circuit below.
j20 Ω
V
40 Ω
120˂15° V
-j30 Ω
6˂30° A
50 Ω
4. Use nodal analysis to find Vo in the circuit
below.
10 mH
40 Ω iO
50 μF
4iO
10 cos 10000t V
20 Ω
+
VO
-
30 Ω
5. Using nodal analysis, find io(t) in the circuit
below.
2H
1H
0.25 F
2Ω
iO
8 sin(2t+30°) V
cos 2t A
0.5 F
1H
3Ω
10 cos(t - 45° )V
+
VO
-
1 F 5 sin(t + 30°) V
6. For each of the circuits below, find Vo/Vi for ω =
0, ω→∞ and ω2 = 1/LC.
C
L
+
R
Vi
a
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
+
+
C VO
Vi
-
-
+
R
L VO
b
10
Electrical Circuits II: AC Circuits
7. Determine Vx in the circuit using any method of
your choice.
-j2 Ω
j6 Ω
8Ω
40 ˂30° V
3Ω
+
VX
-
10 Ω
5˂0° A
8. By nodal analysis, obtain current Io in the circuit.
2iO
iO
10Ω
20 sin1000t A
50μF
20Ω
10mH
9. By nodal analysis, obtain current Io in the circuit.
j4 Ω
1Ω
iO
100˂20° V
2Ω
-j2 Ω
30Ω
10. Obtain Vo in the circuit below using nodal
analysis.
j2 Ω
2Ω
+ 12˂0° V
VO
-
4Ω
-j4 Ω
0.2vO
𝑘𝐼𝑍 = 𝑘𝑉
Homogeneity property
𝑣 = (𝐼1 + 𝐼2 )𝑍 = 𝐼1 𝑍 + 𝐼2 𝑍 = 𝑉1 + 𝑉2
Additive property
NOTE: A linear circuit is one whose output is linearly
related (or directly proportional) to its input.
b. SUPERPOSITION - this principles states that the
voltage across (or current through) an element
in a linear circuit is the algebraic sum of the
voltage across (or current through) that
element due to each independent source acting
alone.
To apply the superposition principle, we must keep
two things in mind:
 We consider one independent source at a time
while all other independent sources are turned
off. This implies that we replace every voltage
source by 0 V (or a short circuit), and every
current source by 0 A (or an open circuit). This
way we obtain a simpler and more manageable
circuit.
 Dependent sources are left intact because they
are controlled by circuit variables.
Steps to Apply Superposition Principle:
1. Turn off all independent sources except one
source. Find the output (voltage or current) due
to that active source using other techniques.
2. Repeat step 1 for each of other independent
sources.
3. Find the total contribution by adding
algebraically all the contributions due to the
independent sources.
NOTE: This theorem becomes important if the
circuit has sources operating at different
frequencies. In this case, since the impedances
depend on frequency, we must have a different
frequency domain circuit for each circuit. The total
response must be obtained by adding the individual
responses in the time domain.
EXAMPLES:
1. Find iO in the circuit using superposition.
4Ω
Circuit Theorems
a. LINEARITY PROPERTY – is the property of an
element describing a linear relationship
between cause and effect.
10 cos 4t V
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
iO 2Ω
1H
8V
11
Electrical Circuits II: AC Circuits
2. Using the superposition principle, find iX in the
circuit below.
iX
1/8 F
3Ω
5cos(2t+10°) A
4H
3. Determine iO in the
superposition principle.
10 cos(2t-60°) V
circuit
24 V
1/6 F
using
the
2H
iO
1Ω
10 sin (t-30°) V
2Ω
4Ω
2cos3t
4. Solve for vO(t) in the circuit using the
superposition principle.
2H
6Ω
1/12 F
12 cos 3t V
+
VO
-
4 sin 2t A
10 V
c. SOURCE TRANSFORMATION or SOURCE
CONVERSION – is the process of replacing a
voltage source in series with an impedance by a
current source in parallel with an impedance, or
vice versa.
EXAMPLES:
1. Using source transformation, find i in the circuit
below.
i
3Ω
20Ω
5mH
8 sin(200 t+30°) A
1mF
2. Use the method of source transformation to
find IX in the circuit below.
j4 Ω
2Ω
60˂0°V
-j2 Ω
4Ω
6Ω
IX
5˂90°A
-j3 Ω
d. THEVENIN’S THEOREM – states that a linear
two – terminal circuit can be replaced by an
equivalent circuit consisting of a voltage source
VTH in series with an impedance ZTH, where VTH is
the open – circuit voltage at the terminals and
ZTH is the input or equivalent impedance at the
terminals when the independent sources are
turned off.
How to Thevenize a Circuit?
1. Temporarily remove the impedance (called load
impedance ZL) whose current is required.
2. Find the open-circuit voltage VOC which appears
across the two terminals from where resistance
has been removed. It is also called Thevenin
Voltage, VTH.
3. Compute the impedance of the whose network
as looked into from these two terminals after all
voltage sources have been removed leaving
behind their internal resistances (if any) and
current sources have been replaced by opencircuit i.e. infinite resistance. It is also called
Thevenin impedance ZTH.
4. Replace the entire network by a single Thevenin
source, whose voltage is VTH or VOC and whose
internal impedance is ZTH.
5. Connect ZL back to its terminals from where it
was previously removed.
6. Finally, calculate the current flowing through ZL
by using the equation,
𝑉𝑇𝐻
𝐼=
𝑍𝑇𝐻 + 𝑍𝐿
To apply the idea in finding the Thevenin
impedance ZTH, we need to consider two cases:
CASE 1: If the network has no dependent source, we
turn off all independent sources. ZTH is the input
impedance of the network looking between the
terminals.
CASE 2: If the network has dependent source, we
turn off all independent sources. As with
superposition, dependent sources are not to be
turned off because they are controlled by circuit
variables. We apply a voltage source VO at terminals
and determine the resulting current iO. Then ZTH is
equal to vO / iO. Alternatively, we may insert a
current source iO at the terminals and find the
terminal voltage vO. Again, ZTH = vO / iO.
𝑉𝑇𝐻
𝑍𝐿
𝐼=
𝑉𝐿 = 𝑍𝐿 𝐼𝐿 =
𝑉
𝑍𝑇𝐻 + 𝑍𝐿
𝑍𝐿 + 𝑍𝑇𝐻 𝑇𝐻
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
12
Electrical Circuits II: AC Circuits
e. NORTON’S THEOREM – states that a linear two
– terminal circuit can be replaced by an
equivalent circuit consisting of a current source
IN parallel with an impedance ZN, where IN is the
short – circuit current through terminals and ZN
is the input or equivalent impedance at the
terminals when the independent sources are
turned off.
𝑉𝑇𝐻
𝑍𝑁 = 𝑍𝑇𝐻
𝐼𝑁 =
𝑍𝑇𝐻
𝑣𝑂𝐶
𝑉𝑇𝐻 = 𝑣𝑂𝐶 𝐼𝑁 = 𝑖𝑆𝐶 𝑍𝑇𝐻 =
= 𝑍𝑁
𝑖𝑆𝐶
How to Nortonize a Circuit?
1. Remove the impedance (if any) across the two
given terminals and put a short – circuit across
them.
2. Compute the short – circuit current ISC.
3. Remove all voltage sources but retain their
internal resistances, if any. Similarly, remove all
current sources and replace them by opencircuits i.e. by infinite resistance.
4. Next, find the impedance ZN of the network as
looked into from the given terminals. It is
exactly the same as ZTH.
5. The current source (ISC) joined in parallel across
ZN between the two terminals gives Norton’s
equivalent circuit.
NOTE: If the circuit has sources operating at
different frequencies, the Thevenin or Norton
equivalent circuit must be determined at each
frequency.
EXAMPLES:
1. Find the Thevenin and Norton equivalent
circuits at terminals a – b for each of the circuits
2. Find the Thevenin and Norton equivalent
circuits of the circuit below.
5Ω
2Ω
-j10 Ω
j20 Ω
60˂120° V
3. Obtain the Norton equivalent of the circuit
depicted below at terminals a – b.
5 μF
a
4cos(200 t+30°) A
10 H
2 kΩ
b
4. Compute iO below using Norton’s Theorem.
2Ω
¼F
iO 5 cos 2t V
½F
4H
5. Find the Thevenin and Norton equivalent
circuits at terminals a – b in the circuit below.
-j5 Ω
4Ω
4Ω
a
b
10 cos 4t V
j6 Ω
4Ω
8Ω
j20 Ω
2Ω
50˂30° V
a
-j10 Ω
b
(a)
a
-j5 Ω
4˂0° A
j10 Ω
8Ω
b
AC Power Analysis
The instantaneous power (in watts) is the power at
any instant of time.
𝑝(𝑡) = 𝑣(𝑡)𝑖(𝑡)
𝑝(𝑡) = 𝑉𝑚 𝐼𝑚 cos(𝜔𝑡 + 𝜃𝑣 ) cos(𝜔𝑡 + 𝜃𝑖 )
1
𝑝(𝑡) = 𝑉𝑚 𝐼𝑚 cos(𝜃𝑣 − 𝜃𝑖 )
2
1
+ 𝑉𝑚 𝐼𝑚 cos(2𝜔𝑡 + 𝜃𝑣 + 𝜃𝑖 )
2
The average power, in watts, is the average of the
instantaneous power over one period.
(b)
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
13
Electrical Circuits II: AC Circuits
1 𝑇
∫ 𝑝(𝑡)𝑑𝑡
𝑇 0
1
1
𝑃 = 𝑅𝑒[𝑉𝐼 ∗ ] = 𝑉𝑚 𝐼𝑚 cos(𝜃𝑣 − 𝜃𝑖 )
2
2
A resistive load (R) absorbs power at all times, while
a reactive load (L or C) absorbs zero average power.
Effective Resistance
Up to now, it is assume that resistance is constant,
independent of frequency. However, this is not
entirely true. For a number of reasons, the
resistance of a circuit to ac is greater than its
resistance to dc. While this effect is small at low
frequencies, it is very pronounced at high
frequencies. AC resistance is known as effective
resistance.
Before looking at why ac resistance is greater than
dc resistance, it is required to reexamine the
concept of resistance itself. Recall from Chapter 3
that resistance was originally defined as opposition
to current, that is, 𝑅 = 𝑉/𝐼 (This is ohmic
resistance.) Building on this, the power is defined
as 𝑃 = 𝐼 2 𝑅. It is this latter viewpoint that allows us
to give meaning to ac resistance. That is, ac or
effective resistance is define as
𝑃
𝑅𝑒𝑓𝑓 = 2 (Ω)
𝐼
where P is dissipated power (as determined by a
wattmeter). From this, it can be seen that anything
that affects dissipated power affects resistance. For
dc and low – frequency ac, both definitions for R,
i.e., 𝑅 = 𝑉/𝐼 and 𝑅 = 𝑃⁄ 2 yield the same value.
𝐼
However, as frequency increases, other factors
cause an increase in resistance. These are Eddy
Currents and Hysteresis, Skin Effect, and Radiation
Losses.
Eddy Currents and Hysteresis
The magnetic field surrounding a coil or other
circuit carrying ac current varies with time and thus
induces voltages in nearby conductive material such
as metal equipment cabinets, transformer cores,
and so on. The resulting currents (called eddy
currents because they flow in circular patterns like
eddies in a brook) are unwanted and create power
losses called eddy current losses. Since additional
power must be supplied to make up for these
losses, the power increases, increasing the effective
resistance of the coil.
𝑃=
If ferromagnetic material is also present, an
additional power loss occurs due to hysteresis
effects caused by the magnetic field alternately
magnetizing the material in one direction, then the
other. Hysteresis and eddy current losses are
important even at low frequencies, such as the 60
Hz power system frequency.
Skin Effect
Magnetically induced voltages created inside a
conductor by its own changing magnetic field force
electrons to the periphery of the conductor,
resulting in a non – uniform distribution of current,
with current density greatest near the periphery
and smallest in the center. This phenomenon is
known as skin effect. Because the center of the wire
carries little current, its cross – sectional area has
effectively been reduced, thus increasing resistance.
While skin effect is generally negligible at power
line frequencies (except for conductors larger than
several hundred thousand circular mils), it is so
pronounced at microwave frequencies that the
center of a wire carries almost no current. For this
reason, hollow conductors are often used instead of
solid wires.
Radiation Resistance
At high frequencies some of the energy supplied to
a circuit escapes as radiated energy. For example, a
radio transmitter supplies power to an antenna,
where it is converted into radio waves and radiated
into space. The resistance effect here is known as
radiation resistance. This resistance is much higher
than simple dc resistance. For example, a TV
transmitting antenna may have a resistance of a
fraction of an ohm to dc but several hundred ohms
effective resistance at its operating frequency.
Maximum Average Power Transfer
For maximum average power transfer, the load
impedance ZL must be equal to the complex
conjugate of the Thevenin impedance ZTH.
𝑍𝐿 = 𝑅𝐿 + 𝑗𝑋𝐿 = 𝑅𝑇𝐻 − 𝑗𝑋𝑇𝐻 = 𝑍𝑇𝐻 ∗
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
14
Electrical Circuits II: AC Circuits
𝑃𝑚𝑎𝑥 =
|𝑉𝑇𝐻 |2
8𝑅𝑇𝐻
Power Formula:
1. True, Real or Active Power (P)
𝑃 = 𝐸𝐼 cos 𝜃 = 𝐼 2 𝑅 watts
2. Reactive or Idle Power (Q)
𝑄 = 𝐸𝐼 sin 𝜃 = 𝐼 2 𝑋vars
where: + capacitive (leading) vars
− inductive (lagging) vars
NOTE: based from voltage conjugate method
3. Apparent Power (S) is the product of the RMS
value of voltage and current.
𝑆 = 𝑉𝑟𝑚𝑠 𝐼𝑟𝑚𝑠 = 𝐸𝐼 = 𝐼 2 𝑍 = √𝑃2 + 𝑄 2 VA (volt
– ampere)
where: θ – angle between E and I
4. Power factor (pf) is the cosine of the phase
difference between voltage and current. It is
also the cosine of the angle of the load
impedance.
𝑃
𝑝𝑓 = = cos(𝜃𝑣 − 𝜃𝑖 )
𝑆
Complex Power (in VA) is the product of the rms
voltage phasor and complex conjugate of the rms
current phasor. As a complex quantity, its real part
is real power P and its imaginary part is reactive
power Q.
1
𝑆 = 𝑃 + 𝑗𝑄 = 𝑉𝐼 ∗ = 𝑉𝑟𝑚𝑠 𝐼𝑟𝑚𝑠 ∠𝜃𝑣 − 𝜃𝑖
2
Determination of Power (In Complex Form)
1. Component Method:
Let 𝐸 = 𝐸1 + 𝑗𝐸2
𝐼 = 𝐼1 + 𝑗𝐼2
𝑃 = 𝐸𝐼
𝑃 = 𝐸1 𝐼1 + 𝐸2 𝐼2
2. Conjugate Method:
Let E*,I*= conjugate of E and I respectively
a. Voltage Conjugate Method
𝑆 = 𝐸 ∗ 𝐼 with respect to the horizontal axis
𝑆 = 𝑃 + 𝑗𝑄 = √𝑃2 + 𝑄 2 ∠ ± 𝜃
where: + capacitive(leading) vars
− inductive (lagging) vars
b. Current Conjugate Method
𝑆 = 𝐸𝐼 ∗ with respect to the horizontal axis
𝑆 = 𝑃 + 𝑗𝑄 = √𝑃2 + 𝑄 2 ∠ ± 𝜃
where:
+
inductive(lagging) vars
− capacitive (leading) vars
POWER in terms of MAGNITUDE
𝑃 = 𝐼 2 𝑅 = 𝐸𝐼𝑝𝑓
𝑄 = 𝐼 2 𝑋 = 𝐸𝐼𝑟𝑓
𝑆 = 𝐼 2 𝑍 = 𝐸𝐼 = √𝑃2 + 𝑄 2
𝑅
𝑃
𝑤𝑎𝑡𝑡𝑠
where: 𝑝𝑓 = cos 𝜃 = 𝑍 = 𝑆 = 𝑉𝐴 = 𝑝𝑜𝑤𝑒𝑟 𝑓𝑎𝑐𝑡𝑜𝑟
𝑋 𝑄 𝑣𝑎𝑟𝑠
𝑟𝑓 = sin 𝜃 = = =
= 𝑟𝑒𝑎𝑐𝑡𝑖𝑣𝑒 𝑓𝑎𝑐𝑡𝑜𝑟
𝑍 𝑆
𝑉𝐴
Conservation of AC Power
The complex, real and reactive powers of the
sources equal the respective sums of the complex,
real and reactive powers of the individual loads.
Power Factor Correction
The process of increasing the power factor without
altering the voltage or current to the original load is
known as power factor correction.
𝑄𝐶
𝑃(tan 𝜃1 − tan 𝜃2 )
𝐶=
=
𝜔𝑉𝑟𝑚𝑠 2
𝜔𝑉𝑟𝑚𝑠 2
𝑄𝐶 = 𝑄1 − 𝑄2
Although the most common situation in practice is
that of an inductive load, it is also possible that the
load is capacitive, that is, the load is operating at a
leading power factor. In this case, an inductor
should be connected across the load for power
factor correction. The required shunt inductance L
can be calculated as such
𝑉𝑟𝑚𝑠 2
𝐿=
𝜔𝑄𝐿
𝑄𝐿 = 𝑄1 − 𝑄2
Examples:
1. If
𝑣(𝑡) = 160 cos 50𝑡 V
and
𝑖(𝑡) =
−20 sin(50𝑡 − 30°)
A,
calculate
the
instantaneous power and average power.
2. A certain load comprises 12 – j8 Ω in parallel
with j4 Ω. Determine the overall power factor.
3. Given the circuit below, find the average power
absorbed by each of the elements.
20 Ω
50˂0° V
10 Ω
j5 Ω
-j10 Ω
4. A relay coil is connected to a 210 V, 50 Hz
supply. If it has a resistance of 30 Ω and an
inductance of 0.5 H, calculate the apparent
power and the power factor.
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
15
Electrical Circuits II: AC Circuits
5. A load draws 5 kVAR at a power factor of 0.86
(leading) from a 220 V rms source. Calculate the
peak current and the apparent power supplied
to the load.
6. Compute the average power absorbed by the 4
Ω resistor in the circuit below.
9. For the circuit below, find the maximum power
delivered to the load ZL.
0.5 VO
2Ω
+
10 cos 4t V
+V
O
VO
1/20 F
1H
ZL
-
4VO
-j1 Ω
-
4Ω
+ 10. The variable resistor R in the circuit is adjusted
until it absorbs the maximum average power.
Find R and the maximum average power
absorbed.
2Ω
4Ω
j2 Ω
4˂60° A
4IO
7. For the network below, assume that the port
impedance is
𝑅
𝑍𝑎𝑏 =
∠ − tan−1 𝜔𝑅𝐶
2
2
2
√1 + 𝜔 𝑅 𝐶
Find the average power consumed by the
network when R = 10 kΩ, C = 200 nF, and i =
2sin (377t + 22°) mA.
i
Linear
Ne twork
+ 3Ω
j1 Ω
4˂0° A
-j10 Ω
R
11. Assuming that the load impedance is to be
purely resistive, what load should be connected
to terminals a – b of the circuit so that the
maximum power is transferred to the load?
100 Ω
+
V
-
-j10 Ω
a
-j10 Ω
40 Ω
120˂60° V
50 Ω
2˂90° A
j30 Ω
b
8. For each of the circuits below, determine the
value of load Z for maximum average power
transferred.
a.
8Ω
Z
-j2 Ω
12. Obtain the power factor for each of the circuits
below. Specify each power factor as leading or
lagging.
a.
j2 Ω
4Ω
4˂0° A
-j2 Ω
b.
-j2 Ω
-j3 Ω
5Ω
b.
j2 Ω
-j1 Ω
10˂30° V
4Ω
1Ω
4Ω
j2 Ω
j1 Ω
Z
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
16
Electrical Circuits II: AC Circuits
13. A source delivers 50 kVA to a load with a power
factor of 65 percent lagging. Find the load’s
average and reactive powers.
14. For the following voltage and current phasors,
calculate the complex power, apparent power,
real power, and reactive power. Specify
whether the pf is leading or lagging.
a. 𝑉 = 220∠30° 𝑉 𝑟𝑚𝑠, 𝐼 = 0.5∠60° 𝐴 𝑟𝑚𝑠
b. 𝑉 = 250∠ − 10° 𝑉 𝑟𝑚𝑠, 𝐼 = 6.2∠ −
25°𝐴 𝑟𝑚𝑠
c. 𝑉 = 120∠0° 𝑉 𝑟𝑚𝑠, 𝐼 = 2.4∠ −
15 °𝐴 𝑟𝑚𝑠
d. 𝑉 = 160∠45° 𝑉 𝑟𝑚𝑠, 𝐼 = 8.5∠90°𝐴 𝑟𝑚𝑠
15. Determine the complex power for the following
cases:
a. P = 269 W, Q = 150 VAR (capacitive)
b. Q = 2000 VAR, pf = 0.9 (leading)
c. S = 600 VA, Q = 450 VAR (inductive)
d. 𝑉𝑟𝑚𝑠 = 220 𝑉, 𝑃 = 1 𝑘𝑊, |𝑍| =
40 Ω (inductive)
16. Obtain the overall impedance for the following
cases
a. P = 1000 W, pf = 0.8 (leading), 𝑉𝑟𝑚𝑠 =
220 𝑉
b. P = 1500 W, Q = 2000 VAR (inductive),
𝐼𝑟𝑚𝑠 = 12 𝐴
c. 𝑆 = 4500∠60° 𝑉𝐴, 𝑉 = 120∠45° 𝑉
17. In the circuit below, load A receives 4 kVA at 0.8
pf leading. Load B receives 2.4 kVA at 0.6 pf
lagging. Box C is an inductive load that
consumes 1 kW and receives 500 VARs.
a. Determine I.
b. Calculate the power factor of the
combination
to make the pf of the system equal to unity.
Calculate the value of the capacitance.
19. Find the complex power absorbed by each of
the five elements in the circuit below.
j10 Ω
-j20 Ω
40˂0° Vrms
20 Ω
5˂90° Vrms
20. For the circuit below, find the average, reactive
and complex power delivered by the dependent
current source.
4Ω
24˂0° V
2Ω
-j1 Ω
1Ω
+
VO
-
2VO
j2 Ω
21. Calculate the reactive power in the inductor and
capacitor in the circuit below.
j30 Ω
50 Ω
24˂0° V
-j20 Ω
4˂0°A
40 Ω
22. Given the circuit, find IO and the overall complex
power supplied.
1.2 kW
0.8 kVAR (cap)
IO
100˂90° V
2 kVA
0.707 pf leading
4 kW
0.9 pf laggin g
A
+
120˂30° V
I
B
C
23. Find IO in the circuit below.
_
IO
18. Two loads are placed in parallel across a 120 V
rms 60 Hz line. The first load draws 150 VA at a
lagging power factor of 0.707, while the second
load draws 50 VAR at a leading power factor of
0.8. A third load is purely capacitive and is
placed in parallel across the 120 V line in order
220˂0° V
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
12 kW
0.866 pf leading
16 kW
0.85 pf lagging
20 kVAR
0.6 pf lagging
17
Electrical Circuits II: AC Circuits
24. Refer to the circuit below.
a. What is the power factor?
b. What is the average power dissipated?
c. What is the value of the capacitance that
will give a unity power factor when connected
to the load?
120 V
60 Hz
C
Z = 10+j12 Ω
25. A 240 V rms 60 Hz supply serves a load that is
10 kW (resistive), 15 kVAR (capacitive), and 22
kVAR (inductive). Find:
a. The apparent power
b. The current drawn from the supply
c. The kVAR rating and capacitance required
to improve the power factor to 0.96 lagging.
d. The current drawn from the supply under
the new power factor conditions.
26. Consider the power system shown. Calculate:
a. The total complex power
b. The power factor
c. The capacitance necessary to establish a
unity power factor
+
240 Vrms, 50 Hz
-
80 – j50 Ω
120 + j70 Ω
60 + j0 Ω
The frequency response of a circuit is the variation
in its behavior with change in signal frequency.
The transfer function H(ω) (also called the network
function) of a circuit is the frequency – dependent
ratio of a phasor output Y(ω) (an element voltage or
current) to a phasor input X(ω) (source voltage or
current).
𝐻(𝜔) =
𝑌(𝜔)
𝑋(𝜔)
𝑉𝑜 (𝜔)
𝑉𝑖 (𝜔)
𝐼𝑜 (𝜔)
𝐻(𝜔) = 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 𝑔𝑎𝑖𝑛 =
𝐼𝑖 (𝜔)
𝑉𝑜 (𝜔)
𝐻(𝜔) = 𝑇𝑟𝑎𝑛𝑠𝑒𝑓𝑒𝑟 𝐼𝑚𝑝𝑒𝑑𝑎𝑛𝑐𝑒 =
𝐼𝑖 (𝜔)
𝐼𝑜 (𝜔)
𝐻(𝜔) = 𝑇𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝐴𝑑𝑚𝑖𝑡𝑡𝑎𝑛𝑐𝑒 =
𝑉𝑖 (𝜔)
The transfer function H(ω) can be also expressed as
𝑁(𝜔)
𝐻(𝜔) =
𝐷(𝜔)
A zero, as a root of the numerator polynomial, is a
value that results in zero value of the function. A
pole, as a root of the denominator polynomial, is a
value for which the function is infinite.
To avoid complex algebra, it is expedient to replace
jω temporarily with s when working with H(ω) and
replace s with jω at the end.
The Decibel Scale
In communications systems, gain is measured in
bels. Historically, the bel is used to measure the
ratio of two levels of power or power gain G; that is,
𝑃2
𝐺 = 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑏𝑒𝑙𝑠 = log10
𝑃1
The decibel (dB) provides us with a unit of less
magnitude. It is 1/10th of a bel and is given by
𝑃2
𝐺𝑑𝐵 = 10 log10
𝑃1
Alternatively, the gain G can be expressed in terms
of voltage or current ratio.
𝑉2
𝐼2
𝐺𝑑𝐵 = 20 log10 = 20 log10
𝑉1
𝐼1
Three things are important to note from the
following equations:
1. That 10 log is used for power, while 20 log is
used for voltage or current, because of the
square relationship between them.
2. That the dB value is a logarithmic measurement
of the ratio of one variable to another of the
same type. Therefore, it applies in expressing
only the voltage and current gain, which are
dimensionless quantities, but not in expressing
the transfer impedance and admittance.
𝐻(𝜔) = 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑔𝑎𝑖𝑛 =
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
18
Electrical Circuits II: AC Circuits
3. It is important to note that we only use voltage
and current magnitudes. Negative signs and
angles will be handled independently.
Application of Decibels
Decibels were originally intended as a measure of
changes in acoustical levels. The human ear is not a
linear instrument; rather, it responds to sounds in a
logarithmic fashion. Because of this peculiar
phenomenon, a ten – fold increase in sound
intensity results in a perceived doubling of sound.
This means that if it is desired to double the sound
heard from a 10 W power amplifier, the output
power must be increased to 100 W.
The minimum sound level which may be detected
by the human ear is called the threshold of hearing
and is usually taken to be 𝐼𝑜 = 1 × 10−12 𝑊/𝑚2.
Sound
Intensity Level
Intensity
(dB)
(𝑾/𝒎𝟐 )
threshold of
0
10−12
hearing
virtual silence
10
10−11
quiet room
20
10−10
watch ticking at
30
10−9
1m
quiet street
40
10−8
quiet
50
10−7
conversation
quiet motor at
60
10−6
1m
busy traffic
70
10−5
door slamming
80
10−4
busy office
90
10−3
room
jackhammer
100
10−2
motorcycle
110
10−1
loud indoor
120
1
rock concert
threshold of
130
10
pain
Bode Plots are semilog plots of the magnitude (in
decibels) and phase (in degrees) of a transfer
function versus frequency.
Rules for Making Bode Plots
1. Constant Term: For a gain K, the magnitude is
20 log10 𝐾 and the phase is 0°; both are
constant with varying frequency. If K is
negative, the magnitude remains the same but
the phase is 180°.
2. Pole/zero at the origin: For the zero (jω) at the
origin, the magnitude is 20 log10 𝜔 and the
phase is 90°. The slope of the magnitude plot is
20 dB/decade, while the phase is constant with
varying frequency.
For the pole (jω)-1 are similar except that the
slope of the magnitude plot is – 20 dB/decade
while the phase is – 90°.
In general, for (jω)N, where N is an integer, the
magnitude plot will have a slope of 20N
dB/decade, while the phase is 90N degrees.
3. Simple pole/zero: For the simple zero (1 +
the magnitude is 20 log10 |1 +
phase
𝜔
is tan−1 𝑧 .
1
𝑗𝜔
|
𝑧1
𝑗𝜔
),
𝑧1
and the
We notice that
𝑗𝜔
𝐻𝑑𝐵 = 20 log10 |1 + | ⇒ 20 log10 1 = 0
𝑧1 𝜔→0
𝑗𝜔
𝜔
𝐻𝑑𝐵 = 20 log10 |1 + | ⇒
20 log10
𝑧1 𝜔→∞
𝑧1
showing that we can approximate the
magnitude as zero (a straight line with zero
slope) for small values of ω and by a straight
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
19
Electrical Circuits II: AC Circuits
line with slope 20 dB/decade for large value of
ω. The frequency 𝜔 = 𝑧1 where the two
asymptotic lines meet is called the corner
frequency or break frequency.
𝜔
The phase tan−1 (𝑧 ) can be expressed as
1
0°, 𝜔 = 0
𝜔
∅ = tan−1 ( ) = {45°, 𝜔 = 𝑧1
𝑧1
90°, 𝜔 → ∞
The straight line plot has a slope of 45° per
decade.
The phase can be expressed as
2𝜉2 𝜔
0°, 𝜔 = 0
𝜔𝑛
−1
−90°,
𝜔 = 𝜔𝑛
∅ = tan (
=
)
{
𝜔 2
−180°, 𝜔 → ∞
1−( )
𝜔𝑛
The phase plot is a straight line with a slope of
𝜔
90° per decade starting at 10𝑛 and ending
at 10𝜔𝑛 .
For the quadratic zero (1 +
The bode plots for the pole (
1
1+
𝑗𝜔
𝑝1
) are similar
to those in the zero except that the corner
frequency is at 𝜔 = 𝑝1 , the magnitude has a
slope of – 20 dB/decade, and the phase has a
slope of – 45° per decade.
4. Quadratic pole/zero: The magnitude of the
quadratic
−20 log10 |
−tan−1 (
1
(
pole
𝑗2𝜉 𝜔
𝑗𝜔 2
1+ 2 +( )
𝜔𝑛
𝜔𝑛
1
𝑗2𝜉 𝜔
𝑗𝜔 2
1+ 2 +( )
𝜔𝑛
𝜔𝑛
2𝜉2 𝜔
𝜔𝑛
𝜔 2
1−( )
𝜔𝑛
)
is
| and the phase is
1
𝑗2𝜉2 𝜔
𝑗𝜔 2
1+
+( )
𝜔𝑛
𝜔𝑛
𝐻𝑑𝐵 = −20 log10 |
𝑗𝜔 2
+ (𝜔 ) ), the
𝑛
plots are inverted because the magnitude plot
has a slope of 40 dB/decade while the phase
plot has a slope of 90° per decade.
5. Time Delay: For a time delay 𝑒 −𝑠𝑇 , the
magnitude remains the same and the phase
drops exponentially.
Examples:
1. For a series RC circuit, obtain the transfer
𝑉
function 𝑉𝑜, and its frequency response.
𝑠
𝑉
𝑉𝑠
2. Obtain the transfer function 𝑜, of a series RL
). But
𝐻𝑑𝐵 = −20 log10 |
𝑗2𝜉2 𝜔
𝜔𝑛
|⇒
𝜔→0
0
circuit and its frequency response.
3. A current source, Ii (t) is connected in parallel to
a series combination 4 Ω and 2 H and in parallel
with 0.5 F capacitor. Find the current gain, and
its poles and zeroes.
1
|
𝑗2𝜉2 𝜔
𝑗𝜔 2
1 + 𝜔 + (𝜔 )
𝑛
𝑛
𝜔
⇒
−40 log10
𝜔→∞
𝜔𝑛
Thus, the amplitude plot consist of two straight
asymptotic lines: one with zero slope for 𝜔 <
𝜔𝑛 and the other with slope – 40 dB/decade
for 𝜔 > 𝜔𝑛 , with 𝜔𝑛 as the corner frequency.
4. Find the transfer impedance and admittance of
a circuit consisting of 5 Ω in series with 0.1 F
capacitor which is in parallel with a series a
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
20
Electrical Circuits II: AC Circuits
combination of 3 Ω and 2 H inductor. Obtain its
zeroes and poles.
or
𝑓𝑟 =
1
𝐻𝑧
2𝜋√𝐿𝐶
Note that at resonance:
1. The impedance is purely resistive, thus, Z = R. In
other words, the LC series combination acts like
a short circuit, and the entire voltage is across
R.
5. Construct the Bode plots for the following
transfer functions:
200𝑗𝜔
a. 𝐻(𝜔) = (𝑗𝜔+2)(𝑗𝜔+10)
5(𝑗𝜔+2)
b. 𝐻(𝜔) = 𝑗𝜔(𝑗𝜔+10)
2. The voltages source and the current is in phase,
so that the power factor is unity.
𝑗𝜔+10
c. 𝐻(𝜔) = 𝑗𝜔(𝑗𝜔+5)2
50𝑗𝜔
d. 𝐻(𝜔) = (𝑗𝜔+4)(𝑗𝜔+10)2
𝑠+1
e. 𝐻(𝑠) = 𝑠2 +60𝑠+100
f.
10
𝐻(𝑠) = 𝑠(𝑠2 +80𝑠+400)
RESONANCE
It is the condition that exist in a circuit containing at
least one resistor, an inductor and capacitor where
in the current behaves as if it is purely resistive.
Resonant circuits (series or parallel) are useful for
constructing filters, as their transfer functions can
be highly frequency selective. They are used in
many applications such as selecting the desired
stations in radio and TV receivers.
Series Resonance
Consider a series RLC circuit. The input impedance is
𝑉
1
𝑍 = 𝐻(𝜔) = = 𝑅 + 𝑗𝜔𝐿 +
𝐼
𝑗𝜔𝐶
Resonance results when the imaginary part of the
transfer function is
zero, or
1
𝐼𝑚(𝑍) = 𝜔𝐿 −
𝜔𝐶
=0
3. The magnitude of the transfer function 𝐻(𝜔) =
𝑍(𝜔) is minimum.
4. The inductor voltage and capacitor voltage can
be much more than the source voltage.
The value of ω that
satisfies this condition
is called the resonant frequency ω0. Thus, the
resonance condition is
1
𝜔0 =
𝑟𝑎𝑑/𝑠
√𝐿𝐶
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
21
Electrical Circuits II: AC Circuits
The highest power dissipated occurs at resonance,
when 𝐼 = 𝑉𝑚 /𝑅, so that
1 𝑉𝑚 2
)
𝑃(𝜔0 =
2 𝑅
At certain frequencies 𝜔 = 𝜔1 , 𝜔2 , the dissipated
power is half the maximum value; that is,
1 𝑉𝑚 2
𝑃(𝜔1 ) = 𝑃(𝜔1 ) =
4 𝑅
Hence, 𝜔1 𝑎𝑛𝑑 𝜔2 are called the half – power
frequencies.
The half – power frequencies are obtained by
setting Z equal to √2𝑅,
1 2
√𝑅 2 + (𝜔𝐿 −
) = √2𝑅
𝜔𝐶
Solving for ω, we obtain
𝑅
𝑅 2
1
√
𝜔1 = − + ( ) +
𝑟𝑎𝑑/𝑠
2𝐿
2𝐿
𝐿𝐶
𝜔2 =
𝑅
𝑅 2
1
+ √( ) +
𝑟𝑎𝑑/𝑠
2𝐿
2𝐿
𝐿𝐶
Or
𝑅
𝐻𝑧
4𝜋𝐿
𝑅
𝑓2 = 𝑓𝑟 +
𝐻𝑧
4𝜋𝐿
𝑓1 = 𝑓𝑟 −
The “sharpness” of the resonance in a resonant
circuit is measured quantitatively by the quality
factor Q. At resonance, the reactive energy in the
circuit oscillates between the inductor and the
capacitor. The quality factor relates the maximum
or peak energy stored to the energy dissipated in
the circuit per cycle of oscillation:
𝑟𝑒𝑎𝑐𝑡𝑖𝑣𝑒 𝑝𝑜𝑤𝑒𝑟
𝑄=
𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑝𝑜𝑤𝑒𝑟
𝑀𝑎𝑥 𝑆𝑡𝑜𝑟𝑒𝑑 𝐸𝑛𝑒𝑟𝑔𝑦
𝑄 = 2𝜋 {
}
𝐸𝑛𝑒𝑟𝑔𝑦 𝐷𝑖𝑠𝑠𝑖𝑝𝑎𝑡𝑒𝑑/𝐶𝑦𝑐𝑙𝑒
It is also regarded as a measure of the energy
storage property of a circuit in relation to its energy
dissipation property. In the series RLC circuit, the
peak energy stored is ½LI2, while the energy
dissipated in one period is ½(I2R)(1/f0). Hence,
1 2
𝐿𝐼
2𝜋𝑓0 𝐿 𝜔0 𝐿
1
𝑄 = 2𝜋 2
=
=
=
1 2
1
𝑅
𝑅
𝜔
𝑠
𝑠
0 𝑅𝑠 𝐶
2 𝐼 𝑅𝑠 (𝑓0 )
Or
2𝜋𝑓𝑟 𝐿
1
1 𝐿
=
= √
𝑅𝑠
2𝜋𝑓𝑟 𝐶𝑅𝑠 𝑅𝑠 𝐶
Q factor is defined as the voltage magnification in
the circuit at the time of resonance.
In series resonance, higher quality factor means
higher voltage magnification as well as higher
selectivity of the tuning coil.
As a series resonance circuit only functions on
resonant frequency, this type of circuit is also
known as an Acceptor Circuit because at resonance,
the impedance of the circuit is at its minimum so
easily accepts the current whose frequency is equal
to its resonant frequency.
𝑄=
The width of the frequency band between the half –
power frequencies is called bandwidth,
𝑓𝑟
𝐵𝑊 = 𝑓2 − 𝑓1 =
𝑄
Or
𝜔0 𝑅
𝐵𝑊 = 𝜔2 − 𝜔1 =
=
𝑄
𝐿
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
22
Electrical Circuits II: AC Circuits
Parallel Resonance
Consider a parallel RLC circuit. The input admittance
is
𝐼
1
1
𝑌 = 𝐻(𝜔) = = + 𝑗𝜔𝐶 +
𝑉 𝑅
𝑗𝜔𝐿
Resonance
results when
the imaginary
part of the
transfer
function
is
zero
1
𝜔0 = 𝐿𝐶 𝑟𝑎𝑑/𝑠 if Q>10
The half – power frequencies are
1
1 2
1
√
𝜔1 =
− (
) +
𝑟𝑎𝑑/𝑠
2𝑅𝐶
2𝑅𝐶
𝐿𝐶
√
𝜔0 =
1
√𝐿𝐶
√1 −
𝑅2 𝐶
𝑟𝑎𝑑/𝑠
𝐿
if Q<10
𝜔2 =
or
1
1
𝑅2
𝑓𝑟 = 2𝜋 √𝐿𝐶 − 𝐿2 𝐻𝑧 if Q<10
𝑓𝑟 = 2𝜋
1
√𝐿𝐶
1
1 2
1
+ √(
) +
𝑟𝑎𝑑/𝑠
2𝑅𝐶
2𝑅𝐶
𝐿𝐶
Or
𝑅
𝐻𝑧
4𝜋𝐿
𝑅
𝑓2 = 𝑓𝑟 +
𝐻𝑧
4𝜋𝐿
𝑓1 = 𝑓𝑟 −
𝐻𝑧 if Q>10
Note that at resonance:
1. The impedance is purely resistive, thus, Z = R. In
other words, the LC parallel combination acts
like an open circuit, and the entire current is
across R.
2. The voltages source and the current is in phase,
so that the power factor is unity.
3. The current is minimum.
The bandwidth is
𝐵𝑊 = 𝑓2 − 𝑓1 =
𝑓𝑟
𝑄
Or
𝐵𝑊 = 𝜔2 − 𝜔1 =
1
𝜔0
=
𝑅𝐶
𝑄
The quality factor is
𝑄=
𝑅𝑃
= 𝜔0 𝑅𝑃 𝐶
𝜔0 𝐿
Or
𝑅𝑃
𝐿
= 2𝜋𝑓𝑟 𝐶𝑅𝑝 = 𝑅𝑝 √
2𝜋𝑓𝑟 𝐿
𝐶
In parallel resonance, higher quality factor means
higher current magnification.
𝑄=
4. The magnitude of the transfer function 𝐻(𝜔) =
𝑍(𝜔) is maximum.
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
23
Electrical Circuits II: AC Circuits
As a parallel resonance circuit only functions on
resonant frequency, this type of circuit is also
known as a Rejector Circuit because at resonance,
the impedance of the circuit is at its maximum
thereby suppressing or rejecting the current whose
frequency is equal to its resonant frequency. The
effect of resonance in a parallel circuit is also called
“current resonance”.
Examples:
1. A source voltage 20 sin ωt connected to a series
RLC with 2 Ω resistance, 1 mH inductance and
0.4 μF capacitor. Find the resonant frequency
and the half – power frequencies. Calculate the
quality factor and bandwidth. Determine the
amplitude of the current at ω0, ω1 and ω2.
2. A series – connected circuit has R = 4 Ω and L =
25 mH. Calculate the value of C that will
produce a quality factor of 50. Find ω0, ω1 and
ω2 and BW. Determine the average power
dissipated at ω0, ω1 and ω2. Take Vm = 100 V.
3. A parallel resonant circuit has R = 100 kΩ, L = 20
mH, and C = 5 nF. Calculate ω0, Q and B.
4. A source voltage 10 sin ωt is connected to a
parallel RLC with 8kΩ, 0.2 mH and 8μF.
Calculate ω0, Q and B. find the half – power
frequencies. Determine the average power
dissipated at ω0, ω1 and ω2.
5. Determine the resonant frequency of a parallel
circuit consisting of 0.1 F, 10 Ω and series
combination of 2 H and 2 Ω.
6. Calculate the resonant frequency of the circuit
consisting of 1 H in series with the parallel
combination of 0.2 F and 10 Ω resistor.
A filter is a circuit that is designed to pass signals
with desired frequencies and reject or attenuate
others.
There are four types of filters whether passive or
active:
1. A low pass filter passes low frequencies and
stops high frequencies. A typical low pass filter
is formed when the output of an RC circuit is
taken off the capacitor. It is designed to pass
only frequencies from dc up to the cutoff
frequency ωC. The cutoff frequency is also called
as roll off frequency.
1
1
𝐻(𝜔𝐶 ) =
=
√2 √1 + 𝜔𝐶 2 𝑅2 𝐶 2
Or
1
𝜔𝐶 =
𝑅𝐶
A low pass filter can also be formed when the
output of an RL circuit it taken off the resistor.
Other examples of low pass filter configuration
are “L – type” (LC), “Pi – type” (CLC), and “T –
type” (LCL).
2. A high pass filter passes high frequencies and
rejects low frequencies. A high pass filter is
formed when the output of an RC circuit is
taken off the resistor. It is designed to pass all
the frequencies above its cutoff frequency ωC.
The cutoff frequency is also called as corner
frequency.
1
𝜔𝐶 =
𝑅𝐶
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
24
Electrical Circuits II: AC Circuits
A high pass filter can also be formed when the
output of an RL circuit is taken off the inductor.
Other examples of high pass filter configuration
are “L – type” (CL), “Pi – type” (LCL), and “T –
type” (CLC).
3. A band pass filter passes frequencies within a
frequency band and blocks or attenuates
frequencies outside the band. A band pass filter
is formed when the output of a RLC series
resonant circuit is taken off the resistor. It is
designed to pass all frequencies within a band
of frequencies, 𝜔1 < 𝜔 < 𝜔2 and is centered
on ω0, the center frequency.
1
𝜔0 =
√𝐿𝐶
A band pass filter can also be formed by
cascading the low pass filter (𝜔2 = 𝜔𝐶 ) with
the high pass filter (𝜔1 = 𝜔𝐶 ).
Another configuration for a band pass filter is
by placing a parallel resonant circuit in parallel
with the signal path and a series resonant
circuit in series with the signal path.
4. A band stop filter passes frequencies outside a
frequency band and blocks or attenuates
frequencies within the band. A band stop filter
is formed when the output RLC series resonant
circuit is taken off the LC series combination. It
is designed to stop or eliminate all frequencies
within a band of frequencies, 𝜔1 < 𝜔 < 𝜔2 and
is centered on ω0, the frequency of rejection. It
is also called as band reject or notch filter.
Another configuration for a band stop filter is by
placing a series resonant circuit in parallel with
the signal path and one or more parallel
resonant circuit in series with the signal path.
Characteristics of Ideal Filters
Type of
H(0)
H(∞)
H(ωc) or
Filter
H(ωo)
Low pass
1
0
1/√2
High pass
0
1
1/√2
Band pass
0
0
1
Band stop
1
1
0
Passive Filters
A filter is a passive filter if it consists of only the
passive elements R, L, and C. Three major
limitations are:
1. They cannot generate gain greater than 1;
passive element cannot add energy to the
network.
2. They may require bulky and expensive
inductors.
3. They perform poorly at frequencies below the
audio frequency range (300 Hz < f < 3000 Hz).
Nevertheless, passive filters are useful at high
frequency.
Active Filters
A filter is an
active filter if it
consists of active
elements (such
as
transistors
and op amps) in
addition
to
passive R, L, and
C.
The
advantages
of
active filter are:
1. They are often smaller and less expensive,
because they do not require inductors.
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
25
Electrical Circuits II: AC Circuits
2. They can provide amplifier gain in addition to
providing the same frequency response as RLC
filters.
3. They can be combined with buffer amplifiers
(voltage followers) to isolate each stage of the
filter from source and load impedance effects.
This isolation allows designing the stages
independently and then cascading them to
realize the desired transfer function. (Bode
plots, being logarithmic, may be added when
transfer functions are cascaded)
However, active filters are less reliable and less
stable. The practical limit of most active filters is
about 100 kHz.
Filters are often classified according to their order
(or number of poles) or their specific design type.
5. Frequency Spectrum used
a. Power Supply
b. Audio
c. RF
Examples:
1. Determine what type of filter is a 2 H inductor
in series with the parallel combination of 2 kΩ
and 2 μF capacitor, if the output voltage is to be
found in the capacitor. Calculate the corner or
cut off frequency.
2. A 100 Ω resistor, R1 is connected in series to the
parallel combination of 2 mH and 100 Ω
resistor, R2. The output voltage is found in R2.
Obtain the voltage gain and identify what type
of filter the circuit represents and determine
the corner frequency.
Classification of Filters
1. Physical Layout Shape
a. L
b. Pi
c. T
2. Usage
a. Bypass
b. Decouple
c. Band pass
d. Band stop
e. Smoothing
3. Component Types
a. LC
b. RC
c. Active
d. Passive
4. Number of sections
a. Single – section
b. Two – Section
c. Multiple – Section
3. If the band stop filter is to reject 200 Hz
sinusoid while passing other frequencies,
calculate the values of L and C. Take R = 150 Ω
and the bandwidth as 100 Hz.
4. Design a band pass filter with a lower cutoff
frequency of 20.1 kHz and upper cutoff
frequency of 20.3 kHz. Take R = 20 kΩ. Calculate
L, C and Q.
5. Design a low pass active filter with a dc gain of 4
and a corner frequency of 500 Hz.
6. Design a high pass filter with a high frequency
gain of 5 and a corner frequency of 2 kHz. Use a
0.1 μF capacitor in your design.
7. Design a band pass filter consisting a low pass
and high pass filter that passes frequencies
between 250 Hz and 3000 Hz and with K = 10.
Select R = 20 kΩ.
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
26
Electrical Circuits II: AC Circuits
8. Design a notch filter consisting of a low pass,
high pass and summing amplifier with ω0 = 20
krad/s, K = 5, and Q = 10. Use R = Ri = 10 kΩ
TWO PORT – NETWORK
A “port” refers to a
pair
of
terminals
through which a single
current flows and
across which there is a
single voltage.
Externally,
either
voltage or current
could be independently specified while the other
quantity would be computed. The Thevenin
equivalent permits a simple model of the linear
network, regardless of the number of components
in the network.
A two – port
network
requires
two
terminal pairs
(total
4
terminals).
Amongst
the
two voltages and two currents shown, generally
two can be independently specified (externally).
Our study of two – port networks is for at least two
reasons. First, such networks are useful in
communications, control systems, power systems,
and electronics. For example, they are used in
electronics to model transistors and to facilitate
cascaded design. Second, knowing the parameters
of a two – port network enables it to treat as a
“black box” when embedded within a larger
network.
Examples of Two Port Networks
To characterize a two – port network requires that
we relate the terminal quantities V1, V2, I1 and I2 out
of which two are independent. The various terms
that relate these voltages and currents are called
parameters.
Impedance Parameters
Impedance and admittance parameters are
commonly used in the
synthesis of filters. They are
also useful in the design and
analysis
of
impedance
matching networks and
power distribution networks.
A two port network may be
voltage driven or current
driven. The terminal voltages
can be related to the
terminal currents as
𝑉1 = 𝑧11 𝐼1 + 𝑧12 𝐼2 𝑉2 = 𝑧21 𝐼1 + 𝑧22 𝐼2
𝑧11 𝑧12 𝐼1
𝑉
𝐼
[ 1 ] = [𝑧
] [ ] = [𝑧] [ 1 ]
𝑉2
𝐼2
21 𝑧22 𝐼2
where the z terms are called the impedance
parameters, or simply z parameters, and have units
of ohms.
The values of the parameters can be evaluated by
setting I1 = 0 (input port open circuited) or I2 = 0
(output port open circuited). Thus,
𝑉1
𝑉2
𝑧11 = | 𝐼2 = 0
𝑧12 = | 𝐼1 = 0
𝐼1
𝐼2
𝑉2
𝑉2
𝑧21 = | 𝐼2 = 0
𝑧22 = | 𝐼1 = 0
𝐼1
𝐼2
Since the z parameters are obtained by open
circuiting the input or output port, they are also
called the open circuit impedance parameters.
Specifically,
𝑧11 – Short Circuit Input Impedance
𝑧12 – Short Circuit Transfer Impedance from port 2
to port 1
𝑧21 – Short Circuit Transfer Impedance from port 1
to port 2
𝑧22 – Short Circuit Output Impedance
Sometimes 𝑧11 and 𝑧22 are called driving point
impedances, while 𝑧12 and 𝑧21 are called transfer
impedances. A driving point impedance is the input
impedance of a two terminal (one port) device.
Thus,
𝑧11 is the input driving point impedance with the
output port open circuited, while 𝑧22 is the output
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
27
Electrical Circuits II: AC Circuits
driving point impedance with the input port open
circuited.
When 𝑧11 = 𝑧22 , the two port network is said to be
symmetrical. This implies that the network has
mirror like symmetry about some center line; that
is, a line can be found that divides the network into
two similar halves.
When the two port network is linear and has no
dependent sources, the
transfer impedances are
equal (𝑧12 = 𝑧21,), and the
two port is said to be
reciprocal. This means that if
the points of excitation and
response are interchanged,
the transfer impedances
remain the same. Any two
port that is made entirely of resistors, capacitors,
and inductors must be reciprocal. A reciprocal
network can be replaced by the T – equivalent
circuit (figure a). If the network is not reciprocal, a
more general equivalent network modeled (figure
b).
Admittance Parameters
Impedance parameters may
not exist for a two port
network. So there is a need
for an alternative means of
describing such a network.
This need may be met by
the
second
set
of
parameters, which we
obtain by expressing the terminal currents in terms
of the terminal voltages. The terminal currents can
be expressed in terms of the terminal voltages as
𝐼1 = 𝑦11 𝑉1 + 𝑦12 𝑉2 𝐼2 = 𝑦21 𝑉1 + 𝑦22 𝑉2
𝑦11 𝑦12 𝑉1
𝐼
𝑉
[ 1 ] = [𝑦
] [ ] = [𝑦] [ 1 ]
𝑦
𝐼2
𝑉2
21
22 𝑉2
The y terms are known as the admittance
parameters (or, simply, y parameters) and have
units of siemens. The values of the parameters can
be determined by setting V1 = 0 (input port short
circuited) or V2 = 0 (output port short circuited).
Thus,
𝐼1
𝐼1
𝑦11 = | 𝑉2 = 0
𝑦12 = | 𝑉1 = 0
𝑉1
𝑉2
𝐼2
𝐼2
𝑦21 = | 𝑉2 = 0
𝑦22 = | 𝑉1 = 0
𝑉1
𝑉2
Since the y parameters are obtained by short
circuiting the input or output port, they are also
called the short circuit admittance parameters.
Specifically,
𝑦11 – Short Circuit Input Admittance
𝑦12 – Short Circuit Transfer Admittance from port 2
to port 1
𝑦21 – Short Circuit Transfer Admittance from port 1
to port 2
𝑦22 – Short Circuit Output Admittance
The impedance and admittance parameters are
collectively referred to as immittance parameters.
For a two port network that is linear and has no
dependent sources, the transfer admittances are
equal ( 𝑦12 = 𝑦21 ). This can be proved in the same
way as for the z parameters. A reciprocal network
(𝑦12 = 𝑦21 ) can be modeled by a pi – equivalent
(Figure a) circuit. If the network is not reciprocal,
the general network equivalent circuit is modeled.
(figure b)
Hybrid Parameters
The z and y
parameters of a
two
–
port
network do not
always exist. So
there is a need
for developing another set of parameters. This third
set of parameters is based on making V1 and I2 the
dependent variables. Thus, we obtain
𝑉1 = ℎ11 𝐼1 + ℎ12 𝑉2 𝐼2 = ℎ21 𝐼1 + ℎ22 𝑉2
𝑉
ℎ
ℎ12 𝐼1
𝐼
[ 1 ] = [ 11
] [ ] = [ℎ] [ 1 ]
𝐼2
𝑉2
ℎ21 ℎ22 𝑉2
The h terms are known as the hybrid parameters
(or, simply, h parameters) because they are a hybrid
combination of ratios. They are very useful for
describing electronic devices such as transistors; it
is much easier to measure experimentally the h
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
28
Electrical Circuits II: AC Circuits
parameters of such devices than to measure their z
or y parameters.
In fact, we have seen that the ideal transformer in
does not have z parameters. The ideal transformer
can be described by the hybrid parameters.
The values of the h parameters are determined as
𝑉1
𝑉1
ℎ11 = | 𝑉2 = 0
ℎ12 = | 𝐼1 = 0
𝐼1
𝑉2
𝐼2
𝐼2
ℎ21 = | 𝑉2 = 0
ℎ22 = | 𝐼1 = 0
𝐼1
𝑉2
Thus the inverse hybrid parameters are specifically
called as,
ℎ11 – Short Circuit Input Impedance
ℎ12 – Open Circuit Reverse Voltage Gain
ℎ21 – Short Circuit Forward Current Gain
ℎ22 – Open Circuit Output Admittance
The procedure for calculating the h parameters is
similar to that used for the z or y parameters. We
apply a voltage or current source to the appropriate
port, short circuit or open circuit the other port,
depending on the parameter of interest, and
perform regular circuit analysis. For reciprocal
networks, ℎ12 = ℎ21 .
Inverse Hybrid Parameters
A set of parameters closely related to the h
parameters are
the g parameters
or inverse hybrid
parameters.
These are used to
describe the terminal currents and voltages as
𝐼1 = 𝑔11 𝑉1 + 𝑔12 𝐼2 𝑉2 = 𝑔21 𝑉1 + 𝑔22 𝐼2
𝑔11 𝑔12 𝑉1
𝐼
𝑉
[ 1 ] = [𝑔
] [ ] = [𝑔] [ 1 ]
𝑉2
𝐼2
21 𝑔22 𝐼2
The values of the g parameters are determined as
𝐼1
𝐼1
𝑔11 = | 𝐼2 = 0
𝑔12 = | 𝑉1 = 0
𝑉1
𝐼2
𝑉2
𝑉2
𝑔21 = | 𝐼2 = 0
𝑔22 = | 𝑉1 = 0
𝑉1
𝐼2
Thus the inverse hybrid parameters are specifically
called as,
𝑔11 – Open Circuit Input Admittance
𝑔12 – Short Circuit Reverse Current Gain
𝑔21 – Open Circuit Forward Voltage Gain
𝑔22 – Short Circuit Output Impedance
The g parameters are frequently used to model field
effect transistors.
Transmission Parameters
Since there are no restrictions on which terminal
voltages and currents should be considered
independent and which should be dependent
variables, we expect to be able to generate many
sets of parameter.
Another set of parameters relates the variables at
the input port to those at the output port. Thus,
𝑉1 = 𝐴𝑉2 − 𝐵𝐼2 𝐼1 = 𝐶𝑉2 − 𝐷𝐼2
𝑉
𝑉
𝐴 𝐵 𝑉2
[ 1] = [
][
] = [𝑇] [ 2 ]
𝐼1
−𝐼2
𝐶 𝐷 −𝐼2
The two – port parameters shown provide a
measure of how a
circuit transmits voltage
and current from a
source to a load. They
are useful in the
analysis of transmission lines (such as cable and
fiber) because they express sending end variables in
terms of the receiving end variables. For this
reason, they are called transmission parameters.
They are also known as ABCD parameters. They are
used in the design of telephone systems, microwave
networks, and radars.
The transmission parameters are determined as
𝑉1
𝑉1
𝐴 = | 𝐼2 = 0
𝐵 = − | 𝑉2 = 0
𝑉2
𝐼2
𝐼1
𝐼1
𝐶 = | 𝐼2 = 0
𝐷 = − | 𝑉2 = 0
𝑉2
𝐼2
Thus the transmission parameters are specifically
called as,
A – Open Circuit Voltage Ratio
B – Negative Short Circuit Transfer Impedance
C – Open Circuit Transfer Admittance
D – Negative Short Circuit Current Ratio
A and D are dimensionless, B is in ohms, and C is in
siemens. Since the transmission parameters provide
a direct relationship between input and output
variables, they are very useful in cascaded
networks.
Inverse Transmission Parameters
Another set of parameters relates the variables at
the output port to those at the input port. Thus,
𝑉2 = 𝑎𝑉1 − 𝑏𝐼1 𝐼2 = 𝑐𝑉1 − 𝑑𝐼1
𝑉
𝑉
𝑎 𝑏 𝑉1
[ 2] = [
][
] = [𝑡] [ 1 ]
𝐼2
−𝐼
−𝐼
𝑐 𝑑
1
1
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
29
Electrical Circuits II: AC Circuits
The parameters a, b, c, and d are called the inverse
transmission, or t parameters. They are determined
as follows:
𝑉2
𝑉2
𝑎 = | 𝐼1 = 0
𝑏 = − | 𝑉1 = 0
𝑉1
𝐼1
𝐼2
𝐼2
𝑐 = | 𝐼1 = 0
𝑑 = − | 𝑉1 = 0
𝑉1
𝐼1
Thus the transmission parameters are specifically
called as,
a – Open Circuit Voltage Gain
b – Negative Short Circuit Transfer Impedance
c – Open Circuit Transfer Admittance
d – Negative Short Circuit Current Gain
a and d are dimensionless, b is in ohms, and c is in
siemens.
In terms of the transmission or inverse transmission
parameters, a network is reciprocal if
𝐴𝐷 − 𝐵𝐶 = 1
𝑎𝑑 − 𝑏𝑐 = 1
Examples:
1. Determine the z parameters of the network.
2. Find the z parameters of the two port network
shown.
5. Obtain the y parameters for the T network
shown.
6. Determine the h parameters of the circuit.
7. Find the impedance at the input port of the
circuit
8. For the ladder network, determine the g
parameters in the s domain.
3. Find I1 and I2 for the two port network shown.
4. Obtain the y parameters for the pi network
shown.
9. The ABCD parameters of the two port network
are
4
20 Ω
[
]
0.1 𝑆
2
The output port is connected to a variable load
for maximum power transfer. Find RL and the
maximum power transferred.
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
30
Electrical Circuits II: AC Circuits
Relationship between Parameters
z
y
𝑧
𝑧
𝑦
𝑦12
11
12
22
z
−
𝑧21 𝑧22
∆𝑦
∆𝑦
𝑦21 𝑦11
−
∆𝑦
∆𝑦
𝑧
𝑧
𝑦
𝑦
22
12
11
12
y
−
𝑦21 𝑦22
∆𝑧
∆𝑧
𝑧21 𝑧11
−
∆𝑧
∆𝑧
∆𝑧
𝑧12
1
𝑦12
h
−
𝑧22 𝑧22
𝑦11
𝑦11
𝑧21 1
∆𝑦
𝑦21
−
𝑧22 𝑧22
𝑦11
𝑦11
1
𝑧12
∆𝑦
𝑦12
g
−
𝑧11
𝑧11
𝑦22 𝑦22
𝑧21
∆𝑧
𝑦21 1
−
𝑧11
𝑧11
𝑦22 𝑦22
𝑧11 ∆𝑧
𝑦22
1
T
−
−
𝑧21 𝑧21
𝑦21
𝑦21
1 𝑧22
∆𝑦
𝑦11
−
−
𝑧21 𝑧21
𝑦21
𝑦21
𝑧22 ∆𝑧
𝑦11
1
t
−
−
𝑧12 𝑧12
𝑦12
𝑦12
1 𝑧11
∆𝑦
𝑦22
−
−
𝑧12 𝑧12
𝑦12
𝑦12
g
T
1
𝑔
𝐴
∆𝑇
z
12
−
𝑔11
𝑔11
𝐶 𝐶
1 𝐷
∆𝑔
𝑔21
𝐶 𝐶
𝑔11
𝑔11
∆
𝐷
∆𝑇
𝑔
y
𝑔
12
−
𝐵
𝐵
𝑔22
𝑔22
1
𝐴
𝑔21 1
−
−
𝐵
𝐵
𝑔22 𝑔22
𝑔
𝑔
22
12
𝐵
∆𝑇
h
−
∆𝑔
∆𝑔
𝐷
𝐷
𝑔21 𝑔11
1 𝐶
−
−
∆𝑔
∆𝑔
𝐷 𝐷
𝑔
𝑔
𝐶
∆𝑇
11
12
g
−
𝑔21 𝑔22
𝐴
𝐴
1
𝐵
𝐴
𝐴
1
𝑔22
𝐴 𝐵
𝑑 𝑏
𝐶
𝐷
𝑔21 𝑔21
∆𝑡 ∆𝑡
𝑐 𝑎
𝑔11 ∆𝑔
∆𝑡 ∆𝑡
𝑔21 𝑔21
∆
𝐷
𝐵
𝑔
𝑎 𝑏
t
𝑔
22
−
−
𝑐 𝑑
∆𝑇 ∆𝑇
𝑔12
𝑔12
𝐶
𝐴
𝑔11
1
−
−
∆𝑇 ∆𝑇
𝑔12
𝑔12
∆𝑧 = 𝑧11 𝑧22 − 𝑧12 𝑧21 ∆𝑦 = 𝑦11 𝑦22 − 𝑦12 𝑦21
∆ℎ = ℎ11 ℎ22 − ℎ12 ℎ21 ∆𝑔 = 𝑔11 𝑔22 − 𝑔12 𝑔21
∆ 𝑇 = 𝐴𝐷 − 𝐵𝐶 ∆𝑡 = 𝑎𝑑 − 𝑏𝑐
Examples:
1. Find the [T] , [g], [h], [t] and [y] of a two port
network if
6 4
[𝑧] = [
]𝛺
4 6
2. Find the [z], [g], [h], [t] and [y] of a two port
network if
10 1.5 Ω
[𝑇] = [
]
2𝑆
4
Interconnection of Networks
A large, complex network may be divided into sub
networks for the purposes of analysis and design.
The sub networks are modeled as two – port
networks, interconnected to form the original
network. The two – port networks may therefore be
regarded as building blocks that can be
interconnected to form a complex network. The
interconnection can be in series, in parallel, or in
cascade. Although the interconnected network can
be described by any of the six parameter sets, a
certain set of parameters may have a definite
advantage. For example, when the networks are in
series, their individual z parameters add up to give
the z parameters of the larger network. When they
are in parallel, their individual y parameters add up
to give the y parameters of the larger network.
When they are cascaded, their individual
transmission parameters can be multiplied together
to get the transmission parameters of the larger
network.
The networks are regarded
as being in series because
their input currents are the
same and their voltages
add. In addition, each
network has a common
reference, and when the
T
h
∆ℎ
ℎ12
ℎ22
ℎ22
ℎ21 1
−
ℎ22 ℎ22
1
ℎ12
−
ℎ11
ℎ11
ℎ21
∆ℎ
ℎ11
ℎ11
ℎ11 ℎ12
ℎ21 ℎ22
ℎ22
∆ℎ
ℎ21
−
∆ℎ
∆ℎ
−
ℎ21
ℎ22
−
ℎ21
1
ℎ12
ℎ22
ℎ12
ℎ12
∆ℎ
ℎ11
∆ℎ
ℎ11
−
ℎ21
1
−
ℎ21
ℎ11
ℎ12
∆ℎ
ℎ12
−
t
𝑑 1
𝑐 𝑐
∆𝑡 𝑎
𝑐 𝑐
𝑎
1
−
𝑏
𝑏
∆𝑡 𝑑
−
𝑏
𝑏
𝑏 1
𝑎 𝑎
∆𝑡 𝑐
𝑎 𝑎
𝑐
1
−
𝑑
𝑑
∆𝑡
𝑏
−
𝑑
𝑑
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
31
Electrical Circuits II: AC Circuits
circuits are placed in series, the common reference
points of each circuit are connected together.
𝑧
+ 𝑧𝑏11 𝑧𝑎12 + 𝑧𝑏12
[𝑧] = [𝑧𝑎 ] + [𝑧𝑏 ] = [ 𝑎11
𝑧𝑎21 + 𝑧𝑏21 𝑧𝑎22 + 𝑧𝑏22 ]
Two networks are in
parallel when their
port voltages are equal
and the port currents
of the larger network
are the sums of the
individual
port
currents. In addition,
each circuit must have
a common reference and when the networks are
connected together, they must all have their
common references tied together.
𝑦
+ 𝑦𝑏11 𝑦𝑎12 + 𝑦𝑏12
[𝑦] = [𝑦𝑎 ] + [𝑦𝑏 ] = [ 𝑎11
𝑦𝑎21 + 𝑦𝑏21 𝑦𝑎22 + 𝑦𝑏22 ]
Two networks are said to be cascaded when the
output of one is the input of the other.
𝐴 𝐴
𝐵𝑎 𝐵𝑏
[𝑇] = [𝑇𝑎 ] + [𝑇𝑏 ] = [ 𝑎 𝑏
]
𝐶𝑎 𝐶𝑏 𝐷𝑎 𝐷𝑏
Examples:
1. Evaluate 𝑉2 /𝑉𝑠 .
3. Find the transmission parameter of the circuit
REFERENCES:
Alexander, Charles K. and Sadiku, Michael N.O.
(2013). Fundamentals of Electric Circuits pp. 367 –
501, pp. 613 – 672, pp. 853 – 904. McGraw Hill
Higher Education.
Bird, John (2007). Electrical and Electronic Principles
and Technology pp. 205 – 271. Elsevier.
Boylestad, Robert L. (2007). Introductory Circuit
Analysis pp. 539 – 988, pp. 1067 – 1093. Pearson
Prentice Hall.
Meade, Russel L. (2007). Foundations of Electronics:
Circuits & Devices (Electron Flow Version) pp. 301 –
354, 381 – 438, 503 – 640. Thomson Delmar
Learning.
Nahvi, Mahmood and Edminister, Joseph (2003).
Schaum’s Outlines: Electric Circuits pp. 191 – 247,
pp. 273 – 333. McGraw Hill Higher Education.
Rajput, R.K. (2006). A Textbook of Electrical
Technology pp. 48 – 128. Laxmi Publications LTD.
Theraja, B.L. and Theraja, A.K. (2003). A Textbook of
Electrical Technology pp. 453 – 654, pp. 753 – 778.
S. Chand & Company LTD.
2. Find the y parameters of the two port network
Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario
32