Electrical Circuits -2
Dr. Mohammed Tawfeeq
Alternating Current Circuit Theory
Electrical Circuit: An electrical circuit is a mathematical model that approximates the
behavior of an actual electrical system.
Types of Electrical circuits
Linear Circuits:
Non Linear Circuits:
Circuits in which its
elements R, L and C
do not change their
values
Circuits in which its
elements R, L and C change
their values
NOT IN OUR TEXT
DC Circuits (Circuit 1)
AC Circuits (Circuit 2)
Circuits Analysis
Circuits Analysis
Transient
Analysis
Steady-State
Analysis
Time
Response
Frequency
Response
Steady-State
Analysis
Transient
Analysis
1
Steady-State
Analysis
Transient
Analysis
Electrical Circuits -2
Dr. Mohammed Tawfeeq
Alternating Current Circuit Theory
Alternating Current (AC) Circuits Theory
1. INTRODUCTION
The analysis in circuit 1 has been limited to dc networks, networks in which the currents
or voltages are fixed in magnitude except for transient effects. In circuit 2 will turn our
attention to the analysis of networks in which the magnitude of the source varies in a set
manner. Of particular interest is the time-varying voltage that is commercially available
in large quantities and is commonly called the ac voltage. (The letters ac are an
abbreviation for alternating current.) To be absolutely rigorous, the terminology ac
voltage or ac current is not sufficient to describe the type of signal we will be analyzing.
Each waveform of Fig.1 is an alternating waveform available from commercial
supplies.
The term alternating indicates only that the waveform alternates between two prescribed
levels in a set time sequence (Fig.1).
Fig.1 Alternating waveforms.
To be absolutely correct, the term sinusoidal, square wave, or triangular must also be
applied. The pattern of particular interest is the sinusoidal ac waveform for voltage of
Fig.1. Since this type of signal is encountered in the vast majority of instances, the
abbreviated phrases ac voltage and ac current are commonly applied without confusion.
For the other patterns of Fig.1, the descriptive term is always present, but frequently the
ac abbreviation is dropped, resulting in the designation square-wave or triangular
waveforms.
2
Definitions
The sinusoidal waveform of Fig.2 with its additional notation will now be used as a
model in defining a few basic terms.
Fig.2 Important parameters for a sinusoidal voltage.
These terms, however, can be applied to any alternating waveform. It is important to
remember as you proceed through the various definitions that the vertical scaling is in
volts or amperes and the horizontal scaling is always in units of time.
Waveform: The variation of an ac voltage or current versus time is called its waveform.
Since waveforms vary with time, they are designated by lowercase letters v(t), i(t), e(t),
and so on, rather than by uppercase letters V, I, and E as for dc. Often we drop the
functional notation and simply use v, i, and e.
Instantaneous value: The magnitude of a waveform at any instant of time; denoted by
lowercase letters (e1, e2).
Peak amplitude Em : The maximum value of a waveform as measured from its average,
or mean, value, denoted by uppercase letters (such as Em for sources of voltage and Vm
for the voltage drop across a load).
Peak value Ep: The maximum instantaneous value of a function as measured from the
zero-volt level. For the waveform of Fig.2, the peak amplitude and peak value are the
same, since the average value of the function is zero volts.
Peak-to-peak value: Denoted by Ep-p or Vp-p, the full voltage between positive and
negative peaks of the waveform, that is, the sum of the magnitude of the positive and
negative peaks.
3
Periodic waveform: A waveform that continually repeats itself after the same time
interval. The waveform of Fig.2 is a periodic waveform.
Period (T): The time interval between successive repetitions of a periodic waveform (the
period T1 _ T2 _ T3 in Fig.2), as long as successive similar points of the periodic
waveform are used in determining T.
Cycle: The portion of a waveform contained in one period of time. The cycles within T1,
T2, and T3 of Fig.2 may appear different in Fig.3, but they are all bounded by one period
of time and therefore satisfy the definition of a cycle.
Fig.3 Defining the cycle and period of a sinusoidal waveform.
Frequency ( f ): The number of cycles that occur in 1 s. The frequency of the waveform
of Fig.4 (a) is 1 cycle per second, and for Fig.4 (b), 2 1⁄2 cycles per second. If a
waveform of similar shape had a period of 0.5 s [Fig.4(c)], the frequency would be 2
cycles per second.
Fig.4 Demonstrating the effect of a changing frequency on the period of a sinusoidal
waveform.
The unit of measure for frequency is the hertz (Hz), where:
1 hertz (HZ) = 1 cycle per second (c/s)
4
Since the frequency is inversely related to the period—that is, as one increase, the other
decreases by an equal amount—the two can be related by the following equation:
𝑓=
1
f in Hz
𝑇
T in second
and
𝑇=
Now
define the angular frequency as:
And the angle θ = ωt
radians
1
𝑓
𝜔=2πf
rad/sec
or degrees
2. THE SINE WAVE
A sine wave is a sinusoidally varying quantity, or sinusoid, which can be expressed
mathematically as
𝑣 = 𝑉𝑚 sin 𝜔𝑡
or
𝑣 = 𝑉𝑚 sin θ
v
v
Sine wave
𝑉𝑚
0
90⁰
180⁰ 270⁰
ωt
360⁰
ωt
0
π/2
π 3π/2
in degrees
ωt
Fig.5 Sine wave.
5
2π
ωt
in radians
Voltages and Currents as Functions of Time
Recall from Equation 1, v = Vm sin 𝜔𝑡 .
Similarly
e = Em sin 𝜔𝑡
and
i = Imsin 𝜔𝑡
Voltages and Currents with Phase Shifts
If a sine wave does not pass through zero at t = 0 s, it has a phase shift. Waveforms may
be shifted to the left or to the right (see Fig.6). For a waveform shifted left as in (a),
while, for a waveform shifted right as in (b),
Fig.6
6
Phase Difference
Phase difference refers to the angular displacement between different waveforms of the
same frequency. Consider Figure 7. If the angular displacement is 0° as in (a), the
waveforms are said to be in phase; otherwise, they are out of phase. When describing a
phase difference, select one waveform as reference. Other waveforms then lead, lag, or
are in phase with this reference.
For example, in (b), for reasons to be discussed in the next paragraph, the current
waveform is said to lead the voltage waveform, while in (c) the current waveform is said
to lag.
Fig. 7 Illustrating phase difference. In these examples, voltage is taken as reference.
Sometimes voltages and currents are expressed in terms of cos qt rather than sin qt. A
cosine wave is a sine wave shifted by +90°, or alternatively, a sine wave is a cosine wave
shifted by - 90°. For sines or cosines with an angle, the following formulas apply.
Average Value and RMS Value of Sine Wave
Because a sine wave is symmetrical, its area below the horizontal axis is the same as its
area above the axis; thus, over a full cycle its net area is zero, independent of frequency
and phase angle. Thus, the average of sin ω t, sin(ωt ± θ), sin 2ωt, cos ω t, cos (ω t ± θ),
cos 2 ω t, and so on are each zero.
7
Fig.8 Vavg = 0 V
RMS Value
RMS values is also called effective value . An effective value is an equivalent dc value:
it tells you how many volts or amps of dc that a time-varying waveform is equal to in
terms of its ability to produce average power.
𝑉𝑟𝑚𝑠 = √
1
𝑇
𝑇
∫0 𝑉𝑚2 ( sin 𝜃 )2 𝑑𝑡
𝑉𝑟𝑚𝑠 =
8
𝑉𝑚
√2
(1)
To compute effective values using this equation, do the following:
Step 1: Square the current (or voltage) curve.
Step 2: Find the area under the squared curve.
Step 3: Divide the area by the length of the curve.
Step 4: Find the square root of the value from Step 3.
9
One cycle of a voltage waveform is shown in Figure
3(a). Determine its effective value.
10
Solution Square the voltage waveform and plot it as in (b). Apply the following
Equation:
𝑣𝑒𝑓𝑓 = √
𝑎𝑟𝑒𝑎 𝑢𝑛𝑑𝑒𝑟 𝑡ℎ𝑒 𝑐𝑢𝑟𝑣𝑒 𝑜𝑓 𝑣 2
𝑏𝑎𝑠𝑒
11
a. Compute the average for the current waveform of Figure 4
b. If the negative portion of Figure 4 is - 3 A instead of -1.5 A, what is the average?
c. If the current is measured by a dc ammeter, what will the ammeter indicate?
d. Determine the effective value of the current .
Fig.4
12
(Answer : 1.73 A )
Determine the averages for Figures 5(a) and (b).
Answers: a. 1.43 A
b. 6.67 V
13
Complex Number Review
By 1893 , Steinmetz had reduced the very complex alternatingcurrent theory to a simple problem
in algebra. The key concept in this simplification was the phasor—a representation based on
complex numbers.
By representing voltages and currents as phasors, Steinmetz was able to define a quantity called
impedance and then use it to determine voltage and current magnitude and phase relationships in
one algebraic operation.
A complex number is a number of the form C = a + jb, where a and b are real numbers and
𝑗 = √−1 . The number a is called the real part of C and b is called its imaginary part. (In
circuit theory, j is used to denote the imaginary component rather than i to avoid confusion with
current i.)
Geometrical Representation
Complex numbers may be represented geometrically, either in rectangular form or in polar form
as points on a two-dimensional plane called the complex plane (Figure 1). The complex number
C = 6 + j8, for example, represents a point whose coordinate on the real axis is 6 and whose
coordinate on the imaginary axis is 8. This form of representation is called the rectangular form.
Complex numbers may also be represented in polar form by magnitude and angle. Thus, C =
(Figure 2) is a complex number with magnitude 10 and angle 53.13°. This magnitude and angle
representation is just an alternate way of specifying the location of the point represented by
C = a + jb.
FIGURE 1 A complex number in
rectangular form.
Conversion between Rectangular and Polar Forms
To convert between forms, note from Figure 16–3 that
C= a + jb (rectangular form) (1)
(polar form) (2)
where C is the magnitude of C. From the geometry of the
triangle,
14
where
and
FIGURE 2 A complex number
in polar form
Fig.3 Polar and rectangular equivalence.
15
EXAMPLE 1 Determine rectangular and polar forms for the complex numbers C, D, V,
and W of Figure 4(a).
Fig.4
16
Solution
Powers of j
Powers of j are frequently required in calculations. Here are some useful
powers:
Addition and Subtraction of Complex Numbers
Addition and subtraction of complex numbers can be performed analytically or
graphically. Analytic addition and subtraction is most easily illustrated in rectangular
form, while graphical addition and subtraction is best illustrated in polar form. For
analytic addition, add real and imaginary parts separately.
Similarly for subtraction. For graphical addition, add vectorially as in Figure 5(a); for
subtraction, change the sign of the subtrahend, then add, as in Figure 5(b).
17
EXAMPLE 2 Given A = 2 + j1 and B + 1 + j3. Determine their sum and difference
analytically and graphically.
FIGURE 5
Multiplication and Division of Complex Numbers
These operations are usually performed in polar form. For multiplication, multiply
magnitudes and add angles algebraically. For division, divide the magnitude of the
denominator into the magnitude of the numerator, then subtract algebraically the angle of
the denominator from that of the numerator.
Thus, given
18
19
CIRCUIT 2
DR. MOHAMMED TAWFEEQ
Series-Parallel ac Networks
1. VOLTAGE DIVIDER RULE
The basic format for the voltage divider rule in ac circuits is exactly the
same as that for dc circuits:
where Vx is the voltage across one or more elements in series that have total
impedance Zx, E is the total voltage appearing across the series circuit, and
ZT is the total impedance of the series circuit.
EXAMPLE 1 Using the voltage divider rule, find the voltage across each
element of the circuit of Fig. 1.
Fig.1
20
2. CURRENT DIVIDER RULE
The basic format for the current divider rule in ac circuits is exactly the
same as that for dc circuits; that is, for two parallel branches with
impedances Z1 and Z2 as shown in Fig. 2,
Fig.2
EXAMPLE 2 Using the current divider rule, find the current through each
impedance of Fig. 3.
Fig.3
21
Series-Parallel
ac Networks
EXAMPLE .1 For the network of Fig.1:
a. Calculate ZT.
b. Determine Is.
c. Calculate VR and VC.
d. Find IC.
e. Compute the power delivered.
f. Find Fp of the network.
Solution
Fig.1
We re-draw the network as shown
In Fig.2
Fig.2
22
c. Referring to Fig.2, we find that VR and VC can be found by a direct
application of Ohm’s law:
d. Now that VC is known, the current IC can also be found using Ohm’s law.
EXAMPLE .2 For the network of Fig.3:
a. If I is 50 A /_30°, calculate I1 using the current divider rule.
b. Repeat part (a) for I2.
c. Verify Kirchhoff’s current law at one node.
Solutions:
a. Redrawing the circuit as in Fig.4, we have
Fig.3
Fig.4
23
EXAMPLE .3 For the network of Fig. 5:
a. Compute I.
b. Find I1, I2, and I3.
c. Verify Kirchhoff’s current law by showing that
I = I1 + I2 + I3
d. Find the total impedance of the circuit.
Fig.5
Solutions:
a. Redrawing the circuit as in Fig.6 reveals a strictly parallel network
where
Fig.6
24
The total admittance is
SOURCE CONVERSIONS
When applying the methods to be discussed, it may be necessary to convert
a current source to a voltage source, or a voltage source to a current source.
This source conversion can be accomplished in much the same manner as
for dc circuits, except now we shall be dealing with phasors and impedances
instead of just real numbers and resistors.
Independent Sources
In general, the format for converting one type of independent source to
another is as shown in Fig. 1.
Fig.1
25
EXAMPLE 1 Convert the voltage source of Fig. 2.(a) to a current
Solution:
Solution:
Fig. 2(b)
EXAMPLE 2 Convert the current source of Fig. 3(a) to a voltage source.
26
Solution
3
3
MESH ANALYSIS
EXAMPLE 1 Using the general approach to mesh analysis, find the current
I1 in Fig. 1.
Fig.1
Solution: When applying these methods to ac circuits, it is good practice to
represent the resistors and reactances (or combinations thereof) by
subscripted impedances. When the total solution is found in terms of these
subscripted impedances, the numerical values can be substituted to find the
unknown quantities.
The network is redrawn in Fig. 2 with subscripted impedances:
Fig.2 Assigning the mesh currents and
Subscripted impedances for the network of Fig. 1.
27
Steps of solution:
1. Assign a current in the clockwise direction to each independent closed loop of the
network. It is not necessary to choose the clockwise direction for each loop current.
2. Indicate the polarities within each loop for each impedance as determined by the
assumed direction of loop current for that loop.
Steps 1 and 2 are as indicated in Fig. 2.
3. Apply Kirchhoff’s voltage law around each closed loop in the clockwise direction.
Using determinants, we obtain
28
Appendix A : DETERMINANTS
Consider the following equations, where x and y are the unknown variables and a1, a2,
b1, b2, c1, and c2 are constants:
(1)
(2)
It is certainly possible to solve for one variable in Eq. (1) and substitute into Eq. (2). That
is, solving for x in Eq. (1),
and substituting the result in Eq. (2),

Using determinants to solve for x and y requires that the following formats be
established for each variable:
(3)
The determinant is :
(4)
Solution for x and y is :
(5)
29
(6)
Example A1 Solve for x and y :

Consider the three following simultaneous equations:
30
D=
Warning: This method of expansion is good only for third-order determinants! It
cannot be applied to fourth- and higher-order systems.
EXAMPLE A2 Evaluate the following determinant:
EXAMPLE A3 Solve for x, y, and z:
Solution
31
To solve for z:
NODAL ANALYSIS
1. Determine the number of nodes within the network.
2. Pick a reference node and label each remaining node with a subscripted value of
voltage: V1, V2, and so on.
3. Apply Kirchhoff’s current law at each node except the reference. Assume that all
unknown currents leave the node for each application of Kirchhoff’s current law.
4. Solve the resulting equations for the nodal voltages.
EXAMPLE 2 Determine the voltage across the inductor for the network of
Fig. 3.
Fig.3
32
Solution 1:
Steps 1 and 2 are as indicated in Fig. 4.
Fig.4
Step 3: Note Fig. 5 for the application of Kirchhoff’s current law to node V1:
Fig.5
Rearranging terms:
Note Fig. 6 for the application of Kirchhoff’s
current law to node V2.
Fig.6
33
Applying Kirchhoff’s current law to the node
V2 of Fig. 4.
34
CIRCUIT 2
DR. MOHAMMED TAWFEEQ
AC Power Calculations
AVERAGE POWER AND POWER FACTOR
For any load in a sinusoidal ac network, the voltage across the load and the current through the load will
vary in a sinusoidal nature. The questions then arise, How does the power to the load determined by the
product v· i vary, and what fixed value can be assigned to the power since it will vary with time?
If we take the general case depicted in Fig. 1 and use the following for v and i:
then the power is defined by
Using the trigonometric identity
Fig.1
A plot of v, i, and p on the same set of axes is shown in Fig. 2. Note that the second factor in the preceding
equation is a cosine wave with an amplitude of VmIm/2 and with a frequency twice that of the voltage or
current. The average value of this term is zero over one cycle, producing no net transfer of energy in any
one direction.
35
Fig.2
The first term in the preceding equation, however, has a constant magnitude (no time dependence) and
therefore provides some net transfer of energy. This term is referred to as the average power, the reason
for which is obvious from Fig. 2. The average power, or real power as it is sometimes called, is the power
delivered to and dissipated by the load. It corresponds to the power calculations performed for dc networks.
The angle (θv - θi) is the phase angle between v and i. Since cos(-a) = cos a,
the magnitude of average power delivered is independent of whether v leads i or i leads v.
Defining as equal to θv -θi, where indicates that only the magnitude is important and the sign is
immaterial, we have
where P is the average power in watts. This equation can also be written
Hence

Resistor
In a purely resistive circuit, since v and i are in phase, θv –θi = θ = 0°, and cosθ = cos 0° = 1, so that
36

Inductor
In a purely inductive circuit, since v leads i by 90°,  θ v - θ i= θ = -°= 90°. Therefore,
The average power or power dissipated by the ideal inductor (no associated resistance) is zero watts.

Capacitor
In a purely capacitive circuit, since i leads v by 90°,  θ v - θ i= θ = -°= 90°. Therefore,
The average power or power dissipated by the ideal capacitor (no associated resistance) is zero watts.
EXAMPLE 1 Find the average power dissipated in a network whose input current and voltage are the
following:
i = 5 sin(ωt + 40°)
v =10 sin(ωt + 40°)
Solution: Since v and i are in phase, the circuit appears to be purely resistive at the input terminals.
Therefore,
EXAMPLE 2: Determine the average power delivered to networks having the following input voltage and
current:
37
Power Factor
In the equation P = (VmIm/2)cos θ, the factor that has significant control over the delivered power level is
the cos θ . No matter how large the voltage or current, if cos θ = 0, the power is zero; if cos θ= 1, the power
delivered is a maximum. Since it has such control, the expression was given the name power factor and is
defined by
Purely resistive load with Fp = 1.
Purely inductive load with Fp = 0.
ADMITTANCE AND SUSCEPTANCE
The discussion for parallel ac circuits will be very similar to that for dc circuits. In dc circuits,
conductance (G) was defined as being equal to 1/R. The total conductance of a parallel circuit was then
found by adding the conductance of each branch. The total resistance RT is simply 1/GT.
In ac circuits, we define admittance (Y) as being equal to 1/Z. The unit of measure for admittance as
defined by the SI system is siemens, which has the symbol S. Admittance is a measure of how well an ac
circuit will admit, or allow, current to flow in the circuit. The larger its value, therefore, the heavier the
38
current flow for the same applied potential. The total admittance of a circuit can also be found by finding
the sum of the parallel admittances. The total impedance ZT of the circuit is then 1/YT; that is, for the
network of Fig. 15.54:
or, since Z =1/Y,
For two impedances in parallel,
or
As pointed out in the introduction to this section, conductance is the reciprocal of resistance, and
The reciprocal of reactance (1/X) is called susceptance and is a measure of how susceptible an element is
to the passage of current through it. Susceptance is also measured in siemens and is represented by the
capital letter B.
For the inductor,
39
For the capacitor,
Complex Power
For any system such as in Fig.1, the power delivered to a load at any instant is defined by
the product of the applied voltage and the resulting current; that is,
p =vi
In this case, since v and i are sinusoidal quantities, let us establish a general case where
The chosen v and i include all possibilities because, if the load is purely
resistive, θ = 0°. If the load is purely inductive or capacitive, θ = 90°
or θ = - 90°, respectively. For a network that is primarily inductive, θ
is positive (v leads i), and for a network that is primarily capacitive, θ is
negative (i leads v).
Fig.1
40
Substituting the above equations for v and i into the power equation
will result in
(1)
where V and I are the rms values.
 RESISTIVE CIRCUIT
For a purely resistive circuit, v and i are in phase, and θ = 0°,
The total power delivered to a resistor will be dissipated in the form of heat.
This is called the real or Active power.
APPARENT POWER
It shown that the power is simply determined by the product of the applied voltage and
current, with no concern for the components of the load; that is, P = VI. It is called the
apparent power and is represented symbolically by S.*
Since it is simply the product of voltage and current, its units are voltamperes,
for which the abbreviation is VA. Its magnitude is determined
41
and the power factor of a system Fp is
The power factor of a circuit, therefore, is the ratio of the average power to the apparent
power. For a purely resistive circuit, we have
P = VI = S
INDUCTIVE CIRCUIT AND
REACTIVE POWER
For a purely inductive circuit , v leads i by 90°, as shown in Fig. . Therefore, Substituting θ = 90° in Eq. (1),
yields
42
The power curve for a purely inductive load.
 The net flow of power to the pure (ideal) inductor is zero over
a full cycle, and no energy is lost in the transaction.
 In general, the reactive power associated with any circuit is define to
be VI sin θ, The symbol for reactive power is Q, and its unit of
measure is
the volt-ampere reactive (VAR)
where θ is the phase angle between V and I.
For the inductor,
43
CAPACITIVE CIRCUIT
For a purely capacitive circuit , i leads v by 90°, . Therefore, Substituting
θ = -90° , in Eq. (1) we obtain
 The net flow of power to the pure (ideal) capacitor is zero
over a full cycle,
The reactive power associated with the capacitor is equal to the peak
value of the pC curve, as follows:
But, since V = IXC and I = V/XC, the reactive power to the capacitor
can also be written
44
The apparent power associated with the capacitor is
THE POWER TRIANGLE
The three quantities average power, apparent power, and reactive power can be
related in the vector domain by

For an inductive load, the phasor power S, as it is often called, is
defined by

For a capacitive load, the phasor power S is defined by
45
Complex power For Series RLC circuit
It is particularly interesting that the equation
(2)
will provide the vector form of the apparent power of a system. Here, V is the voltage
across the system, and I* is the complex conjugate of the current.
Consider, for example, the simple R-L circuit of Fig. , where
The real power (the term real being derived from the positive real axis
of the complex plane) is
Fig.A
and the reactive power is
as shown in Fig. A. Applying Eq. (2) yields
46
as obtained above.
EXAMPLE Find the total number of watts, volt-amperes reactive, and volt-amperes, and
the power factor Fp of the network in Fig.1. Draw the power triangle and find the current
in phasor form.
Fig.1
Solution
The power triangle is shown in Fig. 2.
Since ST = VI = 1000 VA, I = 1000 VA/100 V = 10 A;
and since θ is the angle between the input voltage and current:
Fig.2
The plus sign is associated with the phase angle since the circuit is predominantly
capacitive.
47
14.5 AVERAGE POWER AND POWER FACTOR
For any load in a sinusoidal ac network, the voltage across the load and the current through the load will
vary in a sinusoidal nature. The questions then arise, How does the power to the load determined by the
product v· i vary, and what fixed value can be assigned to the power since it will vary with time?
If we take the general case depicted in Fig. 1 and use the following for v and i:
then the power is defined by
Using the trigonometric identity
Fig.1
A plot of v, i, and p on the same set of axes is shown in Fig. 2. Note that the second factor in the preceding
equation is a cosine wave with an amplitude of VmIm/2 and with a frequency twice that of the voltage or
current. The average value of this term is zero over one cycle, producing no net transfer of energy in any
one direction.
Fig.2
48
The first term in the preceding equation, however, has a constant magnitude (no time dependence) and
therefore provides some net transfer of energy. This term is referred to as the average power, the reason
for which is obvious from Fig. 2. The average power, or real power as it is sometimes called, is the power
delivered to and dissipated by the load. It corresponds to the power calculations performed for dc networks.
The angle (θv - θi) is the phase angle between v and i. Since cos(-a) = cos a,
the magnitude of average power delivered is independent of whether v leads i or i leads v.
Defining as equal to θv -θi, where indicates that only the magnitude is important and the sign is
immaterial, we have
where P is the average power in watts. This equation can also be written
Hence

Resistor

Inductor
In a purely resistive circuit, since v and i are in phase, θv –θi = θ = 0°, and cosθ = cos 0° = 1, so that
In a purely inductive circuit, since v leads i by 90°,  θ v - θ i= θ = -°= 90°. Therefore,
The average power or power dissipated by the ideal inductor (no associated resistance) is zero watts.
49

Capacitor
In a purely capacitive circuit, since i leads v by 90°,  θ v - θ i= θ = -°= 90°. Therefore,
The average power or power dissipated by the ideal capacitor (no associated resistance) is zero watts.
EXAMPLE 1 Find the average power dissipated in a network whose input current and voltage are the
following:
i = 5 sin(ωt + 40°)
v =10 sin(ωt + 40°)
Solution: Since v and i are in phase, the circuit appears to be purely resistive at the input terminals.
Therefore,
EXAMPLE 2: Determine the average power delivered to networks having the following input voltage and
current:
50
Power Factor
In the equation P = (VmIm/2)cos θ, the factor that has significant control over the delivered power level is
the cos θ . No matter how large the voltage or current, if cos θ = 0, the power is zero; if cos θ= 1, the power
delivered is a maximum. Since it has such control, the expression was given the name power factor and is
defined by
Purely resistive load with Fp = 1.
Purely inductive load with Fp = 0.
15.4 VOLTAGE DIVIDER RULE
The basic format for the voltage divider rule in ac circuits is exactly the same as that for dc circuits:
where Vx is the voltage across one or more elements in series that have total impedance Zx, E is the total
voltage appearing across the series circuit, and ZT is the total impedance of the series circuit.
EXAMPLE 15.9 Using the voltage divider rule, find the voltage across each element of the circuit of Fig.
15.40.
51
EXAMPLE 15.10 Using the voltage divider rule, find the unknown voltages VR, VL, VC, and V1 for the
circuit of Fig. 15.41.
9 CURRENT DIVIDER RULE
52
The basic format for the current divider rule in ac circuits is exactly the same as that for dc circuits; that
is, for two parallel branches with impedances Z1 and Z2 as shown in Fig. 15.76,
EXAMPLE 15.15 Using the current divider rule, find the current through each impedance of Fig. 15.77.
EXAMPLE 15.16 Using the current divider rule, find the current through each parallel branch of Fig.
15.78.
53
54
CIRCUIT 2
DR. MOHAMMED TAWFEEQ
SECTION 4 : SELECTED TOPICS IN AC CIRCUITS
Δ-Y, Y- Δ CONVERSIONS
The Δ-Y, Y- Δ conversions for ac circuits will not be derived here since the development
corresponds exactly with that for dc circuits. Taking the Δ -Y configuration shown in
Fig. 1, we find the general equations for the impedances of the Y in terms of those for the
Δ:
(1)
(2)
(3)
Fig.1 Δ -Y configuration
For the impedances of the Δ in terms of those for the Y, the equations are
(4)
(5)
(6)
55
 Note that each impedance of the Y is equal to the product of the impedances in
the two closest branches of the Δ , divided by the sum of the impedances in the
Δ.
 Further, the value of each impedance of the Δ is equal to the sum of the
possible product combinations of the impedances of the Y, divided by the
impedances of theYfarthest from the impedance to be determined.
EXAMPLE 1 Find the total impedance ZT of the network of Fig.1 below,
Fig.1 Converting the upper Δ of a bridge configuration to a Y.
Solution
Recall from the study of dc circuits that if two branches of the Y or Δ were the same, the
corresponding Δ or Y, respectively, would also have two similar branches. In this
example, ZA = ZB. Therefore, Z1 = Z2,
56
And
Replace the Δ by the Y (Fig. 2):
Fig.2 The network of Fig. 1 following the substitution of the Y configuration.
57
AC Network Theorems
1- SUPERPOSITION THEOREM
EXAMPLE 1 Using the superposition theorem, find the current I through the 4-Ω
reactance (XL2 ) of Fig. 1.
Fig.1
Solution
For the re-drawn circuit in Fig.2
Considering the effects of the voltage source E1 (Fig. 3), we have
Fig.2
Fig.3
58
Fig.4
Considering the effects of the voltage source E2 (Fig.4), we have
Fig.5
The resultant current through the 4-Ω reactance XL2 (Fig. 5) is
2. THEVENIN’S THEOREM
Thévenin’s theorem, any two-terminal linear ac network can be replaced with an
equivalent circuit consisting of a voltage source and an impedance in series, as shown
in Fig. 1.
Steps of Solution:
1. Remove that portion of the network across which the Thévenin
equivalent circuit is to be found.
2. Mark (○) the terminals of the remaining two-terminal
network.
3. Calculate ZTh by first setting all voltage and current sources to
zero (short circuit and open circuit, respectively) and then finding
the resulting impedance between the two marked terminals.
4. Calculate ETh by first replacing the voltage and current sources
59
and then finding the open-circuit voltage between the marked
terminals.
5. Draw the Thévenin equivalent circuit with the portion of the
circuit previously removed replaced between the terminals of the
Thévenin equivalent circuit.
Fig.1
EXAMPLE 1 Find the Thévenin equivalent circuit for the network external to resistor R
in Fig. 1.
Fig.1
Solution
Step 1 and 2 (Fig.2)
Fig.2
Step 3 (Fig.3)
Fig.3
60
Step 4 ( Fig.4)
Fig.4
Step 5: The Thévenin equivalent circuit is shown in Fig. 5.
Fig.5
NORTON’S THEOREM
The method described for Thévenin’s theorem will be altered to permit its use with
Norton’s theorem. Since the Thévenin and Norton impedances are the same for a
particular network, certain portions of the discussion will be quite similar to those
encountered in the previous section. Norton’s theorem allows us to replace any twoterminal linear bilateral ac network with an equivalent circuit consisting of a current
source and an impedance, as in Fig. 6.
The Norton equivalent circuit, like the Thévenin equivalent circuit, is applicable at only
one frequency since the reactances are frequency dependent.
Fig.1The Norton equivalent circuit for ac networks.
61
Steps of solution
1. Remove that portion of the network across which the Norton equivalent circuit is to
be found.
2. Mark (○) the terminals of the remaining two-terminal network.
3. Calculate ZN by first setting all voltage and current sources to zero (short circuit and
open circuit, respectively) and then finding the resulting impedance between the two
marked terminals.
4. Calculate IN by first replacing the voltage and current sources and then finding the
short-circuit current between the marked terminals.
5. Draw the Norton equivalent circuit with the portion of the circuit previously
removed replaced between the terminals of the Norton equivalent circuit.
 The Norton and Thévenin equivalent circuits can be found from each other by
using the source transformation shown in Fig. 7. The source transformation is
applicable for any Thévenin or Norton equivalent circuit determined from a
network.
Fig.7
EXAMPLE 18.14 Determine the Norton equivalent circuit for the network external to the 6-Ω resistor of
Fig. 8.
Fig.8
Solution :
Step 1 and 2 : (Fig.9)
62
Fig.9
Step 3 : ( Fig. 10)
Fig.10
Step 4: (Fig.11)
Fig.11
Step 5: The Norton equivalent circuit is shown in Fig. 12.
Fig.12
MAXIMUM POWER TRANSFER THEOREM
When applied to ac circuits, the maximum power transfer theorem states that :
maximum power will be delivered to a load when the load impedance is the conjugate
of the Thévenin impedance across its terminals.
That is, for Fig. 13, for maximum power transfer to the load,
63
Or in rectangular form
Fig.13
The conditions just mentioned will make the total impedance of the circuit appear purely
resistive, as indicated in Fig. 14:
Fig.14
Since the circuit is purely resistive, the power factor of the circuit under maximum power
conditions is 1; that is,
(maximum power transfer)
The magnitude of the current I of Fig. 14 is
The maximum power to the load is
64
EXAMPLE 18.19 Find the load impedance in Fig. 15 for maximum power to the load,
and find the maximum power.
Fig.15
Solution: Determine ZTh [Fig. 16(a)]:
Fig.16
To find the maximum power, we must first find ETh [ Fig. 16(b)], as follows:
65
Resonance
1.SERIES RESONANCE
1.2 SERIES RESONANT CIRCUIT
This section will introduce the very important resonant (or tuned) circuit, which is
fundamental to the operation of a wide variety of electrical and electronic systems in use
today. The resonant circuit is a combination of R,
L, and C elements. The basic configuration for the series resonant circuit appears in Fig. 1
.
The total impedance of this network at any frequency
is determined by
(1)
The resonant conditions described in the introduction
will occur when
(2)
Fig.1
removing the reactive component from the total impedance equation. The total
impedance at resonance is then simply
(3)
representing the minimum value of ZT at any frequency. The subscript s will be
employed to indicate series resonant conditions. The resonant frequency can be
determined in terms of the inductance and capacitance by examining the defining
equation for resonance [Eq. (2)]:
XL = XC
Substituting yields
and
(4)
(5)
66
The current through the circuit at resonance is
which you will note is the maximum current for the circuit of Fig. 1for an applied voltage
E since ZT is a minimum value.
The frequency response characteristic of the circuit is shown in Fig. 2. Note in the figure
that the response is a maximum for the frequency fr (resonant frequency = fs) .
Fig.2
1.2 THE QUALITY FACTOR (Q )
The quality factor Q of a series resonant circuit is defined as the ratio of the reactive
power of either the inductor or the capacitor to the average power of the resistor at
resonance; that is,
Hence at resonant Q factor is given by:
67
or
EXAMPLE 1
a. For the series resonant circuit of Fig. 2, find I, VR, VL, and VC at resonance.
b. What is the Qs of the circuit?
Fig.2
Solution
68
Three phase systems


The voltage induced by a single coil when rotated in a uniform magnetic field is
shown in Figure 1 and is known as a single-phase voltage.
three-phase supply is generated when three coils are placed 120° apart and the
whole rotated in a uniform magnetic field as shown in Figure 2(a). The result is
three independent supplies of equal voltages which are each displaced by 120°
from each other as shown in Figure 2(b).
(a) Basic single –phase generator
(b) Voltage waveform
Fig.1
Fig.2
If the three-phase windings shown in Fig.2 are kept independent then six wires are
needed to connect a supply source to a load (Fig.3) . To reduce the number of wires it is
usual to interconnect the three phases. There are two ways in which this can be done,
these being:
(a) a star, or Y, connection, and (b) a delta, or mesh, connection.
69
Points A/ B/ and C/ may be replaced by N
Fig.3
THE Y-CONNECTED SOURCE ( GENERATOR)
If the three terminals denoted N of Fig. 3 are connected together, the source ( generator)
is referred to as a Y-connected three-phase source ( generator). (Fg.4).
Fig.4
The point at which all the terminals are connected is called the neutral point. If a
conductor is not attached from this point to the load, the system is called a Y-connected,
three-phase, three-wire system. If the neutral is connected, the system is a Y-connected,
three-phase, four-wire system.
The phase voltages and line voltages
The three generated voltages in each phase are:
Where eAN , e BN and eCN are the phase voltages
70
The phasor diagram of the induced voltages is shown
in Fig. 5,where the rms (effective) value of each is
determined by
Fig.5
and
By rearranging the phasors as shown in Fig. 6 and applying a law of vectors which states
that the vector sum of any number of vectors drawn such that the “head” of one is
connected to the “tail” of the next, and that the head of the last vector is connected to the
tail of the first is zero, we can conclude that the phasor sum of the phase voltages in a
three-phase system is zero. That is,
Fig.6
The voltage from one line to another is called a line voltage. On the phasor diagram (Fig.
7) it is the phasor drawn from the end of one phase to another in the counterclockwise
direction.
Fig.7
Fig.8
71
Applying Kirchhoff’s voltage law around the indicated loop of Fig. 7, we obtain
Or
EAB - EAN + EBN = 0
EAB = EAN - EBN = EAN + ENB
The phasor diagram is redrawn to find EAB as shown in Fig. 8. Since each phase voltage,
when reversed (ENB), will bisect the other two, α = 60°. The angle β is 30° since a line
drawn from opposite ends of a rhombus will divide in half both the angle of origin and
the opposite
angle. Lines drawn between opposite corners of a rhombus will also bisect each other at
right angles.
The length x is
Noting from the phasor diagram that angle of EAB = β = 30°, the result is
In words, the magnitude of the line voltage of a Y-connected generator is
the phase voltage:
times
with the phase angle between any line voltage and the nearest phase voltage at 30°.
In sinusoidal notation,
Line Current and Phase Current
The three conductors connected from A, B, and C to the load are called lines. For the Yconnected system, it should be obvious from Fig. 4 that the line current equals the phase
current for each phase; that is,
72
where ϕ is used to denote a phase quantity and g is a generator parameter.

Y-Y connection may take two forms : 3-wire or 4-Wire system as shown in Fig.9

(a) 4-Wire Y-Y connection
(b) 3-Wire Y-Y connection
Fig.9
Phase Sequence
Phase sequence refers to the order in which three-phase voltages are generated. Consider
again Figure 2. As the rotor turns in the counterclockwise direction, voltages are
generated in the sequence eAN, eBN, and eCN and the system is said to have an ABC phase
sequence. On the other hand, if the direction of rotation were reversed, the sequence
would be ACB . Sequence ABC is called the positive phase sequence and is the sequence
generated in practice as shown in Fig.10. It is therefore the only sequence considered in
this section .
Fig.10 Positive sequence ABC
73
EXAMPLE 1 The phase sequence of the Y-connected generator in Fig. 11 is ABC.
a. Find the phase angles θ2 and θ3.
b. Find the magnitude of the line voltages.
c. Find the line currents.
d. Verify that, since the load is balanced, IN = 0.
Fig.11
Solutions:
a. For an ABC phase sequence,
74
THE Y-Δ SYSTEM
There is no neutral connection for the Y-Δ system of Fig. 12. Any variation in the
impedance of a phase that produces an unbalanced system will simply vary the line and
phase currents of the system.
Fig.12
For a balanced load,
The voltage across each phase of the load is equal to the line voltage of the generator for
a balanced or an unbalanced load:
The relationship between the line currents and phase currents of a balanced Δ load is:
75
 For a balanced load, the line currents will be equal in magnitude, as will the phase
currents.
EXAMPLE 2 For the three-phase system of Fig. 12:
a. Find the phase angles θ2 and θ3.
b. Find the current in each phase of the load.
c. Find the magnitude of the line currents.
Fig.12
Solutions:
a. For an ABC sequence,
76
POWER CALCULATION IN 3- PHASE SYSTEM
 Y-Connected Balanced Load
Consider the balanced Y-cnnected
load shown in Fig.13.
Fig.13
Average Power The average power delivered to each phase can be determined by:
Watts
Where
is the phase angle between the voltage and current.
The total power to the balanced load is
77
Reactive Power The reactive power of each phase (in volt-amperes reactive) is
The total reactive power of the load is
or, proceeding in the same manner as above, we have
Apparent Power The apparent power of each phase is
78
EXAMPLE 3 For the Y-connected load of Fig. 14:
Fig.14
a. Find the average power to each phase and the total load.
b. Determine the reactive power to each phase and the total reactive power.
c. Find the apparent power to each phase and the total apparent power.
d. Find the power factor of the load.
Solutions:
a. The average power is
or
79
 Δ-Connected Balanced Load
Consider the Δ-Connected Balanced Load shown in Fig.15
Fig.15
80
81
EXAMPLE 4 For the Δ -Y connected load of Fig. 16, find the total average, reactive,
and apparent power. In addition, find the power factor of the load.
Fig.16
Solution: Consider the Δ and Y separately.
For the Δ:
82
83
Circuit 2
Dr.Mohammed Tawfeeq
AC Power Measurements
Measuring Power in Three-Phase Circuits
1-The Three-Wattmeter Method
Measuring power to a 4-wire Y load requires one wattmeter per phase (i.e., three
wattmeters) as in Figure 1 . As indicated, wattmeter W1 is connected across voltage Van
and its current is Ia. Thus, its reading is
which is power to phase an. Similarly, W2 indicates power to phase bn and W3 to phase
cn. Loads may be balanced or unbalanced. The total power is
If the load of Figure 23–28 could be guaranteed to always be balanced, only one
wattmeter would be needed. PT would be 3 times its reading.
PT = 3 Pϕ
Fig.1 Three-wattmeter connection for a 4-wire load.
84
2-The Two-Wattmeter Method
While three wattmeters are required for a four-wire system, for a three-wire system, only
two are needed. The connection is shown in Figure 2. Loads may be Y or Δ, balanced or
unbalanced. The meters may be connected in any pair of lines with the voltage terminals
connected to the third line. The total power is the algebraic sum of the meter readings.
Fig.2 Two-wattmeter connection. Load may be balanced or unbalanced.
EXAMPLE 1 For Figure 3 , Van = 120 V∠ 0°. Compute the readings of each meter, then
sum to determine total power.
Solution Van=120V∠ 0°. Thus, Vab =208V∠ 30° and Vbc =208V∠ -90°.
Ia = Van/Zan = 120 V∠ 0°/(9 - j12) Ω = 8 A∠ 53.13°. Thus, Ic = 8A∠ 173.13°.
First consider wattmeter 1, Figure 4. Note that W1 is connected to terminals a-b; thus it
has voltage Vab across it and current Ia through it. Its reading is therefore P1 = Vab Ia cos
θ1, where v1 is the angle between Vab and Ia. Vab has an angle of 30° and Ia has an angle
of 53.13°. Thus, θ1 = 53.13° - 30° = 23.13° and P1= (208)(8)cos 23.13°=1530W.
85
Now consider wattmeter 2, Figure 5. Since W2 is connected to terminals c-b, the voltage
across it is Vcb and the current through it is Ic. But Vcb = Vbc = 208 V∠ 90° and
Ic = 8A∠ 173.13°. The angle between Vcb and Ic is thus 173.13° - 90° = 83.13°.
Therefore,
P2 = VcbIccos θ2 = (208)(8)cos 83.13° = 199 W
and PT = P1 + P2 = 1530 + 199 = 1729 W.
Note that one of the wattmeters reads lower than the other. (This is generally the case for
the two-wattmeter method.)
Fig.4
Fig.5
86
Circuit 2
Dr.Mohammed Tawfeeq
Summary of Basic Three-Phase Relationships
Table –1 summarizes the relationships developed so far. Note that in balanced
systems (Y or Δ), all voltages and all currents are balanced.
TABLE –1 Summary of Relationships (Balanced System). All Voltages and Currents
Are Balanced
(a) Y- connection
(b) Δ - connection
87
EXAMPLE 1 For Figure 1, EAN = 120 V∠ 0°.
a. Solve for the line currents.
b. Solve for the phase voltages at the load.
c. Solve for the line voltages at the load.
(a)
(b)
88
PROBLEM 2 For the circuit of Figure , EAB = 208 V∠ 30°.
a. Determine the phase currents.
b. Determine the line currents.
Solution
89
Coupled Circuits
Voltages in Air-Core Coils
To begin, consider the isolated (noncoupled) coil of Figure 1. , the voltage across this coil
is given by vL = L di/dt, where i is the current through the coil and L is its inductance.
Note carefully the polarity of the voltage; the plus sign goes at the tail of the current
arrow. Because the coil’s voltage is created by its own current, it is called a self-induced
voltage.
Fig.1
Now consider a pair of coupled coils (Figure 2). When coil 1 alone is energized as in (a),
it looks just like the isolated coil of Figure 1; thus its voltage is
v11 = L1di1/dt
(self-induced in coil 1)
where L1 is the self-inductance of coil 1 and the subscripts indicate that v11 is the voltage
across coil 1 due to its own current. Similarly, when coil 2 alone is energized as in (b), its
self-induced voltage is
v22 = L2di2/dt
(self-induced in coil 2)
For both of these self-voltages, note that the plus sign goes at the tail of their respective
current arrows.
(a)
(b)
Fig.2 Self and mutual voltages.
90
Mutual Voltages
Consider again Figure 2(a). When coil 1 is energized, some of the flux that it produces
links coil 2, inducing voltage v21 in coil 2. Since the flux here is due to i1 alone, v21 is
proportional to the rate of change of i1. Let the constant of proportionality be M. Then,
v21 = Mdi1/dt
(mutually induced in coil 2)
v21 is the mutually induced voltage in coil 2 and M is the mutual inductance between
the coils. It has units of henries. Similarly, when coil 2 alone is energized as in (b), the
voltage induced in coil 1 is
v12 = Mdi2/dt
(mutually induced in coil 1)
When both coils are energized, the voltage of each coil can be found by superposition; in
each coil, the induced voltage is the sum of its self-voltage plus the voltage mutually
induced due to the current in the other coil. The sign of the self term for each coil is
straightforward: It is determined by placing a plus sign at the tail of the current arrow for
the coil as shown in
Figures 2(a) and (b). The polarity of the mutual term, however, depends on whether the
mutual voltage is additive or subtractive.
Additive and Subtractive Voltages
Whether self- and mutual voltages add or subtract depends on the direction of currents
through the coils relative to their winding directions. This is best described in terms of the
dot convention. Consider Figure 3(a). Comparing the coils here to Figure 4, you can see
that their top ends correspond and thus can be marked with dots. Now let currents enter
both coils at dotted ends. Using the right-hand rule, you can see that their fluxes add. The
total flux linking coil 1 is therefore the sum of that produced by i1 and i2;
Fig.3 When both currents enter dotted terminals, use the + sign for the mutual term in
Equation(1).
Fig.4 Determining dot positions
91
therefore, the voltage across coil 1 is the sum of that produced by i1 and i2.
That is,
(1 a)
For coil 2 :
(1b)
Now consider Figure 5. Here, the fluxes oppose and the flux linking each coil is the
difference between that produced by its own current and that produced by the current of
the other coil. Thus, the sign in front of the mutual voltage terms will be negative.
Fig.5 When one current enters a dotted terminal and the other enters an undotted
terminal, use the - sign for the mutual term in Equation (1).
EXAMPLE Write equations for v1 and v2 of Figure 6 (a).
Fig.6
Solution Since one current enters an undotted terminal and the other enters a dotted
terminal, place a minus sign in front of M. Thus,
92
Write equations for v1 and v2 of Figure 6 (b).
Solution
Coefficient of Coupling
For loosely coupled coils, not all of the flux produced by one coil links the other. To
describe the degree of coupling between coils, we introduce a coefficient of coupling, k.
Mathematically, k is defined as the ratio of the flux that links the companion coil to the
total flux produced by the energized coil.
For iron-core transformers, almost all the flux is confined to the core and links both coils;
thus, k is very close to 1. At the other extreme (i.e., isolated coils where no flux linkage
occurs), k = 0. Thus, 0 ≤ k ˂ 1. Mutual inductance depends on k. It can be shown that
mutual inductance, self-inductances, and the coefficient of coupling are related by the
equation
Inductors with Mutual Coupling
If a pair of coils are in close proximity, the field of each coil couples the other, resulting
in a change in the apparent inductance of each coil. To illustrate, consider Figure 7 (a),
which shows a pair of inductors with selfinductances L1 and L2. If coupling occurs, the
effective coil inductances will no longer be L1 and L2. To see why, consider the voltage
induced in each winding—it is the sum of the coil’s own self-voltage plus the voltage
mutually induced from the other coil. Since current is the same for both coils,
v1 = L1 di/dt + M di/dt = (L1 + M) di/dt,\
which means that coil 1 has an effective inductance of L1 = L1 + M.
Similarly, v2 = (L2 + M) di/dt, giving coil 2 an effective inductance of L2 = L2 + M.
The effective inductance of the series combination [Figure 7 (b)] is then
and
93
Fig.7
EXAMPLE Three inductors are connected in series (Figure 8). Coils 1 and 2 interact,
but coil 3 does not.
a. Determine the effective inductance of each coil.
b. Determine the total inductance of the series connection.
Fig.8
Solution
94
System Analysis: An Introduction
1 Introduction
The growing number of packaged systems in the electrical, electronic, and
computer fields now requires that some form of system analysis appear in
the syllabus of the introductory course. The increasing use of packaged
systems is quite understandable when we consider the advantages
associated with such structures: reduced size, sophisticated and tested
design, reduced construction time, reduced cost compared to discrete
designs, and so forth. The use of any packaged system is limited solely to
the proper utilization of the provided terminals of the system. Entry into
the internal structure is not permitted, which also eliminates the possibility
of repair to such systems.
System analysis includes the development of two-, three-, or multiport
models of devices, systems, or structures. The emphasis in this section will
be on the configuration most frequently subject to modelling techniques:
the two-port system of Fig. 1., and multi port network of Fig.2.
Fig.1 Two port network
Fig.2 Multi-port network
95
Fig,3 Two port transistor
Network.
2 THE IMPEDANCE PARAMETERS Zi AND Zo
For the two-port system of Fig. 4 , Zi is the input impedance between
terminals 1 and 1′, and Zo is the output impedance between terminals 2
and 2′. For multiport networks an impedance level can be defined between
any two (adjacent or not) terminals of the network.
The input impedance is defined by Ohm’s law in the following form:
Fig.4 Defining Zi and Zo.
with Ii the current resulting from the application of a voltage Ei.
The output impedance Zo is defined by
96
with Io the current resulting from the application of a voltage Eo to the
output terminals, with Ei set to zero.
Input and output impedance calculation using Digital Multimeter (DMM)
EXAMPLE 1 Given the DMM measurements appearing in Fig.5, determine
the input impedance Zi for the system if the input impedance is known to
be purely resistive.
Solution:
Sensing resistance
Source
Fig.5
EXAMPLE 2 Using the provided DMM measurements of Fig. 6, determine
the output impedance Zo for the system if the output impedance is known
to be purely resistive.
Solution
Fig.6
97
3 THE VOLTAGE GAINS AvNL, Av, AND AvT
The voltage gain for the two-port system of Fig. 7 is defined by
Fig.7
The capital letter A in the notation was chosen from the term amplification
factor, with the subscript v selected to specify that voltage levels are
involved. The subscript NL reveals that the ratio was determined under noload conditions; that is, a load was not applied to the output terminals
when the gain was determined. The no-load voltage gain is the gain
typically provided with packaged systems since the applied load is a
function of the application.
 In Fig. 8 a load has been introduced to establish a loaded gain that
will be denoted simply as Av and defined by
Fig.8
98
 For all two-port systems the loaded gain Av will always be less than
the no-load gain.
 A third voltage gain can be defined using Fig. 8 since it has an
applied voltage source with an associated internal resistance—a
situation often encountered in electronic systems. The total voltage
gain of the system is represented by AvT and is defined by
 the voltage gain AvT is always less than the loaded voltage gain Av
or unloaded gain AvNL.
The relationship between Ei and Eg can be determined from Fig.8 if we
recognize that Ei is across the input impedance Zi and thus apply the
voltage divider rule as follows:
99
Substituting into the above relationships will result in
EXAMPLE 3 For the system of Fig. 9(a) employed in the loaded amplifier of
Fig. 9 (b):
Fig,9
a. Determine the no-load voltage gain AvNL.
b. Find the loaded voltage gain Av.
c. Calculate the loaded voltage gain AvT.
100
4 THE CURRENT GAINS Ai AND AiT
The current gain of two-port systems is typically calculated from voltage
levels. A no-load gain is not defined for current gain since the absence of RL
requires that Io = Eo /RL = 0 A and Ai = Io / Ii = 0. For the system of Fig.10,
however, a load has been applied, and
Fig.10
Note the need for a minus sign when Io is defined, because the defined
polarity of Eo would establish the opposite direction for Io through RL.
101
The loaded current gain is
and
In general, therefore, the loaded current gain can be obtained directly from
the loaded voltage gain and the ratio of impedance levels, Zi over RL.
If the ratio AiT = Io/Ig were required, we would proceed as follows:
Fig. 11 shows these relations
Fig11
102
 The result is an equation for the loaded current gain of an amplifier
in terms of the nameplate no-load voltage gain and the resistive
elements of the network.
 Note: the larger the level of RL, the less the current gain of a loaded
amplifier.
In the design of an amplifier, therefore, one must balance the desired
voltage gain with the current gain and the resulting ac output power level.
103
THE POWER GAIN AG
For the system of Fig. 11, the power delivered to the load is determined by
Eo2/RL, whereas the power delivered at the input terminals is Ei2 /Ri. The
power gain is therefore defined by
Fig.11
104
EXAMPLE 4 Given the system of Fig. 12 with its nameplate data:
a. Determine Av.
b. Calculate Ai.
c. Increase RL to double its current value, and note the effect on Av and Ai.
d. Find AiT.
e. Calculate AG.
Fig.12
Solution
105
106