Chapter_3_Alternating Voltages and Currents.pdf

School of Electrical and Computer Engineering
Instructor: Mr. Lidetu A.
9 April, 2025
Fundamentals of Electricity and
Electronics Devices
Chapter-Three
Alternating current and voltage
Chapter 3| Alternating current and voltage Year II, Sem II

Outline
• Introduction
• Sinusoidal ac voltage characteristics and definitions
• The sinusoidal waveforms
• General format for the sinusoidal voltage and current
• Phase relations
• Average value
• Effective (rms) value
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3.1 Introduction
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• The time-varying voltage commercially available in large quantities is
commonly called the AC voltage.
• (The letters AC are an abbreviation for alternating current.)
Fig: Alternating waveforms.
• The term alternating indicates that the waveform alternates between two
prescribed levels in a set time sequence.

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• Sinusoidal AC voltages are available from a variety of sources.
• The most common source is the typical home outlet, which provides an
AC voltage that originates at a power plant.
3.2 Sinusoidal AC Voltage Characteristics And Definitions
Generation
• As shown in the figure (a), an AC generator (also called an alternator) is the primary
component in the energy-conversion process.
FIG: Various sources of AC power: (a) generating plant; (b) portable AC generator; (c) wind-power station; (d) solar panel; (e) function generator.

Definitions
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▪ In the sinusoidal waveform shown in the figure, the vertical scaling is measured
in volts or amperes, while the horizontal scaling is represented in units of time.
Waveform: The path traced by a quantity. (time, position, degrees, radians,
temperature, and so on.
Instantaneous value: The magnitude of a waveform at any instant of time; (e1, e2).
Peak amplitude: The maximum value of a waveform as measured from its
average, or mean value (Em)
Peak value: The maximum instantaneous value of a function as measured from the
zero-volt level.
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.
Periodic waveform: A waveform that continually repeats itself after the same time
interval.
Period (T): The time of a periodic waveform.
Cycle: The portion of a waveform contained in one period of time.
Frequency (f): The number of cycles that occur in 1s.

EXAMPLE: For the sinusoidal waveform in Figure
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a. What is the peak value?
b. What is the instantaneous value at 0.3s and 0.6s?
c. What is the peak-to-peak value of the waveform?
d. What is the period of the waveform?
e. How many cycles are shown?
f. What is the frequency of the waveform?
Solutions:
a. 8 V.
b. At 0.3s,-8 V; at 0.6s, 0 V.
c. 16 V.
d. 0.4 s.
e. 3.5 cycles.
f. 2.5 cps, or 2.5 Hz.

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3.3 The Sinusoidal Waveform
• The sinusoidal waveform is the only alternating waveform whose shape is unaffected by the
response characteristics of R, L, and C elements.
• The unit of measurement for the horizontal axis can be time (as appearing in the figures thus far),
degrees, or radians.

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• For comparison purposes, two sinusoidal voltages are plotted in Fig. using
degrees and radians as the units of measurement for the horizontal axis.

3.4 General Format For The Sinusoidal Voltage Or Current
The basic mathematical format for the sinusoidal waveform is
where Am is the peak value of the waveform and α is the unit of measure for the horizontal axis
FIG: Basic sinusoidal function.
• The general format of a sine wave can also be written
α = ωt
where

• For electrical quantities such as current and voltage, the
general format is:
i = Im sin wt = Imsin α
E = Em sin ωt = Em = sin α
• The angle at which a particular voltage level is attained
can be determined by rearranging the equation

EXAMPLE 13.8 Given e = 5 sin α, determine e at α = 40 ° and α = 0.8 π.
Solution: For α = 40 °,
e = 5 sin 40 ° = 5(0.6428) =3.21 V

For α = 0.8 π ,
α( ) ° = (180°)/ 0.8 π = ° 144
and e = 5 sin144 = 5(0.5878) = 2.94V
EXAMPLE:
a. Determine the angle at which the magnitude of the sinusoidal
function v = 10 sin 377t is 4 V.
b. Determine the time at which the magnitude is attained.
EXAMPLE: Given i = 6 × 10-3 sin 1000t , determine i at t = 2 ms.
Ans: 5.46 mA

3.5 Phase Relations
• If the waveform is shifted to the right or left of 0°, the expression becomes
where θ is the angle in degrees or radians that the waveform has been shifted
• If the waveform passes through the horizontal axis with a positive-going (increasing with
time) slope before 0°, as shown in Fig., the expression is

➢If the waveform passes through the horizontal axis with a
positive-going slope after 0 , ° as shown in Fig., the
expression is
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At ωt = α = 0 °, the magnitude is determined by Am sin .
Finally, at ωt =α = 0°, the magnitude is Am sin (-θ), which, by a
trigonometric identity, is -Am sin θ .

• If the waveform crosses the horizontal axis with a positive-going slope
90° (π/ 2) sooner, as shown in Fig., it is called a cosine wave; that is,
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Or

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➢ The terms leading and lagging are used to indicate the
relationship between two sinusoidal waveforms of the
same frequency plotted on the same set of axes.
In Fig., the cosine curve is said to lead the sine curve by 90°, and the sine curve
is said to lag the cosine curve by 90°. The 90° is referred to as the phase angle
between the two waveforms.

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EXAMPLE 13.12 What is the phase relationship between the sinusoidal
waveforms of each of the following sets?
a. υ = + 10 sin (ωt + 30°)
i = + 5 sin ( ωt + 70°)
b. i = 15 sin (ωt+ 60°)
υ = 10 sin (ω t - 20°)
Solutions: a.

3.6 Average Value
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Since the area under the positive (or negative) pulse is,
we can easily determine the average value of the positive (or negative)
region of a sine wave pulse by applying
➢ The average is the
same as for a full pulse.

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EXAMPLE: Determine the average value of the waveform in Fig.
Solution:
The peak-to-peak value of the sinusoidal function is 16 mV + 2 mV = 18 mV. The peak
amplitude of the sinusoidal waveform is, therefore, 18 mV/2 = 9 mV. Counting down 9
mV from 2 mV (or 9 mV up from -16 mV) results in an average or dc level of -7 mV, as
noted by the dashed line

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3.7 EFFECTIVE (Rms) VALUES
• This section begins to relate dc and ac quantities with respect to the
power delivered to a load.
• It determine the amplitude of a sinusoidal ac current required to
deliver the same power as a particular dc current.
➢ the equivalent dc value of a sinusoidal current or voltage is
1/√2 or 0.707 of its peak value.
• The relationship between the peak value and the rms value is the same for voltages,

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EXAMPLE: Find the rms values of the sinusoidal waveform in each part in Fig.

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EXAMPLE: The 120 V dc source in Fig. (a) delivers 3.6 W to the load.
Determine the peak value of the applied voltage ( Em) and the
current (Im) if the ac source Fig. (b) is to deliver the same power to
the load.
Im = 42.42 mA
Vm = 169 68 V

Chapter 1| Basics of electricity Year II, Sem II

Chapter_3_Alternating Voltages and Currents.pdf

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