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Fundamentals of Alternating Current
In this chapter, we lead you through a study of the mathematics and physics of alternating current (AC) circuits. After completing this chapter you should be able to:
Develop a familiarity with sinusoidal functions. Write the general equation for a sinusoidal signal based on its amplitude, frequency, and phase shift. Define angles in degrees and radians. Manipulate the general equation of a sinusoidal signal to determine its amplitude, frequency, phase shift at any time. Compute peak, RMS, and average values of voltage and current. Define root-mean-squared amplitude, angular velocity, and phase angle. Convert between time domain and phasor notation. Convert between polar and rectangular form. Add, subtract, multiply, and divide phasors. Discuss the phase relationship of voltage and current in resistive, inductive, and capacitive loads. Apply circuit analysis using phasors. Define components of power and realize power factor in AC circuits. Understand types of connection in three-phase circuits.
FOCUS ON MATHEMATICS
This chapter relates the application of mathematics to AC circuits, covering complex numbers, vectors, and phasors. All these three concepts follow the same rules.
REFERENCES
- Stephan J. Chapman, Electric Machinery Fundamentals, Third Edition, McGraw-Hill,
- Stephan J. Chapman, Electric Machinery and Power System Fundamentals, McGraw- Hill, 2002.
- Bosels, Electrical Systems Design, Prentice Hall.
- James H. Harter and Wallace D. Beitzel, Mathematics Applied to Electronics, Prentice Hall.
2 Chapter 12
12.1 INTRODUCTION
The majority of electrical power in the world is generated, distributed, and consumed in the form of 50- or 60-Hz sinusoidal alternating current (AC) and voltage. It is used for household and industrial applications such as television sets, computers, microwave ovens, electric stoves, to the large motors used in the industry. AC has several advantages over DC. The major advantage of AC is the fact that it can be transformed, however, direct current (DC) cannot. A transformer permits voltage to be stepped up or down for the purpose of transmission. Transmission of high voltage (in terms of kV) is that less current is required to produce the same amount of power. Less current permits smaller wires to be used for transmission. In this chapter, we will introduce a sinusoidal signal and its basic mathematical equation. We will discuss and analyze circuits where currents i ( t ) and voltages v ( t ) vary with time. The phasor analysis techniques will be used to analyze electric circuits under sinusoidal steady-state operating conditions. Single-phase power will conclude the chapter.
12.2 SINUSOIDAL WAVEFORMS
AC unlike DC flows first in one direction then in the opposite direction. The most common AC waveform is a sine (or sinusoidal) waveform. Sine waves are the signal whose shape neither is nor altered by a linear circuit, therefore, it is ideal as a test signal. In discussing AC signal, it is necessary to express the current and voltage in terms of maximum or peak values, peak-to-peak values, effective values, average values, or instantaneous values. Each of these values has a different meaning and is used to describe a different amount of current or voltage. Figure 12-1 is a plot of a sinusoidal wave. The correspondence mathematical form is
v ( t ) = Vp cos( wt +θ) (12.1)
Where Vp is the peak voltage, ω = 2 π f is the angular speed expressed in
radians per second (rad/s), f is the frequency expressed in Hertz (Hz), t is the time
expressed in second (s), and θ is phase of the sinusoid expressed in degrees.
The function (Figure 12-1) starts at a value of 0 at 0 o, and rise smoothly to a maximum of 1 at 90o. They then fall, just as they rose, back to 0o^ at 180 o. The negative peak is reached three quarters of the way at 270 o. The function then returns symmetrically to 0 o^ at 360o.
4 Chapter 12
The following comparative table of degree and radian provides measure for some angles we often deal with:
Figure 12-2 Radian and arc length.
Table 12-1 Angles in Degree and Radian
Angle (deg) (^) 0 45 90 180 270 360 Angle (rad) (^) (^0) π/4 π/2 π 3 π/2 2 π
12.2.2 Peak and Peak-to-Peak Values
During each complete cycle of AC signal there are always two maximum or peak values, one for the positive half-cycle and the other for the negative half- cycle. The peak value is measured from zero to the maximum value obtained in either the positive or negative direction. The difference between the peak positive value and the peak negative value is called the peak-to-peak value of the sine wave. This value is twice the maximum or peak value of the sine wave and is sometimes used for measurement of ac voltages. The peak value is one-half of the peak-to-peak value.
12.2.3 Instantaneous Value
The instantaneous value of an AC signal is the value of voltage or current at one particular instant. The value may be zero if the particular instant is the time in the cycle at which the polarity of the voltage is changing. It may also be the same as the peak value, if the selected instant is the time in the cycle at which the voltage or current stops increasing and starts decreasing. There are actually an infinite number of instantaneous values between zero and the peak value.
A
B
Fundamentals of Alternating Current 5
12.2.4 Average Value
The average value of an AC current or voltage is the average of all the instantaneous values during one alternation. They are actually DC values. The average value is the amount of voltage that would be indicated by a DC voltmeter if it were connected across the load resistor. Since the voltage increases from zero to peak value and decreases back to zero during one alternation, the average value must be some value between those two limits. It is possible to determine the average value by adding together a series of instantaneous values of the alternation (between 0° and 180°), and then dividing the sum by the number of instantaneous values used. The computation would show that one alternation of a sine wave has an average value equal to 0. times the peak value. The formula for a average voltage is
Vav = 0. 636 V max (12.5)
Where V av is the average voltage for one alteration, and Vmax is the maximum or peak voltage. Similarly, the formula for average current is
I (^) av = 0. 636 I max (12.6)
Where I (^) av is the average current for one alteration, and I (^) max is the maximum or peak current.
12.2.5 Effective Value
This is the value of AC signal that will have the same effect on a resistance as a comparable value of direct voltage or current will have on the same resistance. It is possible to compute the effective value of a sine wave of current to a good degree of accuracy by taking equally spaced instantaneous values of current along the curve and extracting the square root of the average of the sum of the squared values. For this reason, the effective value is often called the “root-mean- square” (RMS) value. Therefore,
I (^) eff = Average ofthesumofthesquaresof I ins (12.7)
The effective or rms value ( I (^) eff ) of a sine wave of current is 0.707 times the maximum value of current ( I (^) max ). Thus, I (^) eff = 0.707 × I (^) ma x. When I (^) eff is known, we may find I (^) max by using the formula I (^) max = 1.414 × I (^) eff. We might wonder where the constant 1.414 comes from. To find out, examine Figure and read the following explanation. Assume that the DC in Figure is maintained at 1 A and the resistor
Fundamentals of Alternating Current 7
Example 12-
Express each of the following frequencies in Hertz
a) 40 cycles in 4.0 seconds b) 80 cycles in 200 milliseconds c) 1000 revolutions in 0.5 seconds d) 600 rotations in 1 minute
Solution:
a) 40/4.0 = 10 cycles per second = 10 Hz b) 80/0.2 = 400 cycles per second = 400 Hz c) 1000/0.5 = 2000 cycles per second = 4000 Hz (4 kHz) d) 600/60 = 10 cycles per second = 10 Hz
Example 12-
Express each of the following as angular velocity in radians per second
a) 80 rad in 10 s b) 2.5 krad in 50 s c) 400 rad in 200 s d) 40 Mrad in 10 s
Solution:
a) ω = 80/10 = 8 rad/s
b) ω = 2500/50 = 500 rad/s
c) ω = 400/200 = 2.0 rad/s
d) ω = (40× 10 6 )/10 = 4.0 rad/s
Example 12-
Express each of the following frequencies as angular velocity in radians per second
a) 60 Hz b) 500 Hz c) 10 kHz d) 1 MHz
8 Chapter 12
Solution:
a) ω = 2π×60 = 377 rad/s
b) ω = 2π×500 = 3141.5 rad/s
c) ω = 2π× (10× 10 3 ) = 62.83 krad/s
d) ω = 2π× (1.0× 10 6 ) = 6.28 Mrad/s
12.2.7 Period
The period of a waveform is the time required for completing one full cycle. It is measured in seconds. In Figure 15-1, the sinusoidal waveform is plotted as a
function of the argument^ ω t , and the periodic nature of the sine wave is evident.
The function repeats itself every 2 π radians, and its period is therefore 2 π radians.
The relationship between time ( T ) and frequency ( f ) is indicated by the formulas
f
T
Example 12-
Express each of the following periods in seconds
a) 500 Hz b) 90 kHz c) 900 MHz d) 5 Hz
Solution: Use Equation (12.10)
a) T^ = 2 ms b) T = 1/(90× 10 3 ) = 11.11 μs c) T = 1/(900× 10 6 ) = 1.11 ps d) T = 0.2 s
12.2.8 Phase
When two sinusoidal waves, such as those represented by Figure 12-3, are precisely in step with one another, they are said to be in phase. To be in phase,
10 Chapter 12
Figure 12-3 The sine wave V P sin ( ω t + θ) leads V P sin ω t.
12.3 PHASORS
We have learnt from the previous section how to define and express in a single equation the magnitude, frequency, and phase shift of a sinusoidal signal. Any linear circuit that contains resistors, capacitors, and inductors do not alter the shape of this signal, nor its frequency. However, the linear circuit does change the amplitude of the signal (amplification or attenuation) and shift its phase (causing the output signal to lead or lag the input). The amplitude and phase are the two important quantities that determine the way the circuit affects the signal. Accordingly, signal can be expressed as a linear combination of complex sinusoids. Phase and magnitude defines a phasor (vector) or complex number. The phasor is similar to vector that has been studied in mathematics. Figure 12-4 shows how AC sinusoidal quantities are represented by the position of a rotating vector. As the vector rotates it generates an angle. The location of the vector on the plane surface is determined by the magnitude (length) of the vector and by the generated angle. Representing sinusoidal signals by phasors is useful since circuit analysis laws such as KVL and KCL and familiar algebraic circuit analysis tools, such as series and parallel equivalence, voltage and current division are applicable in the phasor domain, which have been studied in DC circuits can be applied. We do not need new analysis techniques to handle circuits in the phasor domain. The only difference is that circuit responses are phasors (complex numbers) rather than DC signals (real numbers). In order to work with these complex numbers without drawing vectors, we first need some kind of standard mathematical notation. There are two basic forms of complex number notation: polar and rectangular.
v
ω t
V P sin ( ω t + θ) V P sin^ ω t
θ
V P
Fundamentals of Alternating Current 11
Figure 12-4 (a) Magnitude of a sine wave. (b) A vector with its end fixed at the origin and rotating in a counterclockwise (CCW) direction representing the varying conditions of the AC signal.
12.3.1 Polar Form
Polar form is where the length (magnitude) and the angle of its vector denote a complex number. Standard orientation for vector angles in AC circuit calculations defines 0 o^ as being to the right (horizontal), making 90o^ straight up, 180 o^ to the left, and 270o^ straight down. Vectors angled “down” can have angles represented in polar form as positive numbers in excess of 180 or negative numbers less than 180 (Figure 12-5). For example, a vector angled ∠ 270 o (straight down) can also be said to have an angle of -90o.
Figure 12-5 Standard orientation for vector angles.
0 o
90 o
180 o
270 o
(^3 )
1
2 4
5 1 9
(^9 )
7
6
(^4 )
6
7 (a) (b)
Fundamentals of Alternating Current 13
Figure 12-7 Phasor diagram of Example 12-5.
12.3.2 Rectangular Form
The horizontal and vertical components denote a complex number. The angled vector is taken to be the hypotenuse of a right triangle, described by the lengths of the adjacent and opposite sides. These two dimensional figures (horizontal and vertical) are symbolized by two numerical figures. In order to distinguish the horizontal and vertical dimensions from each other, the vertical is prefixed with a lower-case “ i ” (in pure mathematics) or “ j ” (in electronics). Figure 12-8 shows that a point on a complex plane located by a phasor could be described in rectangular form.
Figure 12-8 A point on the complex plane located by the phasor 4+ j 3 expressed in the rectangular form.
+Real
+Imaginary
-Real
-Imaginary
-45 o
4+ j 3
5
14 Chapter 12
15.3.3 Transforming Forms
Consider the triangle in Figure 12-9. The hypotenuse is labeled as C. The
angle is θ. A represents the real value and B represents the imaginary value of the
rectangular form.
V =C∠ θ =A+ j B (12.15)
A complex number is the sum of a real number and an imaginary number [A = Real (A) + j Imaginary (A)]. We know what real numbers are since we use them very often. What are imaginary numbers? The answer to this question is related to another question. What is the square root of minus one ( − 1 )? The answer is j! Any number of the form j is called imaginary number. Sometimes, the letter i is used to define the imaginary number. Electrical engineers use j because i is used for instantaneous current.
Example 12-
Express − 16 as an imaginary number.
Solution: Write
− 16 = - 1 × 16
Replace − 1 with j , then
− 16 = j 4
Focus on Mathematics
Complex Algebra
16 Chapter 12
Example 12-
Convert each of the following polar phasors into their rectangular form.
a) V = 100 Vrms ∠ 60 o, and
b) V = 100 Vrms ∠- 60 o
Solution:
a) V = 50 Vrms +j86.6Vrms
b) V = 50 Vrms-j86.6Vrms
Example 12-
Convert each of the following polar phasors into their rectangular form.
a) V = 2 Vrms ∠ 45 o
b) V = 240 Vrms ∠- 160 o
Solution:
a) V =1.414Vrms +j1.414Vrms
b) V =-225.526Vrms - j82.084Vrms
12.3.4 Euler’s Identity
Euler’s identity forms the basis of phasor notation. It is named after the Swiss Mathematician Leonard Euler. It states, the identity defines the complex exponential ej^ θ^ as a point in the complex plane. It may be represented by real and imaginary components:
e j θ^^ = cosθ + j sin θ (12.20)
Figure 12-10 shows how the complex exponential may be visualized as a point (or vector, if referenced to the origin) in the complex plane. The magnitude of e j^ θ is equal to 1
Fundamentals of Alternating Current 17
Figure 12-10 Euler’s identity.
e j θ= 1 (12.21)
since
cos θ +sinθ= cos^2 θ^ +sin^2 θ = 1 (12.22)
Remember that writing Euler’s identity corresponds to equating the polar form of a complex number to its rectangular form
Ae j θ^^ = A cosθ+ j Asinθ=A∠ θ (12.23)
Simply, Euler’s identity is a trigonometric relationship in the complex form. To see how complex numbers are used to represent sinusoidal signals, we may rewrite the expression for a generalized sinusoid using Euler’s equation:
A cos ( wt + θ ) = Re( A ej (^ wt + θ )) (12.24)
Equation (12.24) is simplified as
A cos ( wt + θ ) = Re( A ej^ (^ wt +^ θ )) = Re( A ejθejwt ) (12.25)
sin θ
cos θ
Fundamentals of Alternating Current 19
To subtract phasor quantities, express each in rectangular form
- Change the sign of both the real and the imaginary part of the phasor to be subtracted.
- Add the phasors following the steps in the previous box.
Example 12-
Subtract 10 - j 4 from 15 + j
Solution: Change the signs of 10 – j 4. Accordingly the answer is
-(10- j 4) = -10 + j 4, Now add
j
j
j
The answer is 5 + j 12
Focus on Mathematics
Subtracting Phasors
20 Chapter 12
Rectangular Form
To multiply phasor quantities in rectangular form, multiply the numbers as if they were two binomials
- Distribute the real part of the first complex number over the second complex number.
- Distribute the imaginary part of the first complex number over the second complex number.
- Replace j^2 with –1.
- Combine like terms.
- Form the product as a phasor written in rectangular form.
Example 12-
Multiply 3 + j 2 and 4 – j 5
Solution: Follow steps 1 to 5
Distribute (3 + j 2) over (4- j 5). This means
(3 + j 2)(4- j 5) = 12 – j 15 + j 8 – j^210
Replace j^2 with –1. This yields
12 – j 15 + j 8 + 10
Combine like terms to obtain the answer
22 – j
Focus on Mathematics
Multiplying Phasors
