  Introduction to Electrical Engineering - Basic vocational knowledge (Institut f³r Berufliche Entwicklung, 213 p.)  7. Alternating Current  7.1. Importance and Advantages of Alternating Current 7.2. Characteristics of Alternating Current 7.3. Resistances in an Alternating Current Circuit 7.4. Power of Alternating Current

### 7.2. Characteristics of Alternating Current

Alternating current and alternating voltage are produced in generators. In Section 5.3.2. the generator principle has been pointed out. Below, explanations in great detail are given.

When turning a coil (represented by a conductor loop in Fig. 7.1.) in a magnetic field, a voltage which can drive a current is induced in this coil. The direction can be determined with the help of the right-hand rule. At a constant rotational speed, magnitude and direction of the induced voltage is dependent on the position of the conductor loop.

Fig. 7.1. Model of an alternating current generator

1 - North pole
2 - South pole a) Position of the loop of conductor at a certain instant b) Position of the loop of conductor after half a revolution

In horizontal position, the entire magnetic flux penetrates the loop, DF/Dt has the smallest value and the induced voltage is equal to zero. In vertical position, (parallel to the magnetic fields), the rate of flux variation DF/Dt is maximum and the induced voltage has the highest value. Upon further rotation, the voltage again drops and after half a revolution reaches the value of zero. In the further course, the voltage changes its direction and reaches its negative maximum value in vertical position. After one full revolution, the initial condition is again reached, the voltage has dropped to zero and another cycle with exactly the same course can begin. In Fig. 7.2., the described conditions are shown for eight selected positions of the conductor. The behaviour of the curve is a sine function. Fig. 7.2. Development of a sinusoidal voltage or current curve

1 - North pole
2 - South pole

Frequently, the angle is expressed in circular measure; the correlation is easily given with the circumference of a circle having the radius r = 1. We have:

Circumference of a circle C = 2 rp = 2p; the perigon of a circle is a = 360°; this means that and, hence, , , etc.

The voltage and current course represented in Fig. 7.2. is called oscillation, cycle or wave. Each wave is made up of a positive (l) and a negative (2) half wave. The time for a full revolution of the conductor loop is called time of oscillation or duration of a cycle (formula sign T). The angular velocity at which the conductor loop rotates is the angle through which the loop has passed in a certain unit of time. With a perigon of 360° = 2p, the duration of a cycle T is required. The angular velocity t is usually called angular frequency w 1) in electrical engineering and it is written as

1) w Greek letter omega

w = 2p/T

(7.1.)

where:

 w angular frequency T duration of a cycle 2p circular measure of the circle (= circumference of the unit circle)

The product of wt is the angle a at time t and at me angular frequency w; a = wt. This angle is called phase angle or, in short, phase. The phase is the condition of oscillation given at a certain time which is repeated at the same time intervals.

When, for example, t = T/4, then a = w · T/4. Using = 2p/T from equation (7.1), we have a = 2p/T. . This is the phase angle at a quarter of a cycle. The number of cycles or wanes produced in a certain time (e.g. t = 1 s) is called frequency f of the alternating voltage or alternating current. The greater the duration of a cycle T, the smaller the frequency f. Fig. 7.3. shows two different curves for a duration of 1 s. The curve drawn as a solid line is a cycle, its duration T = 1 s. The curve represented by a dashed line covers five cycles (the frequency is higher), and the duration of a cycle T = 1/5 s = 0.2 s

f = 1/T

(7.2.) Fig.7.3 Alternating currents with frequencies of 1 Hz and 5 Hz

where:

 f frequency T duration of a cycle

In honour of the German physicist Heinrich Hertz (1857 - 1894), the unit of frequency is called Hertz = Hz.

The following subunits are frequently required

1 kHz = 1 kilohertz = 103 Hz
1 MHz = 1 megahertz = 106 Hz
1 GHz = 1 gigahertz = 109 Hz

The correlation between the angular frequency and frequency is given by the equations (7.1.) and (7.2.).

w = 2pf

(7.3.)

Another rarely used characteristic for the electrical wave is the wavelength l 1). Wavelength is the length of a wave measured in a unit of length (e.g. m, km, cm, mm). As the electrical energy propagates with light velocity

1) l Greek letter lambda

c = 300,000 km/s

the distance over which a wave extends can be calculated on the basis of a given frequency. We have:

c = lf;

after inversion we obtain for the wavelength

l = c/f

(7.4)

where:

 l wavelength c velocity of propagation (c = 300,000 km/s = 3 · 108 m/s) f frequency

[l] = m/s · 1/s = m

The magnitude of the alternating voltage or the alternating current can be determined on the basis of the sine curve developed in Fig. 7.2. The maximum value occurring at 90° and 270° is called peak value, maximum value or amplitude and is designated by (or . All other values which very continuously and, thus, are different at any time are called instantaneous values and are designated by u (or i).

When the maximum value is known the instantaneous values can be determined at any time. The general equation of a sinusoidal alternating current is

u = sin wt

(7.5.)

where:

 u instantaneous value maximum value, peak value or amplitude sin wt factor of the sine function at angle

Besides an exclusively mathematical treatment, alternating current processes are frequently represented in diagrams which offer a better survey. Particularly suitable for this purpose are vector and line diagrams.

Vector diagram

The alternating voltage or the alternating current is represented by a pointer (vector) capable of rotating whose length corresponds to the peak value. This pointer rotates anticlockwise at the angular velocity w. The pointer position at any time indicates the position of the conductor loop. In order to determine the instantaneous of voltage or current for any desired position, a straight line is drawn from the pointer tip to the horizontal axis which passes through the centre of the circle. The length of the straight line corresponds to the instantaneous value in question (Fig. 7.4.a). The particular advantage of the vector diagram is lucidity; a disadvantage is the fact that the conditions can be represented only for one point of time or for a few selected instants.

Line diagram

The alternating voltage or alternating current is represented by a sine curve from which the values for all instants can be read off (Fig. 7.4.b). The line diagram can be developed from the vector diagram in the following manner. Close by the vector diagram, a horizontal line is drawn. This line is divided into periods, the smallest one being equal to the duration of one cacle T, or into angular degrees up to . Perpendicular lines are drawn from the points of division which resemble the lines on the vector diagram as to size and direction, when connecting the end points of the perpendicular lines, the sine curve is obtained. Fig. 7.4. shows the construction described. For reasons of clearness, the vector diagram (Fig. 7.4.a) only shows the vectors for 30°, 60°, 90°, 180° and 225°. The advantage of the line diagram is the possibility of representing all of the instantaneous values; a disadvantage is the restricted lucidity especially when several curves have to be represented. Fig. 7.4. Graphical representation, of the alternating current and alternating voltage

a) Vector diagram
b) Line diagram

1 Periode = 1 cycle

Example 7.1.

A sinusoidal alternating voltage having a frequency of f = 50 Hz has a peak value = 311 V. Draw the vector diagram for the angles of rotation wt = 30°, 45°, 60° and develop the line diagram on this basis! Further, determine the angular frequency, the duration of a cycle T, and the wavelength l!

Given:

= 311 V
f = 50 Hz
wt = 30°, 45°, 60°

To be found:

vector and line diagrams
w
T
l

Solution: (Fig. 7.5.)

w = 2 f
w = 2 · 3.14 · 3.14 · 50 1/s
w = 314 1/s

T = 1/f
T = 1/50 s = 0.02 s
T = 20 ms

l = c/f l = 6,000 km Fig. 7.5. Vector diagram and line diagram for example 7.1.

Like direct current, alternating current lends itself to the operation of heating and thermal appliances as well as incandescent lamps. Alternating current motors are used for the conversion into mechanical energy. Since alternating current, in contrast to direct current, continuously changes its magnitude and direction, a mean value must be found which has the same effect as a corresponding direct current. This mean value is called effective value (or root mean square value = r.m.s. value).

In Section 4.1., the energy conversion and the power of the current have been represented in general and the relation P = I2Rt =U2/R · t has been derived. Evidently, it is the square of the current intensity and of the voltage that matters. In case of alternating current, we have to square all instantaneous values. Of all squared values of a cycle, the arithmetic mean must be formed. In this way, the square of the effective value is obtained. This is illustrated by Fig. 7.6. The sine curve has been squared. Fig. 7.6. Determination of the effective value of the alternating current

In this way, all values, even those of the negative half wave, become positive. The squared sine cure is also sinusoidal but has double the frequency of the original curve. The arithmetic mean is I/2.

This value is the square of the effective value I, hence,

SUP>2 = SUP>2/2; and, with respect to voltage, SUP>2 = SUP>2/2

By extraction of roots in the equation, we obtain or Since and , we obtain for the current

I = 0.707 or = 1.414 I

(7.6.a)

and for the voltage

U = 0.707 or = 1.414 U

(7.6.b)

A sinusoidal alternating current causes the same thermal effect as a direct current of intensity I if its peak value I is 1.414 times the current intensity of the direct current. Analogous conditions hold for the sinusoidal alternating voltage.

Example 7.2.

Two usual mains voltages have the following values:

a) U = 220 V;
b) = 535 V. Determine the peak voltage for a) and the r.m.s. voltage for b)

Given:

a) U = 220 V
b) = 535 V

To be found:

a) BR>b) U

 Solution: a) = 1.414 U b) U = 0.707 = 1.414 U · 220 V U = 0.707 · 535 V = 311 V U = 380 V

In the general use of alternating voltages and alternating currents, always effective (r.m.s.) values are involved.

When turning a coil in a magnetic field, a voltage is induced in the coil which changes periodically with respect to magnitude and direction. The voltage and current path produced during one revolution is called oscillation, cycle or wave. The most important characteristics of a wave are duration of a cycle, frequency, angular frequency, wavelength, phase, instantaneous value, peak value or maximum value or amplitude. Any sinusoidal quantity can be described mathematically, namely, analytically by an equation, graphically by a vector diagram or a line diagram.

The value of an alternating quantity (voltage or current) is called effective value (or r.m.s. value) if the same thermal effect is produced as caused by a corresponding direct quantity (voltage or current). Current and voltage data without special designation are always effective values in alternating current engineering.

Questions and problems:

1. Determine the duration of a cycle, angular frequency, and wavelength of the oscillations with the following frequencies

 a) technical alternating current f =50 Hz b) test tone for electrical paths f = 16 · 2/3 Hz c) test tone for telecommunication installations f = 1 kHz d) transmitter frequency of a long-wave transmitter f = 182 kHz e) transmitter frequency of a short-wave transmitter f = 6115 kHz f) transmitter frequency of a VHF transmitter f =97.15 MHz

2. What are the advantages and disadvantages of the vector diagram as compared, with the line diagram?

3. Draw the vector and line diagrams of an alternating voltage whose frequency is 50 Hz, peak value 156 V and zero-phase angle 30°! Select a suitable scale!

4. Determine the peak values of the following sinusoidal quantities:

a) 6 V
b) 380 V
c) 15 kV
d) 200 µA
e) 10 A
f) 25 A

5. The peak values of the following sinusoidal quantities are given s

a) 311 V
b) 70.7 mV
c) 8.5 A
d) 4.25 mA

Find the effective values!