|Introduction to Electrical Engineering - Basic vocational knowledge (Institut für Berufliche Entwicklung, 213 p.)|
|1. Importance of Electrical Engineering|
|2. Fundamental Quantities of Electrical Engineering|
|2.3. Resistance and Conductance|
|3. Electric Circuits|
|3.1. Basic Circuit|
|3.2. Ohms Law|
|3.3. Branched and Unbranched Circuits|
|3.3.1. Branched Circuits|
|3.3.2. Unbranched Circuits|
|3.3.3. Meshed Circuits|
|4. Electrical Energy|
|4.1. Energy and Power|
|4.3. Conversion of Electrical Energy into Heat|
|4.4. Conversion of Electrical Energy into Mechanical Energy|
|4.5. Conversion of Electrical Energy into Light|
|4.5.1. Fundamentals of Illumination Engineering|
|4.5.2. Light Sources|
|4.5.3. Illuminating Engineering|
|4.6. Conversion of Electrical Energy into Chemical Energy and Chemical Energy into Electrical Energy|
|5. Magnetic Field|
|5.1. Magnetic Phenomena|
|5.2. Force Actions in a Magnetic Field|
|5.3. Electromagnetic Induction|
|5.3.1. The General Law of Induction|
|5.3.2. Utilisation of the Phenomena of Induction|
|6. Electrical Field|
|6.1. Electrical Phenomena in Non-conductors|
|6.2.1. Capacity and Capacitor|
|6.2.2. Behaviour of a Capacitor in a Direct Current Circuit|
|6.2.3. Types of Capacitors|
|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|
|8. Three-phase Current|
|8.1. Generation of Three-phase Current|
|8.2. The Rotating Field|
|8.3. Interlinking of the Three-phase Current|
|8.4. Power of Three-phase Current|
|9. Protective Measures in Electrical Installations|
|9.1. Danger to Man by Electric Shock|
|9.2. Measures for the Protection of Man from Electric Shock|
|9.2.1. Protective Insulation|
|9.2.2. Extra-low Protective Voltage|
|9.2.3. Protective Isolation|
|9.2.4. Protective Wire System|
|9.2.5. Protective Earthing|
|9.2.6. Connection to the Neutral|
|9.2.7. Fault-current Protection|
|9.3. Checking the Protective Measures|
· Inductance and coil
A wire usually wound on ferromagnetic core is called coil. This component stores energy at a certain current. The storage capacity for magnetic energy is called inductance of a coil.
L = N · F/I
inductance (more precisely self-inductance)
number of turns
L = Vs/A = H
The following subunits are most frequently used:
1 mH = 1 milihenry = 10-3 H
1 µH = 1 microhenry = 10-6 H
The storage capacity of the coil is dependent on the number of turns, the dimensions and the permeability of the core. From the equations (5.5.) and 5.15.) we have
L = N2 · µ · A/I
number of turns
permeability (material constant)
coil (core) cross-section
length of coil
Like resistors, coils can be connected in series or in parallel. In series connection according to fig. 5.27., the same current passes through the coils with the individual inductances of L1 and L2. In case of a current variation, voltage proportional to the individual inductances of the coils in induced in the latter. The equivalent inductance of this arrangement is
Lequ = L1 + L2
This equation has the same structure as the equation for the determination of Requ of a series connection of resistors.
Fig. 5.27. Series connection of two coils (Lers = Lequ)
Fig. 5.28. Parallel connection of two coils (Lers = Lequ)
The parallel connection of two coils is shown in Fig. 5.28. The same voltage is applied to the two coils and the equivalent inductance is analogous to the equivalent resistance of resistors connected in parallel.
1/Lequ = 1/L1 + 1/L2
From the equations (5.22.) and (5.23.), the following general statement can be derived: In a series connection of coils, the equivalent inductance is always greater than the greatest individual inductance and in a parallel connection of coils, the equivalent inductance is always smaller than the smallest individual inductance.
Two coils having the inductances of 1.5 H and 5 H have to be connected in series and then in parallel. Determine the equivalent inductances for these two types of connections!
L1 = 1.5 H
L2 = 3 H
To be found:
Lequ in series connection and in parallel connection
series connection of L1 and L2
Lequ = L1 + L2
Lequ = 1.5 H + 3 H
Lequ = 4.5 H
parallel connection of L1 and L2
1/Lequ = 1/L1 + 1/L2 = (L2 + L1)/(L1L2)
Lequ = L1L2/(L1 + L2)
Lequ = (1.5 H · 3H)/(1.5 H + 3 H) = 4.5 H/4.5
Lequ = 1 H
· Behaviour of a coil in a direct-current circuit
A coil is connected to a direct voltage source according to Fig. 5.29. (switch position 1). At the instant of switching on (time t1), current starts flowing. The maximum current limited by R cannot flow immediately because self-induction counteracts any current change. After a short time, the current has reached a certain value and the magnetic flux the value proportional to the current. The current causes a voltage drop at the resistor R; consequently, the voltage across the coil is reduced. In the following time, the current is not allowed to rise as quickly as immediately after the instant of switching on. All this shows that, after switching on, the current first increases rapidly and then more and more slowly while the coil voltage first drops rapidly and then more and more slowly.
Fig. 5.29. Circuit for switching on and off of a direct voltage in a coil
Now, the voltage source is to be switched off from the coil (switch position 2). At the instant of switching off (time t2), the current passing through the coil is not immediately interrupted because self-induction opposes any current change. The starting change in current causes a self-induced voltage which, according to the Lena rule, is so directed that it counteracts the cause of origin. An induced current is driven in the same direction as before when the voltage source was connected. Now, the magnetic field gradually dies out and the stored magnetic energy is converted into heat energy in resistor R.
The course taken by current and voltage during switching on and off is shown in Fig. 5.30. It is evident that at the instant of switching on and at the instant of switching off the coil voltage reaches its highest value rapidly and, after some time, drops to zero. The current, however, changes its value only slowly in switching. In coils, there are no sudden current changes.
Fig. 5.30. Behaviour of current and voltage in a coil when a direct voltage is being switched on and off
When switching off a coil, the following should be observed: The energy stored in the magnetic field is only maintained by a current flow. In case of an interruption (instant of switching off), the field must disappear and the energy be converted into another form of energy. An instantaneous interruption (Dt = 0) according to the law of induction leads to a high induced voltage which can attain values of such a magnitude that connected components and the insulation of the coil winding may be destroyed.
When circuits include coils, caution is imperative at any time. In switching off, dangerous overvoltages can occur. They are prevented by closing the current path for the induced current. For this purpose, a resistor, a capacitor or a semiconductor diode is connected in parallel to the coil.
The coil is a storage element. The energy stored by a coil in the form of magnetic energy is
W = L/2 · I2
W = (V · s)/A ··A2 = V · A · s = W · s
In a magnetic field considerably higher energies can be stored than in a dielectric field (see Section 6.2.2.). Therefore, large force actions can be achieved with magnetic fields.
A coil having an inductance of L = 10 H carries a current of 5 A. Calculate the energy stored!
I = 5 A
L = 10 H
To be found:
W = L/2 · I2
W = 10/2 · (V·· s)/A · 52A2 = 5·· 25 W·· s
W = 125 W·· s
Any magnetic flux variation causes an electromotive force (electromagnetic induction). It is directed, in such a way that the magnetic field caused by the induced current counteracts the cause of its origin. A distinction is made between induction of rest and induction of motion. The electromagnetic induction forms the basis of a large number of technical applications including generators, motors, transformers and measuring instruments.
When a coil carries a current, the latter is associated with a magnetic flux. When the current varies, the magnetic flux also varies inducing an electromotive force. When this takes place in another, galvanically separated coil, this is called mutual induction; when it takes place in the same coil, it is called self-induction. In any case, the magnitude of the induced electromotive force is proportional to the rate of current variation. The proportionality factor in mutual induction is called mutual inductance M, that in self-induction is called inductance L.
The characteristic circuit parameter of a coil is the inductance; its unit is henry. The equivalent inductance in series and parallel connections of coils is expressed by the equations (5.22.) and (5.23.).
Loss-free coils (in practice, tow-loss coils are only possible) allow a direct current to pass without any restriction. In switching on and off, however, a certain sluggishness is imparted to the current by the self-induced voltage, that is to say, there are no sudded current changes in a coil. When switching off a coil, very high over-voltages may occur which have to be limited in the circuit.
A current-carrying coil stores energy in the form of magnetic energy by means of which great force actions can be attained.
Questions and problems:
1. Describe in which way induced voltages are brought about!
2. In which way are self-induction and mutual induction physically related?
3. Compare generator principle and motor principle and explain the relations!
4. Explain the mode of action of an eddy-current brake with the belt of the law of induction!
5. Which property of a coil is described by inductance?
6. Explain the course taken by current and voltage in a coil when it is switchen on a off a direct voltage source!
7. Why can a very high overvoltage occur in a coil when it is switched off a voltage source? By which measures can this overvoltage be limited or avoided?
8. The inductance of a coil is 4 H. The current flowing through the coil changes uniformly by - 150 mA within 5 ms. Calculate the self-induced voltage!
9. Calculate the inductance of a coil when a self-induced voltage of 100 V is brought about with a rate of current change of 50 A/s
10. Calculate the energy that is required for the building up of the magnetic field of a coil having an inductance of 500 mH when the coil carries a current of 2 A!