Cover Image
close this bookIntroduction to Electrical Engineering - Basic vocational knowledge (Institut für Berufliche Entwicklung, 213 p.)
View the document(introduction...)
View the documentPreface
View the document1. Importance of Electrical Engineering
close this folder2. Fundamental Quantities of Electrical Engineering
View the document2.1. Current
View the document2.2. Voltage
View the document2.3. Resistance and Conductance
close this folder3. Electric Circuits
View the document3.1. Basic Circuit
View the document3.2. Ohm’s Law
close this folder3.3. Branched and Unbranched Circuits
View the document3.3.1. Branched Circuits
View the document3.3.2. Unbranched Circuits
View the document3.3.3. Meshed Circuits
close this folder4. Electrical Energy
View the document4.1. Energy and Power
View the document4.2. Efficiency
View the document4.3. Conversion of Electrical Energy into Heat
View the document4.4. Conversion of Electrical Energy into Mechanical Energy
close this folder4.5. Conversion of Electrical Energy into Light
View the document4.5.1. Fundamentals of Illumination Engineering
View the document4.5.2. Light Sources
View the document4.5.3. Illuminating Engineering
View the document4.6. Conversion of Electrical Energy into Chemical Energy and Chemical Energy into Electrical Energy
close this folder5. Magnetic Field
View the document5.1. Magnetic Phenomena
View the document5.2. Force Actions in a Magnetic Field
close this folder5.3. Electromagnetic Induction
View the document5.3.1. The General Law of Induction
View the document5.3.2. Utilisation of the Phenomena of Induction
View the document5.3.3. Inductance
close this folder6. Electrical Field
View the document6.1. Electrical Phenomena in Non-conductors
close this folder6.2. Capacity
View the document6.2.1. Capacity and Capacitor
View the document6.2.2. Behaviour of a Capacitor in a Direct Current Circuit
View the document6.2.3. Types of Capacitors
close this folder7. Alternating Current
View the document7.1. Importance and Advantages of Alternating Current
View the document7.2. Characteristics of Alternating Current
View the document7.3. Resistances in an Alternating Current Circuit
View the document7.4. Power of Alternating Current
close this folder8. Three-phase Current
View the document8.1. Generation of Three-phase Current
View the document8.2. The Rotating Field
View the document8.3. Interlinking of the Three-phase Current
View the document8.4. Power of Three-phase Current
close this folder9. Protective Measures in Electrical Installations
View the document9.1. Danger to Man by Electric Shock
close this folder9.2. Measures for the Protection of Man from Electric Shock
View the document9.2.1. Protective Insulation
View the document9.2.2. Extra-low Protective Voltage
View the document9.2.3. Protective Isolation
View the document9.2.4. Protective Wire System
View the document9.2.5. Protective Earthing
View the document9.2.6. Connection to the Neutral
View the document9.2.7. Fault-current Protection
View the document9.3. Checking the Protective Measures

3.2. Ohm’s Law

A natural law found by Georg Simon Ohm is called Ohm’s law and it has fundamental importance to electrical engineering. It correlates the quantities current, voltage and resistance. When two quantities are know, the third can be found with the help of this law.

Proof of this fact is given here experimentally by means of an arrangement whose wiring diagram is shown in Fig. 3.4. The following devices and components are required:

· An adjustable voltage source; this may be an accumulator with six cells each of which provides 2 V voltage so that a total of 12 V are available;

· a voltage selector switch with six switch steps;

· two resistors with resistance values of, for example, 10W and 20W;

· two measuring instruments for the measurement of current and voltage.


Fig. 3.4. Circuit for the proof of Ohm’s law

1 - Accumulator with 6 cells of 2 V each
2 - Voltage selector switch
3 - Resistor
4 - Voltmeter
5 - Ammeter

Measuring instruments are indispensable auxiliaries in electrical engineering; they cannot be dealt with in this book, however. Here, it will suffice to know that the current can be measured by means of ammeters and the voltage by means of voltmeters with adequate accuracy.

At first, one switches on the 10W resistance and then, using the voltage selector switch, stepwise applies a voltage of 2 V, 4 V, 6 V, 8 V, 10 V, and 12 V. The current flowing at each switch step is measured, by means of the ammeter and the quantities are recorded in a table of values.

Table of values 1: (for R = 10 W)

Switch step

Voltage U

Current I

1

2 V

0.2 A

2

4 V

0.4 A

5

6 V

0.6 A

4

8 V

0.8 A

5

10 V

1.0 A

6

12 V

1.2 A

A similar test with the 20 resistance results in the following table of values:

Table of values 2: (for R = 20 W)

Switch step

Voltage U

Current I

1

2 V

0.1 A

2

4 V

0.2 A

5

6 V

0,5 A

4

8 V

0.4 A

5

10 V

0.5 A

6

12 V

0.6 A

The experiment is indicative of the following facts:

1. with a given voltage U, the current I is the smaller, the greater the resistance R is. (The resistance offers resistance to the passage of current).

E.g. with U = 6V

measurement series 1: I = 0.6 A with R = 10 W
measurement series 2: I = 0.3 A with R = 20 W

I ~ 1/R

2. With given current I, the voltage U (more exactly: the voltage drop) is the greater, the greater the resistance R is. (When a higher resistance is offered., the charge carriers must give more energy to it!)

E.g. with I = 0.4 A:

measurement series 1:

U = 4 V with
R = 10 W

measurement series 2:

U = 8 V with
R = 20 W

U ~ R

3. The quotient U/I yields the same value for all values of a measurement series:

for measurement series 1: U/I = 10 V/A = 10 W
for measurement series 2: U/I = 20 V/A = 20 W

Obviously, the resistance is constant, its magnitude is thus independent of the current passing through it and of the voltage applied to it.

The result is the important Ohm’s law.

It has the following expression in signs

R = U/I

[R] = V/A = W


(3.1)

where

R

resistance

U

voltage

I

current intensity

When two quantities are known, the third one can be determined; equation must be rearranged in such a way that the unknown is alone, namely,

U = RI

(3.1.a)

I = U/R

(3.1.b)

[U] = W · A = V

[I] = V/W = A

When the measuring results are interpreted graphically, the U-I diagram shown in Fig. 3.5 is obtained. The functional correlation between U and I is linear. The slope of the straight lines is the steeper, the higher the resistance R is.


Fig. 3.5. Current/voltage diagram

U = (I) R1 = 10; R2 = 20W

The correlation between the various units of the quantities current, voltage, resistance and conductance is represented in Table 3.1.

We know that V/A = W; A/V = S = 1/W; V/W = A; W · A = V


Table 3.1. The Units of the Quantities Current, Voltage, Resistance, and Conductance

When sub-units are used in calculations, the unit of the quantity to be determined may also be a sub-unit. When taking a closer look at the 3rd column under V (= volt), you read the unit of the resistance or conductance in the first to fourth lines:

V/mA = MW

V/mA = kW

V/A = W

V/kA = mW

mA/V = mS

mA/V = mS

a/V = S

kA/V = ks

The other units are handled in a similar manner.

Example 3.1

Calculate the resistance and conductance of an electrical soldering iron through which a current of 0.4 A passes when connected to a voltage of 220 V.

Given:

U = 220 V
I = 0.4 A

To be found:

R
G

Solution:

R = U/I
R = 220V/0.4A = 550 V/A
R = 550W

G = 1/R = 1/550W = 0.00182 S
G = 1.82 ms

Example 3.2.

An ammeter has a resistance of 30 mW. Calculate the voltage drop across the instrument when a current of 5 A passes.

Given:

R = 30 m
I = 5 A

To be found:

U

Solution:

R = U/I inverted: U
U = R · I
U = 30 mW · 5 A
U = 0.03 V/A · 5 A = 0.15 V
U = 150 mV

Example 3.3

A voltage of 4.5 V is applied, to a resistance of 15 kW. Calculate the intensity of the current passing through the component.

Given:

R = 15 kW
U = 4.5 V

To be found:

I

Solution:

R = U/I inverted: I
I = U/R
I = 4.5V/15kW
I = 4.5V/15000kW = 0.0003 A
I = 0.3 mA

If a current flows through a consumer and a voltage drops across it because of the current passing through it, then this consumer has a resistance.

The correlation between current, voltage and resistance is described by Ohm’s law. It has fundamental importance to electrical engineering and is written as

R = U/I

An inversion of this law for U or I gives

U= R · I or I = U/R

Questions and problems:

1. A voltage of 220 V is applied to a flat iron while a current of 2.75 A flows through it. Calculate the resistance and conductance of the electrical flat iron.

2. What is the current passing through a resistor of 10 kW at a voltage of 200 V?

3. The power supply system of a workshop is protected by a fuse of 10 A. The voltage is 220 V. Calculate the resistance which the connected loads must have at least.

4. A small horsepower motor has a resistance of 15W during operation. The current intensity is 0.4 A. Calculate the voltage required at the terminals of the motor.