Introduction to Electrical Engineering  Basic vocational knowledge (Institut für Berufliche Entwicklung, 213 p.) 
3. Electric Circuits 

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
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 UI 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) R_{1} = 10; R_{2} = 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 subunits are used in calculations, the unit of the quantity to be determined may also be a subunit. 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.