Electrical Engineering BMT & UFC › Physical Fundamentals

Calculating with Units

We start with the basics of physical units. Without a proper understanding of units we cannot work, because every physical quantity only makes sense together with a unit.

Units are a convention, i.e. a voluntary agreement. Their aim is the international standardisation of measures. The length of a table, for example, can be given in the units metre, centimetre or inch. The numerical value in front of the unit then has a different meaning in each case. We use SI units exclusively, i.e. metres for length.

The length of an 85 cm long example table is given as \(l = 0.85\,\mathrm{m}\). This contains

\(l\): symbol for a length
\(0.85\): numerical value
\(\mathrm{m}\): SI unit metre

These three pieces of information are the minimum required to describe the length of a table correctly according to the convention. The answer "0.85" is therefore wrong, because the symbol and the unit are missing. The statements \(l = 0.85\,\mathrm{m}\) and \(l = 85\,\mathrm{cm}\) are both correct for lengths. However, giving the value in metres is common in technical systems. The metre is the so-called "SI unit", i.e. the officially used unit.

Whenever new electrical quantities are introduced in this text, you will get their symbol and unit listed in a table. So for each physical quantity you learn three items of vocabulary by heart: the quantity (e.g. time), the symbol (e.g. \(t\)) and the unit (e.g. second). This allows you to write a time of 3 seconds correctly as \(t = 3\,\mathrm{s}\).

Prefixes are used to describe very large and very small numerical values clearly. You have already met the prefix centi (c) above for the length of the table. A very small length could be described as \(l = 0.00003\,\mathrm{m}\), as \(l = 3 \cdot 10^{-5}\,\mathrm{m}\) or as \(l = 30\,\text{µm}\). Here the prefix \(\text{µ} = 10^{-6}\) is used. Prefixes save writing many zeros. The following prefixes are commonly used in electrical engineering:

Prefix	Power of ten
Femto (f)	\(10^{-15}\)
Pico (p)	\(10^{-12}\)
Nano (n)	\(10^{-9}\)
Micro (µ)	\(10^{-6}\)
Milli (m)	\(10^{-3}\)
Centi (c)	\(10^{-2}\)
Kilo (k)	\(10^{3}\)
Mega (M)	\(10^{6}\)
Giga (G)	\(10^{9}\)
Tera (T)	\(10^{12}\)

Prefixes act like a multiplication by the associated power of ten. The weight force \(F = 1\,\mathrm{kN}\) (kilonewton) corresponds to \(1 \cdot 10^{3}\,\mathrm{N}\), since kilo is assigned to the power of ten \(10^{3}\).

Why should I learn this?

In electrical engineering, many values are given with prefixes. You can therefore only solve many problems if you master the prefixes. In a problem, for example, a voltage is given as \(U = 5\,\mathrm{kV}\). This means: the voltage \(U\) has the value 5000 volts. Without knowing the meaning and the associated numerical value of the "k", you cannot solve this problem. In the solutions to problems you sometimes have to give the values with prefixes as well.

Many students underestimate this problem. In exams I see many solutions that are wrong because the prefix was handled incorrectly. So please take this topic seriously. Of course, you also need it for many other subjects in your studies.

SI units

In addition to calculating with prefixes, you must also be able to convert units. There are 7 base units, the SI units. The base units are:

m: metre for measuring length
kg: kilogram for mass
s: second for measuring time
A: ampere for measuring electric current
K: kelvin for measuring temperature
cd: candela for measuring luminous intensity
mol: mole for the amount of substance in chemistry

All other units are composed of these units by multiplication and division. This is described using the example of the unit of force: force in mechanics has the unit newton (N). The following applies:

\[ \begin{gathered} \text{Acceleration } [a] = \frac{\mathrm{m}}{\mathrm{s}^2} \\[4pt] \text{Mass } [m] = \mathrm{kg} \\[4pt] F = m \cdot a \;\text{ with }\; [F] = [m] \cdot [a] = \mathrm{kg} \cdot \frac{\mathrm{m}}{\mathrm{s}^2} = \mathrm{N} = [F] \\[4pt] 1\,\mathrm{N} = 1\,\frac{\mathrm{kg} \cdot \mathrm{m}}{\mathrm{s}^2} \end{gathered} \]

When we write a formula for units, the symbols are put in square brackets. We write the statement "The unit of force is the newton" as the formula \([F] = \mathrm{N}\). The force \(F\) is calculated from the product of mass and acceleration. The unit of force is calculated from the product of the unit of mass and the unit of acceleration. In this way, derived units such as the newton N for force are formed from SI units.

If at the end of a force calculation you get a unit such as

\[ [F] = \frac{\mathrm{kg} \cdot \mathrm{m}}{\mathrm{s}} \quad (\text{Caution: wrong!}) \]

then a second is missing in the denominator. So your formula for the calculation was wrong. You probably also have an error in the numerical value. That is why units are often used to check a calculation. If the unit in a calculation is correct, the formula is probably correct too, and then hopefully the numerical value is also correct.

Only quantities with the same unit can be added or subtracted. If in a force calculation you get the result \(F_{\mathrm{Ges}} = 3\,\mathrm{N} + 2\,\mathrm{s}\), you have definitely made a mistake. The second term must also have the unit newton, otherwise the terms cannot be added.

Terms with different units can be multiplied or divided.

As an example, I will show you how problems with units and prefixes can be solved. Experience shows that you make fewer mistakes with this calculation scheme.

Problem 1: Calculate the force F (worked example).

\[ \begin{gathered} \text{Given: } a = 130\,\frac{\mathrm{km}}{\mathrm{s}^2} \;\text{ and }\; m = 33\,\mathrm{g} \\[6pt] \text{Converting the prefixes into powers: } a = 130 \cdot 10^{3}\,\frac{\mathrm{m}}{\mathrm{s}^2};\; m = 33 \cdot 10^{-3}\,\mathrm{kg} \\[6pt] F = m \cdot a = 130 \cdot 10^{3}\,\frac{\mathrm{m}}{\mathrm{s}^2} \cdot 33 \cdot 10^{-3}\,\mathrm{kg} = (130 \cdot 33) \cdot \left(10^{3} \cdot 10^{-3}\right) \frac{\mathrm{m}}{\mathrm{s}^2}\,\mathrm{kg} \\[6pt] F = 4290 \cdot 10^{(3-3)}\,\mathrm{kg}\,\frac{\mathrm{m}}{\mathrm{s}^2} = 4290 \cdot 10^{0}\,\mathrm{N} \\[6pt] F = 4290\,\mathrm{N} = 4.29 \cdot 10^{3}\,\mathrm{N} = 4.29\,\mathrm{kN} \end{gathered} \]

First, the prefixes of the input values \(a\) and \(m\) are converted into powers. The units are given as SI units. For this, for example, the gram g must be converted into kilograms kg. Then the numerical values 130 and 33 are multiplied. The exponents of the powers are added. The brackets in the middle line are not mathematically necessary; they only indicate which values are combined in the next step. At the end, the result can be given in exponential notation or with a prefix. The last three results are all equally correct. You will encounter all three forms in practice.

Aside: Our predecessors were asleep at one point: the metre is denoted by a small "m". So is the prefix milli. If you see an "m" in a formula, is it a metre or a milli? We solve the problem with a convention: the prefix always comes before all units. So for 1000 metres we always write km instead of mk. If you see a small "m" in front of the units, it stands for milli. If the small "m" comes after the prefixes, it stands for "metre". To be on the safe side, we always put the metre at the end of all units.

Please solve the following problems according to this scheme:

Problem 2: Calculate the distance s of the motion

\[ \text{Given: speed } v = 0.2\,\frac{\mathrm{km}}{\mathrm{h}},\; t = 30\,\mathrm{min} \]

Solution:

\[ \begin{gathered} v = 0.2\,\frac{\mathrm{km}}{\mathrm{h}} = 0.2 \cdot 10^{3}\,\frac{\mathrm{m}}{\mathrm{h}} = \frac{0.2}{3600} \cdot 10^{3}\,\frac{\mathrm{m}}{\mathrm{s}} = 55.6 \cdot 10^{-6} \cdot 10^{3}\,\frac{\mathrm{m}}{\mathrm{s}} = 55.6 \cdot 10^{-3}\,\frac{\mathrm{m}}{\mathrm{s}} \\[6pt] t = 30\,\mathrm{min} = 1800\,\mathrm{s} = 1.8 \cdot 10^{3}\,\mathrm{s} \\[6pt] \text{Constant motion: } s = v \cdot t \\[6pt] s = v \cdot t = 55.6 \cdot 10^{-3}\,\frac{\mathrm{m}}{\mathrm{s}} \cdot 1.8 \cdot 10^{3}\,\mathrm{s} = 100 \cdot 10^{3} \cdot 10^{-3}\,\frac{\mathrm{m}}{\mathrm{s}}\,\mathrm{s} = 100\,\mathrm{m} \end{gathered} \]

Problem 3: What is wrong with this formula?

\[ F + t = 3\,\mathrm{N} + 25\,\mathrm{s} \]

Solution: Terms with different units cannot be added.

Further information

YouTube: converting physical units (in German)
Problems with solutions (in German)