1.1 Mathematics, Units and Metric Prefixes
Key Takeaways
- The ACMA Standard syllabus states that a formula sheet is supplied in the examination and may be used to answer any question, so transposition matters more than memorisation.
- The syllabus declares the symbols E and V interchangeable, because E is an abbreviation of electromotive force (EMF).
- Each named prefix step in the ladder pico, nano, micro, milli, unit, kilo, mega, giga is a factor of 1,000, which is three decimal places.
- A pass on the Standard theory paper is 70% of the 50 questions, which is 35 correct answers in 60 minutes.
- Worked conversions to know cold: 3.525 MHz is 3,525 kHz or 3,525,000 Hz; 0.000047 F is 47 µF; 2,200 pF is 2.2 nF.
1.1 Mathematics, Units and Metric Prefixes
ACMA Exam Focus: Syllabus item 3.1 requires you to understand addition, subtraction, multiplication and division, understand fractions, percentage and decimal notation, recall units and sub-units (mega, kilo, UNIT, micro and pico), and understand how to calculate using simple formulae. Almost no question on the theory paper is labelled as a maths question, yet nearly every calculation in Parts 4 to 7 falls apart if you drop a factor of one thousand.
1. What the Standard paper actually expects
The theory paper is 50 multiple-choice questions in 60 minutes, and the syllabus states that a recognition certificate is issued to candidates who correctly answer 70% of the questions in both the theory and regulations papers. Seventy per cent of 50 is 35 correct answers.
Two statements in the official syllabus make the arithmetic far less intimidating than candidates fear.
A formula sheet is supplied. The syllabus says plainly that the formula sheet "will be provided to candidates in the examination and may be used to answer any question", and that reference materials supplied to candidates may also include look-up tables, diagrams, photographs and the relevant class licence. You are therefore not being tested on your memory of formulae. You are being tested on choosing the right formula, substituting consistent units and transposing it into the form you need.
The symbols $E$ and $V$ are interchangeable. The syllabus carries an explicit note: the supplied formulas frequently use $E$ for voltage because $E$ abbreviates electromotive force (EMF), while many educational documents use $V$. For the purposes of this assessment the two symbols mean the same thing. So $E = I \times R$ and $V = I \times R$ are the same law, and $P = E \times I$ is the same statement as $P = V \times I$. Do not lose a mark because a formula-sheet expression looked unfamiliar.
2. Quantities, symbols and units used across the paper
| Quantity | Usual symbol | Unit | Unit symbol |
|---|---|---|---|
| Voltage / EMF / potential difference | $E$ or $V$ | volt | V |
| Current | $I$ | ampere | A |
| Resistance | $R$ | ohm | $\Omega$ |
| Reactance / impedance | $X$ / $Z$ | ohm | $\Omega$ |
| Power | $P$ | watt | W |
| Capacitance | $C$ | farad | F |
| Inductance | $L$ | henry | H |
| Frequency | $f$ | hertz | Hz |
| Period (time for one cycle) | $T$ | second | s |
| Wavelength | $\lambda$ | metre | m |
| Charge | $Q$ | coulomb | C |
| Energy | $W$ | joule | J |
Note the two collisions that trip candidates up: the letter C is both the quantity capacitance and the unit coulomb, and E serves as both voltage and (in some texts) energy. Context decides. In an Ohm's law or power expression, $E$ is always voltage.
3. The metric prefix ladder
| Prefix | Symbol | Multiplier | Power of ten | Typical amateur use |
|---|---|---|---|---|
| giga | G | 1 000 000 000 | $10^{9}$ | 1.296 GHz on the 23 cm band |
| mega | M | 1 000 000 | $10^{6}$ | 14.100 MHz, a 1 M$\Omega$ resistor |
| kilo | k | 1 000 | $10^{3}$ | 3 525 kHz, a 4.7 k$\Omega$ resistor |
| (the unit itself) | — | 1 | $10^{0}$ | 100 W, 50 $\Omega$, 13.8 V |
| milli | m | 0.001 | $10^{-3}$ | 250 mA of receive current, 5 mV |
| micro | µ | 0.000 001 | $10^{-6}$ | 47 µF, 10 µH |
| nano | n | 0.000 000 001 | $10^{-9}$ | a 2.2 nF bypass capacitor |
| pico | p | 0.000 000 000 001 | $10^{-12}$ | a 100 pF trimmer capacitor |
The syllabus names only mega, kilo, UNIT, micro and pico — "UNIT" simply meaning the unprefixed quantity, such as a plain volt, ohm or farad. In practice milli and nano appear constantly on Australian component markings and in worked questions, so learn the whole ladder.
The golden rule: each named step above is a factor of 1,000, which is three decimal places. Moving down the ladder towards smaller prefixes, the number gets bigger and the decimal point moves right. Moving up towards larger prefixes, the number gets smaller and the point moves left.
4. Worked conversions of the type the paper uses
Worked Example 1 — Frequency
Express 3.525 MHz in kilohertz and in hertz.
Mega to kilo is one step down the ladder, so multiply by 1,000 (move the point three places right): Kilo to the plain unit is one more step: This is a real 80 m frequency, and it is the reason band edges are quoted three different ways in different documents.
Worked Example 2 — Capacitance
Express 0.000047 F in microfarads, then in nanofarads.
Farads to microfarads is one step down, so multiply by $10^{6}$ (six places right): Microfarads to nanofarads is one further step:
Worked Example 3 — Small capacitance
Express 2 200 pF in nanofarads and microfarads.
Pico to nano is one step up the ladder, so divide by 1,000: Nano to micro is one more step up: All three markings describe the same physical capacitor. Older schematics label it 0.0022 µF; a modern parts list calls it 2.2 nF.
Worked Example 4 — Current
A transceiver draws 250 mA on receive. Express this in amperes.
Milli to the plain unit is one step up, so divide by 1,000:
5. Fractions, decimals and percentages
A percentage is simply a fraction with a denominator of 100. To find $x%$ of a number, multiply by $x/100$. To reduce a number by $x%$, multiply by $(100-x)/100$.
- Pass mark. $70% \times 50 = 0.70 \times 50 = 35$ questions correct.
- Optimum working frequency. Syllabus item 7.6 states that the OWF is 15% lower than the maximum usable frequency. If the MUF over a path is 21 MHz, then which points you at the 17 m or 20 m band rather than 15 m.
- Amplifier efficiency. Syllabus item 5.6 asks you to determine amplifier efficiency from the DC input power and the RF output power: A power amplifier drawing 13.8 V at 15 A takes $13.8 \times 15 = 207$ W of DC. If it delivers 100 W of RF, efficiency is $(100 \div 207) \times 100 \approx 48%$, and the remaining 107 W leaves as heat. That figure is why heatsinks and fans exist.
6. Transposing simple formulae
Transposition means rearranging an equation to make a different quantity the subject. The rule is symmetrical: whatever operation you apply to one side, apply to the other.
| Given | Rearranged forms |
|---|---|
| $E = I \times R$ | $I = \dfrac{E}{R}$ and $R = \dfrac{E}{I}$ |
| $P = E \times I$ | $I = \dfrac{P}{E}$ and $E = \dfrac{P}{I}$ |
| $f = \dfrac{1}{T}$ | $T = \dfrac{1}{f}$ |
| $\lambda = \dfrac{300}{f_{\text{MHz}}}$ | $f_{\text{MHz}} = \dfrac{300}{\lambda}$ |
A useful memory aid is the triangle: write the product term on top and the two factors side by side underneath. Cover the quantity you want, and the remaining layout tells you whether to divide or multiply.
Worked Example 5 — Power from supply figures
A 13.8 V power supply feeds a transceiver that draws 22 A on transmit.
Worked Example 6 — Frequency from period
An AC waveform has a period of 20 ms. Convert to base units first: 20 ms = 0.020 s. That is the Australian mains frequency, and syllabus item 4.1 expects you to recognise it.
7. Significant figures and sensible rounding
- Convert first, calculate second. Put every quantity into base units (or keep prefixes rigorously consistent) before you substitute. Mixing milliamperes with volts and expecting watts is the single most common arithmetic error on this paper.
- Carry precision through, round at the end. Rounding an intermediate result then feeding it into the next step compounds the error.
- Two or three significant figures is normally enough. The distractors on a multiple-choice paper are widely separated, usually by a factor of ten or by a sign, so an answer of 17.85 MHz and one of 17.9 MHz will point to the same option.
- Sanity-check the order of magnitude. A dipole for 40 m is about 20 m long, not 20 cm. A receive current is milliamperes, not kiloamperes. A bypass capacitor is nanofarads, not farads.
8. Exam checklist for item 3.1
- Know that the formula sheet is supplied and that you may use it for any question.
- Treat $E$ and $V$ as the same quantity — voltage in volts.
- Recite the ladder: pico, nano, micro, milli, unit, kilo, mega, giga, each step a factor of 1,000.
- Convert MHz to kHz to Hz, and µF to nF to pF, without hesitation.
- Be able to take a percentage of a number and to reduce a number by a percentage (the 15% MUF-to-OWF step).
- Transpose $E = IR$, $P = EI$ and $f = 1/T$ in both directions.
A capacitor is marked as 0.000047 F. What is this value expressed in microfarads?
The maximum usable frequency (MUF) over an HF path is calculated as 21 MHz. Applying the relationship given in the syllabus, what is the optimum working frequency (OWF)?
The formula sheet supplied in the ACMA Standard theory examination expresses Ohm's law as E = I x R. Which statement about the symbol E is correct?