Associate CET · Competency 9.0

Mathematics

Competency 9.0 is the arithmetic every other competency leans on: Ohm's law and the other formulas on ETA's sheet, wavelength and frequency, number systems, Boolean algebra, decibels, graphs, SI prefixes and powers of ten. This page follows ETA's list in order, with section 10 as job practice. It teaches each idea once and links the site's tools for practice: the converters, calculators and trainers do the drilling, so this page does not repeat them.

Competency
9.0
Sections
10 + self-check
Then
Drill comp 9

Ohm's law: solving for circuit values

Competency 9.1 asks you to quote Ohm's law formulas for power, voltage, current and resistance and solve for circuit values. ETA's sheet prints all twelve as a wheel, with E for voltage; the full table, with a worked example for every form, is in section 13 of the Electrical Theory guide, and the Ohm's Law Wheel solves any of them. The method is always the same: find the two quantities you know, pick the form that uses them, and put every value in base units first.

E = I × R  ·  I = E / R  ·  R = E / I  ·  P = E × I

Worked example

A resistor with a prefix: 12 V is across a 4.7 kΩ resistor. In base units the resistance is 4700 Ω, so I = 12 / 4700 ≈ 0.002553 A, which is about 2.55 mA. The power is P = 12 × 0.002553 ≈ 0.0306 W, or about 30.6 mW; the other form agrees, 12² / 4700 ≈ 0.0306 W.

Worked example

Solving for the load: a 13.8 V supply delivers 100 mA to a load. With the current in amperes (0.1 A), the load is R = 13.8 / 0.1 = 138 Ω and it takes P = 13.8 × 0.1 = 1.38 W.

Trap

Exam trap: a formula given milliamperes or kilohms without conversion is off by a factor of 1000. Dividing 12 V by "4.7" instead of 4700 gives 2.55 A, a thousand times too much, and that wrong answer is usually one of the choices.

On the exam

Exam point: convert every value to base units (volts, amperes, ohms and watts) before you use an Ohm's law formula, then convert the answer back to a convenient prefix. Section 8 shows the prefix steps.

Rearranging formulas step by step is drilled in the Electronics Algebra trainer, and the numeric drills in the Study Hub generate new Ohm's law problems every time.

Other common formulas

Competency 9.2 asks you to list other common basic electronic formulas. ETA lets you use its Common Formulas sheet on the exam, so the skill is knowing which formula fits the question and using it correctly. Beyond Ohm's law, the sheet gives: resistors, inductors and capacitors in series and parallel; RC and RL time constants; resonant frequency; inductive and capacitive reactance; series and parallel impedance; the sine-wave conversions; decibel gain (section 6); inverting and non-inverting op-amp gain; frequency and wavelength (section 3); capacitor charge and energy; inductor voltage and energy; battery internal resistance; the resistor color code; and the SI prefixes (section 8). The Formula Cheat Sheet and the Formula Matrix list them, and the Formula Trainer drills them.

RT = R1 + R2 + R3  ·  1 / RT = 1 / R1 + 1 / R2 + 1 / R3 (inductors combine the same way)
CT = C1 + C2 + C3 in parallel  ·  1 / CT = 1 / C1 + 1 / C2 + 1 / C3 in series

The sheet also prints the PEMDAS rule (parentheses, exponents, multiplication, division, addition, subtraction). In the reciprocal formulas it means: work out each 1/R, add them, and only then take the reciprocal of the sum. For exactly two parts the same formula reduces to product over sum, R1 × R2 / (R1 + R2), which is algebra rather than a separate formula on the sheet.

Worked example

Parallel resistors and series capacitors: 1 kΩ and 1.5 kΩ in parallel: 1 / 1000 + 1 / 1500 ≈ 0.001667, and 1 / 0.001667 = 600 Ω (product over sum gives 1500000 / 2500 = 600 Ω too). Capacitors of 10 µF and 15 µF in series use the same reciprocal form, 1 / (1/10 + 1/15) = 6 µF; in parallel they simply add to 25 µF.

Trap

Exam trap: capacitors combine the opposite way to resistors and inductors. Capacitors add in parallel and use the reciprocal formula in series, so two capacitors in series always total less than the smaller one.
τ = R × C  ·  τ = L / R  ·  fkHz = 159.2 / √(L × C) (L in µH, C in µF)

Worked example

Time constant and resonance from the sheet: 10 kΩ with 100 µF gives a time constant of τ = 10000 × 0.0001 = 1 s, and after 5τ = 5 s the capacitor is over 99% charged (1 − e−5 ≈ 0.993). A 25 µH coil with a 0.01 µF capacitor resonates at f = 159.2 / √(25 × 0.01) = 159.2 / 0.5 = 318.4 kHz with the sheet's units; the full formula, 1 / (2π√(LC)) in henries and farads, gives 318.3 kHz, the difference being the sheet's rounding of 1/(2π) to 159.2.
AV = −(R_f / R_i) inverting  ·  AV = 1 + (R_f / R_i) non-inverting

With Rf = 47 kΩ and Ri = 4.7 kΩ, the inverting amplifier's gain is −(47 / 4.7) = −10 (the minus sign means the output is inverted) and the non-inverting amplifier's is 1 + 10 = 11. The Calculator Suite works each of these formulas.

Wavelength, frequency and power

Competency 9.3 asks you to explain how to calculate wavelength, frequency and power values. NEETS defines the period as the time of one complete cycle, the frequency as the number of cycles in one second, and the wavelength as the distance the wave travels during one period; the period is the reciprocal of the frequency, and the wavelength is the velocity divided by the frequency. For a radio wave in free space the velocity is the speed of light, c. ETA's sheet gives c = f × λ in three working forms:

λ = c / f  ·  f = 1 / T  ·  λm = 300 / f_MHz  ·  λft = 984 / f_MHz

The sheet writes the metric form as λ = (3 × 108) / f with f in hertz; dividing both by 106 gives the 300 / fMHz shortcut. The exact value of c is 299 792 458 m/s, fixed by definition (NIST), so ETA's 3 × 108 is rounded up by about 0.07%. In feet per second c is 299792458 / 0.3048 ≈ 983571056 (the foot is exactly 0.3048 m, per NIST), which is where ETA's 984 comes from, rounded from 983.6. The sheet's 1 m = 3.28 ft is rounded the same way (1 / 0.3048 ≈ 3.2808). NEETS's warning applies to every form: keep the units consistent, meters with hertz or megahertz with the matching shortcut, never a mixture.

Worked example

A 150 MHz signal: λ = 3 × 108 / 1.5 × 108 = 2 m, or 300 / 150 = 2 m with the shortcut (the exact c gives 1.9986 m). In feet, 984 / 150 = 6.56 ft. A half wave, which the sheet equates with 180°, is 1 m. The period is 1 / 150000000 ≈ 6.67 ns. Going the other way, a 20 m wavelength is 300 / 20 = 15 MHz.

Power values come from Ohm's law with the impedance the power is delivered into. RF power is commonly stated into 50 Ω (Kuphaldt references RF dBm levels to a 50 Ω load), so the square-root forms give the voltage and current a power produces:

E = √(P × R)  ·  I = √(P / R)

Worked example

100 W into 50 Ω: E = √(100 × 50) = √5000 ≈ 70.7 V RMS and I = √(100 / 50) = √2 ≈ 1.41 A. Check: 70.7 × 1.41 ≈ 100 W.

On the exam

Exam point: wavelength is the speed of light divided by the frequency: in meters, 300 divided by the frequency in megahertz; in feet, ETA's sheet uses 984 divided by the frequency in megahertz. Higher frequency means a shorter wavelength.

Antenna lengths, velocity factor and the frequency bands are in section 2 of the RF & Communications guide; the RF / Antenna calculator converts frequency and wavelength.

Binary, decimal, octal and hexadecimal

Competency 9.4 asks you to convert binary, decimal, octal and hexadecimal numbers. All four are place-weighted systems (Kuphaldt): each digit is multiplied by the weight of its place, and each place is worth the base times the place to its right. NEETS calls the base the radix: the number of symbols the system uses.

SystemBase and digitsPlace weights, right to leftBits per digit
Binary2: 0 and 11, 2, 4, 8, 16, 32, 64, 1281 (one bit)
Octal8: 0 to 71, 8, 64, 5123
Decimal10: 0 to 91, 10, 100, 1000No whole number
Hexadecimal16: 0 to 9, then A to F for 10 to 151, 16, 256, 40964 (one nibble)

Worked example

Decimal 200 four ways: 200 = 128 + 64 + 8, so in binary it is 11001000. Grouped in threes from the right, 011 001 000 gives octal 310; grouped in fours, 1100 1000 gives hex C8. Check by division: 200 / 8 = 25 remainder 0, 25 / 8 = 3 remainder 1, 3 / 8 = 0 remainder 3, so octal 310 again; and C8 = 12 × 16 + 8 = 200.

Trap

Exam trap: "10" means the base, not ten. Hex 10 is sixteen, octal 10 is eight and binary 10 is two. Octal has no digit 8 or 9, and hex A to F are the values 10 to 15.

On the exam

Exam point: each octal digit stands for exactly three bits and each hexadecimal digit for exactly four, so you convert to or from binary by grouping the bits from the right. Go through binary whenever octal and hex meet.

The Number Systems lesson in the Study Hub adds binary addition, BCD and register overflow; the Binary Converter and Digital / Logic convert any value, and section 2 of the Computer Applications guide shows where bits and bytes meet storage.

Boolean algebra

Competency 9.5 asks you to explain Boolean algebra and its use in digital circuitry. Kuphaldt: George Boole gave symbolic form to Aristotle's system of logic, built on two kinds of proposition, true and false, in 1854, and Claude Shannon's 1938 thesis applied it to relay and switching circuits, where every signal is either high (1) or low (0). Every Boolean quantity is a single 1 or 0, and there are three operations:

There is no subtraction or division. NEETS gives the purpose: a Boolean expression describes the inputs needed for an output, and its laws let a designer get the same output from the fewest gates, saving space, weight and cost. Kuphaldt adds reliability: fewer components, fewer failures. NEETS's laws:

Law (NEETS)OR formAND form
Union and intersectionA + 0 = A, A + 1 = 1A · 1 = A, A · 0 = 0
IdempotentA + A = AA · A = A
ComplementaryA + A' = 1A · A' = 0
Double negative(A')' = A—
Commutative and associativeA + B = B + AAB = BA, A(BC) = ABC
DistributiveA + BC = (A + B)(A + C)A(B + C) = AB + AC
AbsorptionA + AB = AA(A + B) = A
Common identitiesA + A'B = A + BA(A' + B) = AB
DeMorgan's theorem(A + B)' = A' · B'(AB)' = A' + B'

DeMorgan's theorem in hardware. Kuphaldt: inverting a gate's output gives the same function as the opposite type of gate with inverted inputs. A NAND gate behaves as an OR gate with both inputs inverted, and a NOR gate as an AND gate with both inputs inverted. Every law in the table can be checked with a truth table; each one was checked that way for this page.

Worked example

Two simplifications: AB + AB' = A(B + B') by the distributive law, = A · 1 by the complementary law, = A. The circuit needs no gates at all: the output is just input A, whatever B does. Similarly, A + AB = A(1 + B) = A · 1 = A: if A is 1 the output is 1, and if A is 0 the AB term is 0 too.

Trap

Exam trap: in Boolean algebra 1 + 1 = 1. It is not 2, and it is not binary 10 either: Kuphaldt points out that a Boolean quantity is a single bit, whereas a binary number can have many bits that add up by place weight.

On the exam

Exam point: in Boolean algebra the plus sign means OR, the dot means AND and the complement means NOT; by DeMorgan's theorem, NOT (A AND B) equals (NOT A) OR (NOT B). That is why a NAND gate equals an OR gate with inverted inputs.

The Logic Gates lesson in the Study Hub teaches the seven gates and their truth tables; try the expressions in the Logic Gates lab and Digital / Logic.

Decibels

Competency 9.6 asks you to explain decibels and show the reasons for using them in signal level, power and audio calculations. Kuphaldt: the bel, named for Alexander Graham Bell, was devised to express power loss in telephone lines and is the base-10 logarithm of a power ratio; it proved too large, so the decibel, one-tenth of a bel, is used. ETA's sheet gives both forms:

dB = 10 × log₁₀(P2 / P1)  ·  dB = 20 × log₁₀(V2 / V1)

The decibel is fundamentally a unit of power. Power goes with voltage squared, so for a constant load impedance a voltage gain of 2 is a power gain of 4, and the 20 in the voltage form absorbs that square (Kuphaldt). The voltage form therefore gives the same answer as the power form only when both voltages are measured across equal impedances; ETA's sheet prints it without that condition, but the Decibel tool and the glossary state it, and so does this page. A gain is a positive figure, a loss negative, and no change is 0 dB.

Why decibels are used.

DecibelsPower ratioVoltage ratio (equal impedances)
+3 dBAbout 2 (10 log 2 ≈ 3.01)About 1.41
+6 dBAbout 4About 2 (20 log 2 ≈ 6.02)
+10 dB10About 3.16
+20 dB10010
−3 dBAbout 0.5 (half power)About 0.707

Worked example

Power and voltage gain: an amplifier takes 0.5 W in and delivers 20 W: 10 log (20 / 0.5) = 10 log 40 ≈ 16.0 dB. Kuphaldt's audio line amplifier takes 10 mV in across 600 Ω and gives 1 V across a 600 Ω load: the impedances are equal, so 20 log (1 / 0.01) = 20 log 100 = 40 dB.

Trap

Exam trap: +3 dB doubles the power but multiplies the voltage by only about 1.41; doubling the voltage is +6 dB. And 20 log of a voltage ratio is a power gain only when both voltages are across the same impedance.

On the exam

Exam point: a power ratio in decibels is 10 log of the ratio and a voltage ratio is 20 log, which give the same figure only when both voltages are across equal impedances; +3 dB is about double the power. A negative figure is a loss.

The Decibel tool converts in both directions and the dB Spectrum Analyzer shows levels on a dB scale; section 4 of the RF & Communications guide applies dB to gain and loss along a signal path.

Reading graphs

Competency 9.7 asks you to explain how graphs are used to demonstrate electronics functions. A graph shows how one quantity responds to another. Read the axes first: what each measures, its unit and prefix, and whether the scale is linear or logarithmic. Then read the point, slope or shape the question asks about.

Linear and logarithmic axes. On a linear axis, equal distances are equal differences: 0, 2, 4, 6 V. On a logarithmic axis, equal distances are equal ratios: 10 Hz, 100 Hz, 1 kHz, 10 kHz, each step a factor of ten (a decade). A value's position within a decade is its logarithm, so the gridlines crowd together toward the top of each decade. The audio range, 20 Hz to 20 kHz, is a ratio of 1000, exactly three decades, which a log axis shows evenly; a linear axis would squeeze everything below 2 kHz into the first tenth of the width. Frequency-response plots use a log frequency axis with gain in decibels, which is already logarithmic.

Value within a decadePosition along the decade (log10)
10% (start of the decade)
230.1%
347.7%
3.16 (√10)50%, the midpoint
569.9%
10100% (start of the next decade)

Trap

Exam trap: halfway between the 1 kHz and 10 kHz lines of a log axis is about 3.16 kHz, not 5.5 kHz. The 5 kHz point sits about 70% of the way along the decade.

Characteristic curves. A characteristic curve plots a component's current against its voltage. Kuphaldt: a function that plots as a straight line is linear, and a straight current-versus-voltage line means a stable resistance, which is how a resistor behaves (I = E/R at every point). A lamp's curve rises steeply and then flattens, because its filament's resistance increases as it heats. An air gap or a varistor passes almost no current until a breakdown voltage, then conducts heavily, and a few devices show a falling curve, called negative resistance. With current on the vertical axis, a steeper line means a lower resistance.

Worked example

Reading resistance off a curve: a straight line through zero passes 6 V at 2 mA and 12 V at 4 mA: R = 6 / 0.002 = 3000 Ω and 12 / 0.004 = 3000 Ω, one fixed 3 kΩ. A lamp's curve read at 2 V and 0.5 A, then at 12 V and 1.0 A (example readings), gives 2 / 0.5 = 4 Ω cold and 12 / 1.0 = 12 Ω hot: the curve bends because the resistance changes.

The Mathematics lesson in the Study Hub has a Reading graphs section that works the other curves the exam uses: a sine wave against time, an RC charge curve (63% at one time constant), a frequency response with its −3 dB cutoff, a resonance curve and a diode's knee. The bank's graph questions drill them. See live plots in the Scope Lab and the Visual Learning Lab.

SI prefixes and conversions

Competency 9.8 asks you to explain the SI prefixes and conversions. A prefix is a power of ten written as a letter (Kuphaldt): 2.5 GB is 2.5 × 109 bytes. ETA's sheet lists the nine you will use:

PrefixSymbolPower of tenMultiplier
teraT1012trillion
gigaG109billion
megaM106million
kilok103thousand
(none)—100one
millim10−3thousandth
microµ10−6millionth
nanon10−9billionth
picop10−12trillionth

NIST recognizes 24 prefixes, from quetta (1030) to quecto (10−30); the newest four, ronna, quetta, ronto and quecto, were added in 2022. Converting means moving the decimal point (Kuphaldt): each step between the prefixes above is three places, left toward a larger prefix and right toward a smaller one. NIST also asks for a leading zero before a decimal point (0.47, not .47), so the quantity is read correctly.

Competency 9.8.1 points to NIST Special Publication 811, section 9.3, on spelling unit names with prefixes. When a prefixed unit name is spelled out, there is no space or hyphen between prefix and unit (milligram, kilopascal), and the name is lower case like any noun (section 9.1). Three names drop the prefix's final vowel: megohm (not megaohm), kilohm (not kiloohm) and hectare. In every other name both vowels stay, so 1 GΩ is a gigaohm (SP 811 spells it so in section 6.2.1) and 1 mA a milliampere.

Worked example

Three conversions: 0.0047 µF is 4.7 nF (three places right) and 4700 pF (six places right). Kuphaldt's 0.000023 A is 23 µA (six places right). 2200000 Ω is 2.2 MΩ (six places left), spoken as 2.2 megohms.

On the exam

Exam point: each step between the common SI prefixes is a factor of 1000, three decimal places; the symbols from mega up are capitals, and kilo and everything smaller are lower case. Check the case of an m before you convert.

Drill conversions on the Decimal Ladder and the Unit Conversion Hub; component markings are decoded in Component Values.

Powers of ten

Competency 9.9 asks you to define exponents (powers of ten) and their calculations. 10n means n tens multiplied together, a 1 followed by n zeros; 10−n is its reciprocal, 1 / 10n; and NEETS reminds you that any number to the power 0 is 1, so 100 = 1. Kuphaldt's rules for calculating:

Worked example

Exponents in circuit problems: 2 mA through 4.7 kΩ is E = (2 × 10−3) × (4.7 × 103) = 9.4 × 100 = 9.4 V, because the exponents −3 and 3 add to 0. Dividing, 12 / (4.7 × 103) ≈ 2.55 × 10−3 A. And (6 × 108) / (3 × 10−2) = 2 × 108 − (−2) = 2 × 1010.

Trap

Exam trap: dividing by a negative exponent adds: 108 / 10−2 = 1010, not 106. And multiplying powers of ten adds the exponents, never multiplies them: 103 × 103 = 106, not 109.

The Engineering Notation converter and the Scientific Calculator practise both notations; the Mathematics lesson in the Study Hub works a reactance problem with exponents.

On the job: numbers at the radio site

Job practice, not exam content. These are the same calculations as they turn up in the field. The equipment manual and your employer's procedures come first.

  1. Turn watts into dBm and dBW. A 100 W transmitter is 10 log (100 / 0.001) = 50 dBm, or 10 log 100 = 20 dBW. Because 1 W is 1000 mW, dBW is always dBm minus 30.
  2. Add up losses in decibels, then convert once. Losses in cable, connectors and combiners add in dB. A total of 3 dB between a 50 dBm transmitter and the antenna leaves 47 dBm, about 50.1 W: half the power, gone before the antenna.
  3. Read tiny levels in dBm. A received signal of −100 dBm is 10−10 mW, or 10−13 W. Writing it in dBm keeps such numbers manageable; Kuphaldt notes that signal generators for testing receivers are set in dBm too.
  4. Read a hex value. Where a programming tool or service menu shows a value in hex, 3E8 is 3 × 256 + 14 × 16 + 8 = 1000 in decimal, and 0011 1110 1000 in binary, one nibble per digit.
  5. Read binary switch settings. If a bank of switches is weighted 1, 2, 4 and 8 (check the manual for the order), switches 1, 3 and 4 on set 1 + 4 + 8 = 13.

Check yourself

Answer in your head first, then reveal.

  1. What do you do with prefixed values before using Ohm's law?

    convert every value to base units (volts, amperes, ohms and watts) before you use an Ohm's law formula, then convert the answer back to a convenient prefix

  2. How do you find a wavelength from a frequency in meters and in feet?

    wavelength is the speed of light divided by the frequency: in meters, 300 divided by the frequency in megahertz; in feet, ETA's sheet uses 984 divided by the frequency in megahertz

  3. How do octal and hexadecimal digits relate to binary?

    each octal digit stands for exactly three bits and each hexadecimal digit for exactly four, so you convert to or from binary by grouping the bits from the right

  4. What do +, the dot and the complement mean in Boolean algebra, and what does DeMorgan's theorem say?

    in Boolean algebra the plus sign means OR, the dot means AND and the complement means NOT; by DeMorgan's theorem, NOT (A AND B) equals (NOT A) OR (NOT B)

  5. When do the 10 log and 20 log decibel forms agree, and what is +3 dB?

    a power ratio in decibels is 10 log of the ratio and a voltage ratio is 20 log, which give the same figure only when both voltages are across equal impedances; +3 dB is about double the power

  6. How far apart are the common SI prefixes, and which symbols are capitals?

    each step between the common SI prefixes is a factor of 1000, three decimal places; the symbols from mega up are capitals, and kilo and everything smaller are lower case

Practice this topic: Math questions are in the Study Hub quiz bank, and the key terms are in Flashcards.