Section 5: Basic Electronics and Theory
Canadian Amateur Radio Basic Qualification (B-005)
This section is the mathematical heart of the exam. It covers the fundamental laws that govern every circuit in your radio. The good news: there are only a handful of core formulas, and nearly every exam question is a direct application of one of them. Master the concepts here and the rest of the qualification becomes far easier.
The Big Picture -- Electricity as Water: Imagine water flowing through pipes. Voltage is the water pressure pushing the flow. Current is how much water moves past a point each second. Resistance is how narrow the pipe is -- a skinny pipe resists flow. And power is like how hard the water can spin a turbine at the end. Later you will meet capacitors (think of a spring that stores and releases energy) and inductors (a flywheel that resists changes in flow). Keep these pictures in mind and the formulas will feel natural.
5.1 Unit Conversions B-005-001
Electronics uses an enormous range of values -- from picofarads (trillionths of a farad) to gigahertz (billions of hertz). The metric prefix system lets us express these values without writing out long strings of zeros. You must be able to convert between units quickly on the exam, because many problems bury a unit conversion inside a larger calculation.
Metric Prefix Reference Table
| Prefix | Symbol | Multiplier | Example |
| Giga | G | \(10^9\) = 1,000,000,000 | 1 GHz = 1,000,000,000 Hz |
| Mega | M | \(10^6\) = 1,000,000 | 1 MHz = 1,000,000 Hz |
| Kilo | k | \(10^3\) = 1,000 | 1 kHz = 1,000 Hz |
| (base unit) | - | \(10^0\) = 1 | Hz, V, A, Ω, H, F |
| Milli | m | \(10^{-3}\) = 0.001 | 1 mA = 0.001 A |
| Micro | μ | \(10^{-6}\) = 0.000001 | 1 μF = 0.000001 F |
| Nano | n | \(10^{-9}\) | 1 nF = 0.001 μF |
| Pico | p | \(10^{-12}\) | 1 pF = 0.000001 μF |
"King Henry Died By Drinking Chocolate Milk, My Niece Plays" -- Kilo, Hecto, Deca, Base, Deci, Centi, Milli, Micro, Nano, Pico. For radio, focus on: Mega, Kilo, (base), Milli, Micro, Pico.
How to Convert
The conversion rule is simple: each step between neighbouring prefixes is a factor of 1,000 (three decimal places).
- Going to a smaller unit (e.g., MHz to kHz): Multiply by 1,000. The number gets bigger because you are counting in smaller pieces.
- Going to a larger unit (e.g., Hz to kHz): Divide by 1,000. The number shrinks because each piece is bigger.
A quick sanity check: if your answer seems absurdly large or small, you probably multiplied when you should have divided, or vice versa.
Worked Examples
Worked Example: 3.525 MHz to kHz (B-005-001-001)
Q: A dial marked in megahertz shows a reading of 3.525 MHz. What would it show if it were marked in kilohertz?
Step 1: Identify the direction. MHz to kHz is going to a smaller unit, so we multiply.
Step 2: How far apart are MHz and kHz? One step of 1,000.
Step 3: \(3.525 \times 1{,}000 = 3{,}525\) kHz
Answer: (D) 3 525 kHz
Worked Example: 3000 milliamperes to amperes (B-005-001-002)
Q: An ammeter marked in amperes is used to measure a 3000-milliampere current. What reading would it show?
Step 1: Milliamperes to amperes is going to a larger unit, so we divide.
Step 2: 1,000 mA = 1 A, so divide by 1,000.
Step 3: \(3{,}000 \div 1{,}000 = 3\) amperes
Answer: (A) 3 amperes
Worked Example: 7040 kHz to MHz (B-005-001-007)
Q: How many megahertz is 7040 kHz?
Step 1: kHz to MHz is going to a larger unit, so we divide.
Step 2: 1 MHz = 1,000 kHz.
Step 3: \(7{,}040 \div 1{,}000 = 7.040\) MHz
Answer: (A) 7.040 MHz
Worked Example: 1,000,000 picofarads to microfarads (B-005-001-004)
Q: How many microfarads is 1,000,000 picofarads?
Step 1: Pico to micro is going up by two prefix steps: pico → nano → micro. Each step divides by 1,000.
Step 2: Two steps = divide by \(1{,}000 \times 1{,}000 = 1{,}000{,}000\).
Step 3: \(1{,}000{,}000 \div 1{,}000{,}000 = 1\) μF
Answer: (C) 1 microfarad
A hand-held transceiver puts out 500 milliwatts. How many watts is that?
0.5 watts (divide by 1,000)
Exam shortcut: When converting, count the zeros. MHz to kHz = 3 zeros (multiply by 1,000). MHz to Hz = 6 zeros (multiply by 1,000,000). If the answer choices seem too big or too small, you went the wrong direction.
A few more facts the exam expects you to know:
- A kilohm is 1,000 ohms.
- One quarter ampere = 0.25 A = 250 milliamperes.
- 2 volts = 2,000 millivolts.
- To convert megahertz to gigahertz, divide by 1,000.
- 10,000 microhenries = 10 millihenries (divide by 1,000).
Practice Flash Cards -- B-005-001 (Unit Conversions)
B-005-001-001: If a dial marked in megahertz shows a reading of 3.525 MHz, what would it show if it were marked in kilohertz?
- A) 35.25 kHz
- B) 3 525 000 kHz
- C) 352.5 kHz
- D) 3 525 kHz
D) 3 525 kHz -- 1 MHz = 1,000 kHz, so multiply by 1,000.
B-005-001-002: If an ammeter marked in amperes is used to measure a 3000-milliampere current, what reading would it show?
- A) 3 amperes
- B) 0.003 amperes
- C) 0.3 amperes
- D) 0.03 amperes
A) 3 amperes -- 1,000 mA = 1 A, so divide by 1,000.
B-005-001-003: How many hertz is 1 kHz?
- A) 10 000 Hz
- B) 1 000 Hz
- C) 10 Hz
- D) 100 Hz
B) 1 000 Hz -- 1 kHz = 1,000 Hz by definition.
B-005-001-004: How many microfarads is 1 000 000 picofarads?
- A) 0.01 microfarads
- B) 0.001 microfarads
- C) 1 microfarad
- D) 10 microfarads
C) 1 microfarad -- 1 μF = 1,000,000 pF.
B-005-001-005: If you have a hand-held transceiver that puts out 500 milliwatts, how many watts would this be?
- A) 50 watts
- B) 0.05 watts
- C) 0.5 watts
- D) 5 watts
C) 0.5 watts -- 500 / 1,000 = 0.5 W.
B-005-001-006: A kilohm is:
- A) 10 ohms
- B) 1000 ohms
- C) 0.1 ohms
- D) 0.001 ohms
B) 1000 ohms -- Kilo means 1,000.
B-005-001-007: How many megahertz is 7040 kHz?
- A) 7.040 MHz
- B) 0.740 MHz
- C) 70.40 MHz
- D) 0.074 MHz
A) 7.040 MHz -- 7040 / 1,000 = 7.040.
B-005-001-008: A current of one quarter ampere may be written as:
- A) 2.5 milliamperes
- B) 0.25 milliamperes
- C) 250 microamperes
- D) 250 milliamperes
D) 250 milliamperes -- 1/4 = 0.25 A = 250 mA.
B-005-001-009: How many millivolts equal two volts?
- A) 0.002 mV
- B) 2 000 mV
- C) 0.000 002 mV
- D) 2 000 000 mV
B) 2 000 mV -- 1 V = 1,000 mV, so 2 V = 2,000 mV.
B-005-001-010: How can a frequency in megahertz be stated in gigahertz?
- A) Multiply by 1 000 000
- B) Divide by 1 000
- C) Multiply by 1 000
- D) Divide by 1 000 000
B) Divide by 1 000 -- 1 GHz = 1,000 MHz.
B-005-001-011: How many millihenries equal 10 000 microhenries?
- A) 100 millihenries
- B) 1 millihenry
- C) 1000 millihenries
- D) 10 millihenries
D) 10 millihenries -- 1 mH = 1,000 μH, so 10,000 / 1,000 = 10.
5.2 Conductors and Insulators B-005-002
Every material in the universe falls somewhere on a spectrum from perfect conductor to perfect insulator. Understanding where common materials land on that spectrum -- and the vocabulary the exam uses to describe it -- is essential.
Conductors
A conductor is a material that readily allows the flow of electric current. Metals are the best conductors because their atoms share free electrons that can drift through the material when voltage is applied. Think of the water analogy: a conductor is a wide-open pipe.
| Conductor | Relative Conductivity | Notes |
| Silver | Best overall | Too expensive for most uses |
| Copper | 2nd best | Most common wiring material |
| Gold | 3rd best | Used for connectors (corrosion-resistant) |
| Aluminum | 4th | Used in power lines (lightweight) |
On the exam, when asked "What is the best conductor among silicon, aluminum, copper, and carbon?" the answer is always copper. Silver is technically better but is rarely listed as a choice. The exam also asks you to identify groups of three good conductors -- gold, silver, and aluminum is a valid trio.
Insulators
An insulator resists the flow of electric current. In the water analogy, an insulator is a sealed cap on the pipe -- no flow gets through. Good insulators include glass, air, and porcelain, as well as rubber, plastic, Teflon, and mica. Note that wood and carbon are not good insulators (carbon actually conducts to some degree).
Key Electrical Properties
The exam tests several vocabulary terms related to conductors and insulators:
- The flow of electric charge in a circuit is called current.
- The letter "R" is the symbol for resistance.
- An object conducts electricity well when it has low resistance.
- Conductance is the inverse of resistance -- high conductance means low resistance.
- A voltage drop is the loss of voltage caused by the flow of current through a circuit.
- The resistance of a conductor changes with temperature.
- Polarity describes the direction of current in a DC circuit.
What is the inverse of resistance?
Conductance
Practice Flash Cards -- B-005-002 (Conductors & Insulators)
B-005-002-001: Which of these groups lists three good electrical conductors?
- A) Gold, silver and wood
- B) Copper, aluminum and paper
- C) Copper, gold and mica
- D) Gold, silver and aluminum
D) Gold, silver and aluminum -- All three are metals with free electrons.
B-005-002-002: Which of these groups lists three good electrical insulators?
- A) Plastic, wood and carbon
- B) Teflon, mica and aluminum
- C) Wood, copper and porcelain
- D) Glass, air and porcelain
D) Glass, air and porcelain -- None of these conduct electricity readily.
B-005-002-003: What do we call the flow of electric charge in a circuit?
- A) Capacitance
- B) Current
- C) Voltage
- D) Resistance
B) Current -- Current is the movement of charge through a circuit.
B-005-002-004: What is the best conductor among the following materials?
- A) Silicon
- B) Aluminum
- C) Copper
- D) Carbon
C) Copper -- Among common exam choices, copper is the best conductor.
B-005-002-005: Which of these types of materials readily allows the flow of electric current?
- A) Conductor
- B) Insulator
- C) Semiconductor
- D) Dielectric
A) Conductor -- By definition, conductors allow current flow.
B-005-002-006: What electrical property causes an object to conduct electricity very well?
- A) Low capacitance
- B) Low admittance
- C) Low resistance
- D) Low reluctance
C) Low resistance -- Less opposition means better conduction.
B-005-002-007: The letter "R" is the symbol for:
- A) Resistance
- B) Impedance
- C) Reluctance
- D) Reactance
A) Resistance
B-005-002-008: What is the inverse of resistance?
- A) Reluctance
- B) Permeability
- C) Conductance
- D) Reactance
C) Conductance -- Conductance = 1/Resistance.
B-005-002-009: What is a voltage drop?
- A) The voltage output of a step-down transformer
- B) The loss of voltage caused by the flow of current through a circuit
- C) Any point in a circuit that has zero voltage
- D) The difference in voltage at the output terminals of a transformer
B) The loss of voltage caused by the flow of current through a circuit
B-005-002-010: The resistance of a conductor changes with:
- A) Humidity
- B) Temperature
- C) Voltage
- D) Current
B) Temperature -- Most metals increase in resistance as they heat up.
B-005-002-011: Which term describes the direction of current in a DC circuit?
- A) Phase
- B) Polarity
- C) Polarization
- D) Directivity
B) Polarity -- Polarity tells you which way current flows in DC.
5.3 Basic Electrical Concepts B-005-003
Water Pipe Analogy -- The Core Relationships:
- Voltage (V) = Water pressure -- the force pushing electrons through the wire
- Current (I) = Flow rate -- how many electrons pass a point per second
- Resistance (R) = Pipe narrowness -- how much the wire opposes flow
- Power (P) = Work done -- like how much the water can spin a turbine
Power
Power is the rate at which electrical energy is used. The faster energy is consumed, the higher the power. The unit of power is the watt. A 100-watt light bulb consumes electrical energy at a higher rate than a 40-watt bulb -- so if you are asked which bulb among 40W, 50W, 60W, and 100W consumes energy at the highest rate, the answer is always the 100-watt bulb.
The fundamental power formula ties voltage and current together:
$$P = V \times I$$
Power (watts) = Voltage (volts) × Current (amperes)
This means that to find power, you multiply the two quantities voltage and current. Equivalently, multiplying volts and amperes gives watts.
What is the basic unit of electrical power?
The watt
Circuit Malfunctions
Two kinds of circuit malfunctions come up repeatedly on the exam: open circuits and short circuits. They are opposites of each other.
OPEN CIRCUIT SHORT CIRCUIT
(broken wire) (bypass path)
+--++--[R]--+ +--[R]--+
| gap | | |
[V] [V] [V] +---+ <-- short
| | | | across R
+-----------+ +---+
No current flows. Excessive current flows.
Fuse does NOT blow. Fuse BLOWS.
Open circuit vs. short circuit
- Open circuit: A break in the path. No current flows at all. If a battery and resistor form a circuit and the wire breaks, no current is drawn from the battery -- that is an open circuit.
- Short circuit: An unintended low-resistance bypass that draws excessive current. When you connect a transceiver to a power supply and the fuse blows immediately, a short circuit is the cause.
- When a resistor gets very hot and starts to burn, it is dissipating too much power (not voltage, current, or resistance -- power is the rate of energy release as heat).
- Continuity means the circuit is a closed (complete) circuit -- there are no breaks in the path.
Open circuit = no current, no fuse blow. Short circuit = excessive current, fuse blows. A burning resistor is dissipating too much power.
Practice Flash Cards -- B-005-003 (Basic Electrical Concepts)
B-005-003-001: What term describes the rate at which electrical energy is used?
- A) Resistance
- B) Power
- C) Current
- D) Voltage
B) Power -- Power is energy per unit time.
B-005-003-002: If you have light bulbs marked 40 watts, 50 watts, 60 watts and 100 watts, which one will consume electrical energy at the highest rate?
- A) The 100-watt bulb
- B) The 60-watt bulb
- C) The 50-watt bulb
- D) The 40-watt bulb
A) The 100-watt bulb -- Higher wattage = faster energy consumption.
B-005-003-003: What is the basic unit of electrical power?
- A) The ohm
- B) The watt
- C) The ampere
- D) The volt
B) The watt
B-005-003-004: A circuit consists of a battery and load resistor. What circuit malfunction would cause no current to be drawn from the battery?
- A) A reactive circuit
- B) A closed circuit
- C) An open circuit
- D) A short circuit
C) An open circuit -- A break means no path for current.
B-005-003-005: Which electrical circuit draws too much current?
- A) A closed circuit
- B) An open circuit
- C) A short circuit
- D) A dead circuit
C) A short circuit -- A low-resistance bypass draws excessive current.
B-005-003-006: Power is expressed in:
- A) Amperes
- B) Ohms
- C) Watts
- D) Volts
C) Watts
B-005-003-007: Which of the following two quantities should be multiplied together to find power?
- A) Voltage and current
- B) Inductance and capacitance
- C) Voltage and inductance
- D) Resistance and capacitance
A) Voltage and current -- P = V × I.
B-005-003-008: Which two electrical units multiplied together give the unit "watts"?
- A) Amperes and henries
- B) Volts and amperes
- C) Volts and farads
- D) Farads and henries
B) Volts and amperes -- Watts = Volts × Amperes.
B-005-003-009: A resistor in a circuit becomes very hot and starts to burn. This is because the resistor is dissipating too much:
- A) Voltage
- B) Resistance
- C) Current
- D) Power
D) Power -- Power is the rate of energy release (as heat).
B-005-003-010: When speaking of electrical circuits, what does the term "continuity" mean?
- A) The circuit is supplied with backup power
- B) The circuit is rated for continuous operation
- C) The circuit is a closed circuit
- D) The circuit is designed for direct current (DC)
C) The circuit is a closed circuit -- Complete, unbroken path.
B-005-003-011: You have acquired a transceiver and connected it to a power supply. When you switch on the power supply, its fuse blows immediately. What circuit malfunction caused the fuse to blow?
- A) A closed circuit
- B) A short circuit
- C) An open circuit
- D) A resonant circuit
B) A short circuit -- Immediate fuse blow = excessive current from a short.
5.4 Ohm's Law B-005-004
Ohm's Law is the single most important formula in electronics and the foundation for most calculations on the exam. It describes the relationship between voltage, current, and resistance in any circuit. If you know any two of these three quantities, you can always find the third.
$$V = I \times R \qquad I = \frac{V}{R} \qquad R = \frac{V}{I}$$
The variable E (for electromotive force) is sometimes used instead of V for voltage. On the exam you may see \(R = E / I\) -- this is identical to \(R = V / I\).
THE OHM'S LAW TRIANGLE
+-----+
/ V \ Cover the quantity you want:
/---------\
/ I | R \ Want V? See I x R
/______|______\ Want I? See V / R
Want R? See V / I
V = Voltage (volts) Also written as E (electromotive force)
I = Current (amperes)
R = Resistance (ohms)
The Ohm's Law triangle -- cover the letter you need to find
"VIR" -- Voltage = I × R. Think: "Vultures In Rome." Or simply remember: V on top, I and R on the bottom.
Worked Examples
Worked Example: What is R if V=12V, I=0.25A? (B-005-004-005)
Q: What is the resistance of a circuit that draws 0.25 amperes from a 12-volt source?
Step 1: Write Ohm's Law for resistance: \(R = \frac{V}{I}\)
Step 2: Plug in: \(R = \frac{12}{0.25}\)
Step 3: \(R = 48\) ohms
Answer: (C) 48 ohms
Worked Example: Voltage across a 2-ohm resistor with 0.5A (B-005-004-001)
Q: What is the voltage across a 2-ohm resistor if a current of 0.5 amperes flows through it?
Step 1: Write Ohm's Law for voltage: \(V = I \times R\)
Step 2: Plug in: \(V = 0.5 \times 2 = 1.0\) volt
Answer: (D) 1.0 volts
Worked Example: Resistance to drop 9V at 10mA (B-005-004-006)
Q: What value of resistance is required to drop 9 volts with a current of 10 milliamperes?
Step 1: Convert milliamperes to amperes first! 10 mA = 0.010 A
Step 2: Apply \(R = \frac{V}{I} = \frac{9}{0.010}\)
Step 3: \(R = 900\) ohms
Answer: (D) 900 ohms
Worked Example: Voltage across 50-ohm resistor at 0.44A (B-005-004-007)
Q: If the current flowing through a 50-ohm resistor is 0.44 amperes, what voltage would you measure across the resistor?
Step 1: \(V = I \times R = 0.44 \times 50\)
Step 2: \(V = 22\) volts
Answer: (C) 22 volts
Worked Example: Current through 30-ohm resistor from 6V battery (B-005-004-008)
Q: A 30-ohm resistor is connected across a 6-volt battery. What current does it draw?
Step 1: \(I = \frac{V}{R} = \frac{6}{30}\)
Step 2: \(I = 0.2\) amperes
Answer: (C) 0.2 amperes
Worked Example: Voltage for 200mA through 25-ohm relay (B-005-004-009)
Q: What voltage is needed to supply a current of 200 milliamperes to operate a relay that has a resistance of 25 ohms?
Step 1: Convert 200 mA = 0.2 A
Step 2: \(V = I \times R = 0.2 \times 25\)
Step 3: \(V = 5\) volts
Answer: (D) 5 volts
Worked Example: Resistance from 300mA and 3V (B-005-004-011)
Q: What is the resistance of a circuit if it draws 300 milliamperes from a 3-volt battery?
Step 1: Convert 300 mA = 0.3 A
Step 2: \(R = \frac{V}{I} = \frac{3}{0.3}\)
Step 3: \(R = 10\) ohms
Answer: (A) 10 ohms
Exam strategy: Always convert milliamperes to amperes FIRST (divide by 1,000). Then plug into the formula. The most common mistake is forgetting this conversion.
The formula R = E / I uses E for what quantity?
Voltage (electromotive force) -- it is the same as V
Practice Flash Cards -- B-005-004 (Ohm's Law)
B-005-004-001: What is the voltage across a 2-ohm resistor if a current of 0.5 amperes flows through it?
- A) 0.25 volts
- B) 2.5 volts
- C) 1.5 volts
- D) 1.0 volts
D) 1.0 volts -- V = I × R = 0.5 × 2 = 1.0
B-005-004-002: How is the current in a DC circuit calculated when the voltage and resistance are known?
- A) Current equals voltage divided by resistance
- B) Current equals resistance multiplied by voltage
- C) Current equals resistance divided by voltage
- D) Current equals power divided by voltage
A) Current equals voltage divided by resistance -- I = V/R
B-005-004-003: How is the resistance in a DC circuit calculated when the voltage and current are known?
- A) Resistance equals current divided by voltage
- B) Resistance equals voltage divided by current
- C) Resistance equals current multiplied by voltage
- D) Resistance equals power divided by voltage
B) Resistance equals voltage divided by current -- R = V/I
B-005-004-004: How is the voltage in a DC circuit calculated when the current and resistance are known?
- A) Voltage equals power divided by current
- B) Voltage equals current multiplied by resistance
- C) Voltage equals current divided by resistance
- D) Voltage equals resistance divided by current
B) Voltage equals current multiplied by resistance -- V = I × R
B-005-004-005: What is the resistance of a circuit that draws 0.25 amperes from a 12-volt source?
- A) 12 ohms
- B) 0.25 ohms
- C) 48 ohms
- D) 3 ohms
C) 48 ohms -- R = 12/0.25 = 48
B-005-004-006: What value of resistance is required to drop 9 volts with a current of 10 milliamperes?
- A) 90 ohms
- B) 9 ohms
- C) 9000 ohms
- D) 900 ohms
D) 900 ohms -- 10 mA = 0.01 A; R = 9/0.01 = 900
B-005-004-007: If the current flowing through a 50-ohm resistor is 0.44 amperes, what voltage would you measure across the resistor?
- A) 2.2 volts
- B) 0.22 volts
- C) 22 volts
- D) 222 volts
C) 22 volts -- V = 0.44 × 50 = 22
B-005-004-008: A 30-ohm resistor is connected across a 6-volt battery. What current does it draw?
- A) 0.5 amperes
- B) 0.005 amperes
- C) 0.2 amperes
- D) 2 amperes
C) 0.2 amperes -- I = 6/30 = 0.2
B-005-004-009: What voltage is needed to supply a current of 200 milliamperes to operate a relay that has a resistance of 25 ohms?
- A) 8 volts
- B) 0.5 volts
- C) 50 volts
- D) 5 volts
D) 5 volts -- 200 mA = 0.2 A; V = 0.2 × 25 = 5
B-005-004-010: What formula calculates the resistance of a circuit when the voltage and current are known?
- A) R = I / E
- B) R = E squared / I
- C) R = E x I
- D) R = E / I
D) R = E / I -- Same as R = V / I.
B-005-004-011: What is the resistance of a circuit if it draws 300 milliamperes from a 3-volt battery?
- A) 10 ohms
- B) 9 ohms
- C) 5 ohms
- D) 3 ohms
A) 10 ohms -- 300 mA = 0.3 A; R = 3/0.3 = 10
5.5 Series and Parallel Resistors B-005-005
Real circuits rarely have just one resistor. When you combine resistors, the arrangement determines how they share voltage and current. There are two fundamental arrangements: series and parallel.
Series Circuits
In a series circuit, resistors are connected end-to-end so that current has only one path to follow. The same current flows through every resistor, while voltage divides among them.
SERIES CIRCUIT: Current has only ONE path
+--[R1]--[R2]--[R3]--+
| |
[V] |
| |
+---------------------+
Same current flows through ALL resistors.
Voltage DIVIDES among the resistors.
R_total = R1 + R2 + R3 + ...
Resistors in series -- resistances ADD up
The total resistance is simply the sum: \(R_{total} = R_1 + R_2 + R_3 + \ldots\). This means the total resistance of resistors in series is always greater than the resistance of any single resistor. For example, five 10-ohm resistors in series give \(5 \times 10 = 50\;\Omega\).
Which series combination replaces a single 120-ohm resistor?
Five 24-ohm resistors: 5 × 24 = 120 Ω
Parallel Circuits
In a parallel circuit, resistors are connected across each other so that current has multiple paths. The same voltage appears across every resistor, while current divides among the branches. The total current equals the sum of all branch currents.
PARALLEL CIRCUIT: Current has MULTIPLE paths
+--[R1]--+
| |
+----+--[R2]--+----+
| | | |
[V] +--[R3]--+ |
| |
+------------------+
Same voltage across ALL resistors.
Current DIVIDES among the branches.
1/R_total = 1/R1 + 1/R2 + 1/R3 + ...
Resistors in parallel -- use reciprocal formula
$$\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots$$
The total resistance of resistors in parallel is always less than the smallest individual resistor. Adding more parallel paths makes it easier for current to flow.
Equal-Value Shortcut
For \(n\) equal resistors of value \(R\) in parallel, the formula simplifies to \(R_{total} = R / n\). This shortcut is heavily tested. For example, four 68-ohm resistors in parallel give \(68/4 = 17\;\Omega\). Ten equal resistors of value R in parallel give \(R/10\).
Two-Resistor Shortcut
For exactly two resistors in parallel: \(R_{total} = \frac{R_1 \times R_2}{R_1 + R_2}\).
Worked Examples
Worked Example: Four 100-ohm resistors in parallel across 12V (B-005-005-002)
Q: You connect four 100-ohm resistors in parallel across a 12-volt battery. How many milliamperes of current are drawn from the battery?
Step 1: Equal-value shortcut: \(R_{total} = \frac{100}{4} = 25\;\Omega\)
Step 2: Apply Ohm's Law: \(I = \frac{V}{R} = \frac{12}{25} = 0.48\) A
Step 3: Convert to milliamperes: \(0.48 \times 1{,}000 = 480\) mA
Answer: (D) 480 mA
Worked Example: Two 1000-ohm resistors in parallel across 12V (B-005-005-004)
Q: Two 1000-ohm resistors in parallel across a 12-volt battery. What is the total current?
Step 1: Equal-value shortcut: \(R_{total} = \frac{1{,}000}{2} = 500\;\Omega\)
Step 2: \(I = \frac{12}{500} = 0.024\) A = 24 mA
Answer: (C) 24 milliamperes
An important conceptual point: if two resistors are in parallel and resistor "A" carries twice the current of "B," then by Ohm's Law (since both see the same voltage), "A" must have half the resistance of "B."
What is the total resistance of four 68-ohm resistors wired in parallel?
17 ohms (68 / 4 = 17)
Practice Flash Cards -- B-005-005 (Series & Parallel Resistors)
B-005-005-001: In a parallel circuit with a voltage source and several branch resistors, how is the total current related to the current in the branch resistors?
- A) It equals the average of the branch current through each resistor
- B) It decreases as more parallel resistors are added to the circuit
- C) It is the sum of each resistor's voltage drop multiplied by the total number of resistors
- D) It equals the sum of the branch current through each resistor
D) It equals the sum of the branch current through each resistor
B-005-005-002: You connect four 100-ohm resistors in parallel across a 12-volt battery. How many milliamperes of current are drawn from the battery?
- A) 48 mA
- B) 120 mA
- C) 240 mA
- D) 480 mA
D) 480 mA -- R=100/4=25; I=12/25=0.48A=480mA.
B-005-005-003: Several resistors of various values are connected in parallel. How does the total resistance of the combination compare to the individual resistors?
- A) It is greater than the largest resistor
- B) It equals the average of the resistors
- C) It equals the square root of the sum of the resistors
- D) It is less than the smallest resistor
D) It is less than the smallest resistor
B-005-005-004: Two 1000-ohm resistors are connected in parallel across a 12-volt battery. What is the total current?
- A) 60 milliamperes
- B) 120 milliamperes
- C) 24 milliamperes
- D) 12 milliamperes
C) 24 milliamperes -- R=1000/2=500; I=12/500=0.024A=24mA.
B-005-005-005: The total resistance of resistors connected in series is:
- A) Less than the resistance of any one resistor
- B) Equal to the highest resistance present
- C) Equal to the lowest resistance present
- D) Greater than the resistance of any one resistor
D) Greater than the resistance of any one resistor
B-005-005-006: What is the total resistance of five 10-ohm resistors in series?
- A) 10 ohms
- B) 2 ohms
- C) 50 ohms
- D) 5 ohms
C) 50 ohms -- 5 × 10 = 50.
B-005-005-007: Which of these series combination of resistors would replace a single 120-ohm resistor?
- A) Five 24-ohm resistors
- B) Six 22-ohm resistors
- C) Two 240-ohm resistors
- D) Five 100-ohm resistors
A) Five 24-ohm resistors -- 5 × 24 = 120.
B-005-005-008: If ten resistors of equal value "R" are wired in parallel, what formula yields the total resistance?
- A) 10 / R
- B) 10 x R
- C) 10 + R
- D) R / 10
D) R / 10
B-005-005-009: What is the total resistance of four 68-ohm resistors wired in parallel?
- A) 12 ohms
- B) 34 ohms
- C) 272 ohms
- D) 17 ohms
D) 17 ohms -- 68 / 4 = 17.
B-005-005-010: Two resistors are in parallel. Resistor "A" carries twice the current of resistor "B," which means that:
- A) "B" has half the resistance of "A"
- B) "A" has half the resistance of "B"
- C) The voltage across "B" is twice that across "A"
- D) The voltage across "A" is twice that across "B"
B) "A" has half the resistance of "B" -- Same voltage, twice the current = half the resistance (Ohm's Law).
B-005-005-011: The total current in a parallel circuit is equal to the:
- A) Source voltage divided by the value of one of the resistive elements
- B) Source voltage divided by the sum of the resistive elements
- C) Current in any one of the parallel branches
- D) Sum of the currents through all the parallel branches
D) Sum of the currents through all the parallel branches
5.6 Power Calculations B-005-006
Power is the rate at which electrical energy is consumed or dissipated. You already know \(P = V \times I\). By substituting Ohm's Law into that formula, you get two additional forms that let you calculate power from any two known quantities:
$$P = V \times I \qquad P = I^2 \times R \qquad P = \frac{V^2}{R}$$
THE POWER FORMULA WHEEL
P
/ | \
V | I P = V x I
/ | \ V = P / I
I R V I = P / V
/ | \
R V R P = I² x R P = V² / R
Choose the formula based on what values you know
Power Ratings and Physical Size
A physically larger resistor can dissipate more heat without burning up. That is why a large-size resistor would be used instead of a smaller one of the same resistance -- for greater power dissipation. Resistor wattage ratings are determined by their heat dissipation qualities.
When two resistors are combined (series or parallel), each one can still handle its rated power. Two 1-watt resistors can handle a total of 2 watts, whether they are in series or parallel.
Worked Examples
Worked Example: Minimum transformer rating for 12V at 5A (B-005-006-002)
Q: A load requires 12 volts DC at 5 amperes. What is the minimum required power transformer rating?
Step 1: Use \(P = V \times I\)
Step 2: \(P = 12 \times 5 = 60\) watts
Answer: (C) 60 watts
Worked Example: DC input power at 12V, 500mA (B-005-006-003)
Q: What is the DC input power of a transmitter operating at 12 volts and drawing 500 milliamperes?
Step 1: Convert 500 mA = 0.5 A
Step 2: \(P = V \times I = 12 \times 0.5 = 6\) watts
Answer: (D) 6 watts
Worked Example: What happens when voltage is doubled? (B-005-006-006)
Q: If the voltage applied to two resistors in series is doubled, how will the total power change?
Step 1: Use the formula \(P = \frac{V^2}{R}\)
Step 2: If V doubles, then \(V^2\) quadruples (because \((2V)^2 = 4V^2\))
Step 3: R stays the same, so power increases four times
Answer: (C) Increase four times
Frequently tested concept! Power is proportional to \(V^2\). Doubling voltage means 4 times the power, not 2 times. This trips up many test-takers.
Worked Example: Current drawn by a 12V, 30W bulb (B-005-006-008)
Q: How much current is drawn by a 12-volt, 30-watt light bulb?
Step 1: Rearrange P = V × I to get \(I = \frac{P}{V}\)
Step 2: \(I = \frac{30}{12} = 2.5\) amperes
Answer: (A) 2.5 amperes
Worked Example: Power of two 10-ohm resistors in series with 10V (B-005-006-009)
Q: What is the power consumption of two 10-ohm resistors connected in series with a 10-volt battery?
Step 1: Total resistance: \(R_{total} = 10 + 10 = 20\;\Omega\)
Step 2: Apply \(P = \frac{V^2}{R} = \frac{100}{20} = 5\) watts
Answer: (D) 5 watts
Worked Example: Building a 50-ohm dummy load rated 5W (B-005-006-007)
Q: Which combination of resistors could make up a 50-ohm dummy load capable of safely dissipating 5 watts?
Step 1: We need total resistance = 50 Ω and total power capacity ≥ 5 W.
Step 2: Try four 2-watt 200-ohm resistors in parallel: \(R = 200/4 = 50\;\Omega\). Total power = \(4 \times 2 = 8\) W ≥ 5 W.
Answer: (B) Four 2-watt 200-ohm resistors in parallel
Replacing a single 50-ohm resistor with two 100-ohm resistors in parallel gives the same resistance but greater power rating, because each resistor handles its own rated power and the total capacity doubles.
Two 500-ohm 1-watt resistors in parallel: what is the maximum total power?
2 watts (each handles 1W, total = 1+1 = 2W)
Practice Flash Cards -- B-005-006 (Power Calculations)
B-005-006-001: Why would a large size resistor be used instead of a smaller one of the same resistance?
- A) For greater power dissipation
- B) For better response time
- C) For higher conductance
- D) For less impedance in the circuit
A) For greater power dissipation
B-005-006-002: A load requires 12 volts DC at 5 amperes. What is the minimum required power transformer rating?
- A) 2.4 watts
- B) 6 watts
- C) 60 watts
- D) 17 watts
C) 60 watts -- P = 12 × 5 = 60.
B-005-006-003: What is the DC input power of a transmitter operating at 12 volts and drawing 500 milliamperes?
- A) 24 watts
- B) 60 watts
- C) 600 watts
- D) 6 watts
D) 6 watts -- 500mA=0.5A; P=12×0.5=6.
B-005-006-004: When two 500-ohm 1-watt resistors are connected in series, the maximum total power that can be dissipated by the resistors is:
- A) 2 watts
- B) 1 watt
- C) 0.5 watts
- D) 4 watts
A) 2 watts -- Each handles 1W, total = 2W.
B-005-006-005: When two 500-ohm 1-watt resistors are connected in parallel, they can dissipate a maximum total power of:
- A) 2 watts
- B) 0.5 watts
- C) 1 watt
- D) 4 watts
A) 2 watts -- Each still handles 1W, total = 2W.
B-005-006-006: If the voltage applied to two resistors in series is doubled, how much will the total power change?
- A) Double
- B) Decrease to one quarter
- C) Increase four times
- D) Decrease to half
C) Increase four times -- P = V²/R; doubling V quadruples P.
B-005-006-007: Which of these combinations of resistors could make up a 50-ohm dummy load capable of safely dissipating 5 watts?
- A) Ten quarter-watt 500-ohm resistors in parallel
- B) Four 2-watt 200-ohm resistors in parallel
- C) Two 5-watt 100-ohm resistors in series
- D) Two 2-watt 25-ohm resistors in series
B) Four 2-watt 200-ohm resistors in parallel -- 200/4=50Ω; 4×2=8W≥5W.
B-005-006-008: How much current is drawn by a 12-volt, 30-watt light bulb?
- A) 2.5 amperes
- B) 18 amperes
- C) 4.8 amperes
- D) 0.4 amperes
A) 2.5 amperes -- I = P/V = 30/12 = 2.5.
B-005-006-009: What is the power consumption of two 10-ohm resistors connected in series with a 10-volt battery?
- A) 2 watts
- B) 20 watts
- C) 0.5 watts
- D) 5 watts
D) 5 watts -- R=20Ω; P=100/20=5.
B-005-006-010: What is the advantage of replacing a 50-ohm resistor with a parallel combination of two 100-ohm resistors of the same power rating?
- A) Same resistance but greater power rating
- B) Same resistance but lesser power rating
- C) Greater resistance and same power rating
- D) Lesser resistance and same power rating
A) Same resistance but greater power rating -- 100/2=50Ω; power capacity doubles.
B-005-006-011: Resistor wattage ratings are:
- A) Determined by heat dissipation qualities
- B) Calculated according to physical size and tolerance rating
- C) Expressed in joules
- D) Variable in steps of one hundred
A) Determined by heat dissipation qualities
5.7 AC Theory B-005-007
Everything discussed so far applies to direct current (DC), where current flows in one direction. In alternating current (AC), the direction of current reverses many times per second. Most radio signals are AC, and household electricity is AC, so understanding these concepts is essential.
Frequency and Wavelength
Frequency is the number of times per second an AC signal completes a full cycle (from positive peak to negative peak and back). It is measured in hertz (Hz). For example, 60 Hz means 60 complete cycles per second -- this is the North American AC power frequency.
Wavelength is the distance a signal travels during one complete cycle. Frequency and wavelength are linked by the speed of light:
$$\text{Wavelength} = \frac{\text{Speed of light}}{\text{Frequency}} = \frac{300{,}000{,}000 \text{ m/s}}{f \text{ (Hz)}}$$
This gives us a critical relationship: as frequency increases, wavelength decreases, and vice versa. They are inversely proportional. If a signal's wavelength gets shorter, its frequency is increasing.
Audio Frequencies
The approximate range of frequencies most humans can hear is 20 Hz to 20,000 Hz (20 Hz to 20 kHz). This range is called "audio frequencies" because the human ear can sense sound in this range.
Frequency Bands
| Band | Abbreviation | Frequency Range |
| Low Frequency | LF | 30 kHz - 300 kHz |
| Medium Frequency | MF | 300 kHz - 3 MHz |
| High Frequency | HF | 3 MHz - 30 MHz |
| Very High Frequency | VHF | 30 MHz - 300 MHz |
| Ultra High Frequency | UHF | 300 MHz - 3 GHz |
The exam will test whether you can place a frequency into its band. For example, 7125 kHz = 7.125 MHz, which falls in the HF range (3-30 MHz).
What is the shape of the waveform of electricity supplied from a household receptacle?
Sine wave (sinusoidal)
Other AC Terms
- Sine wave (sinusoidal): The shape of household AC electricity.
- Phase: The timing difference between two AC waveforms of the same frequency that do not begin at the same instant.
- Harmonic: A signal at a whole-number multiple of the fundamental frequency. If the fundamental is 2 kHz, the 4 kHz component is the second harmonic (also simply called a "harmonic").
Practice Flash Cards -- B-005-007 (AC Theory)
B-005-007-001: What is the term for the number of times per second an alternating current completes a positive to negative cycle?
- A) Phase
- B) Frequency
- C) Speed
- D) Pulse rate
B) Frequency
B-005-007-002: What approximate range of frequencies can most humans hear?
- A) 20 Hz to 20 000 Hz
- B) 20 Hz to 30 000 Hz
- C) 200 Hz to 200 000 Hz
- D) 300 Hz to 3 000 Hz
A) 20 Hz to 20 000 Hz
B-005-007-003: Why is the range of frequencies from 20 Hz to 20 kHz termed audio frequencies?
- A) Because the human ear can sense sound in this range
- B) Because this is the speaker response range of a modern SSB receiver
- C) Because sound can be in this range but it's too low for RF signals
- D) Because RF signals in this range can be directly converted to sound
A) Because the human ear can sense sound in this range
B-005-007-004: Electrical energy at a frequency of 7125 kHz is in what frequency range?
- A) Low Frequency (LF)
- B) Medium frequency (MF)
- C) Very High Frequency (VHF)
- D) High Frequency (HF)
D) High Frequency (HF) -- 7125 kHz = 7.125 MHz, which is in the 3-30 MHz HF band.
B-005-007-005: What is the name for the distance an AC signal travels during one complete cycle?
- A) Wave speed
- B) Waveform
- C) Wave spread
- D) Wavelength
D) Wavelength
B-005-007-006: What happens to a signal's wavelength as its frequency increases?
- A) It decreases
- B) It increases
- C) It decreases proportionally to frequency squared
- D) It increases proportionally to frequency squared
A) It decreases -- Inverse relationship.
B-005-007-007: What happens to a signal's frequency as its wavelength gets shorter?
- A) It increases proportionally to frequency squared
- B) It increases
- C) It decreases
- D) It decreases proportionally to frequency squared
B) It increases
B-005-007-008: What does 60 hertz (Hz) mean?
- A) 60 cycles per second
- B) 6000 metres per second
- C) 60 metres per second
- D) 6000 cycles per second
A) 60 cycles per second
B-005-007-009: Two AC waveforms have the same frequency, but their cycles do not begin at the same instant. What term describes that timing difference?
- A) Polarity
- B) Offset
- C) Delta
- D) Phase
D) Phase
B-005-007-010: What is the shape of the waveform of the electricity supplied from a household receptacle?
- A) Pulse wave
- B) Sine wave (sinusoidal)
- C) Complex wave
- D) Modified square wave
B) Sine wave (sinusoidal)
B-005-007-011: A signal is composed of a fundamental frequency of 2 kHz and another of 4 kHz. What name is given to the 4 kHz signal?
- A) Alias
- B) Sub-harmonic
- C) Intermodulation
- D) Harmonic
D) Harmonic -- 4 kHz = 2 × 2 kHz (second harmonic).
5.8 Decibels (dB) B-005-008
The decibel (dB) is used to measure the ratio between two signal levels. Rather than saying "the signal is 10 times stronger," we say "the signal is 10 dB stronger." This logarithmic scale compresses huge power ratios into manageable numbers. The formula is:
$$\text{dB} = 10 \times \log_{10}\left(\frac{P_2}{P_1}\right)$$
You do not need to calculate logarithms on the exam. Instead, you need to memorize a small table of dB-to-power-ratio conversions and combine them.
Essential dB Reference Table
Memorize this table! These values appear repeatedly on the exam. You will be tested on them directly and in word problems.
| dB Change | Power Ratio | What Happens |
| +3 dB | ×2 | Power doubles |
| +6 dB | ×4 | Power quadruples |
| +9 dB | ×8 | Power × 8 |
| +10 dB | ×10 | Power × 10 |
| +20 dB | ×100 | Power × 100 |
| +30 dB | ×1,000 | Power × 1,000 |
| -3 dB | ×0.5 | Power halves |
| -6 dB | ×0.25 | Power ÷ 4 |
| -10 dB | ×0.1 | Power ÷ 10 |
| -20 dB | ×0.01 | Power ÷ 100 |
| -30 dB | ×0.001 | Power ÷ 1,000 |
"3-6-10 Rule": 3 dB = double, 6 dB = 4×, 10 dB = 10×. You can combine these: 9 dB = 3+3+3 = 2×2×2 = 8×. And 20 dB = 10+10 = 10×10 = 100×.
S-Meter Readings
S-meter readings on a receiver are often expressed as "dB over S9." When the received power changes, you can calculate the new S-meter reading by converting the power ratio to dB. For example, if power is reduced to one-tenth (a factor of 10), the S-meter drops by 10 dB. If reduced to one-hundredth, it drops by 20 dB.
Worked Examples
Worked Example: 9 dB amplifier on a 2W hand-held (B-005-008-009)
Q: You add a 9 dB gain amplifier to your 2-watt hand-held. What is the power output of the combination?
Step 1: Break down 9 dB = 3 dB + 3 dB + 3 dB
Step 2: Each +3 dB doubles power: 2 × 2 × 2 = 8×
Step 3: \(2 \times 8 = 16\) watts
Answer: (D) 16 watts
Worked Example: 6 dB line loss on 100W (B-005-008-010)
Q: The power of your transmitter is 100 watts and your transmission line introduces a loss of 6 dB. How much power is delivered to the antenna?
Step 1: -6 dB = divide by 4 (from the table)
Step 2: \(100 \div 4 = 25\) watts
Answer: (B) 25 watts
Worked Example: S-meter drops from "10 dB over S9" (B-005-008-004)
Q: A signal transmitted with a power of 200 watts is received with an S-meter reading of "10 dB over S9." What would be the new reading if power was reduced to 20 watts?
Step 1: Power ratio: \(200 / 20 = 10\times\) reduction
Step 2: A factor of 10 = 10 dB
Step 3: 10 dB over S9 minus 10 dB = S9
Answer: (A) S9
Worked Example: 100W at "30 dB over S9" -- what power for S9? (B-005-008-011)
Q: Your 100-watt 2-metre simplex signal reads "30 dB over S9." What power would reduce the reading to S9?
Step 1: Need to reduce by 30 dB = divide by 1,000
Step 2: \(100 \div 1{,}000 = 0.1\) W
Answer: (A) 0.1 W
Exam strategy for dB problems: Break complex dB values into 3s and 10s. For example, 9 dB = 3+3+3 = 2×2×2 = 8×. For 13 dB = 10+3 = 10×2 = 20×.
Power increases from 5W to 50W. How many dB of gain is that?
10 dB (50/5 = 10× = 10 dB)
Practice Flash Cards -- B-005-008 (Decibels)
B-005-008-001: A two-times increase in power results in a change of how many dB?
- A) 6 dB higher
- B) 12 dB higher
- C) 1 dB higher
- D) 3 dB higher
D) 3 dB higher
B-005-008-002: What change in transmitter power results in a 3 dB decrease?
- A) Divide the original power by 1.5
- B) Divide the original power by 3
- C) Divide the original power by 4
- D) Divide the original power by 2
D) Divide the original power by 2 -- -3 dB = half power.
B-005-008-003: What change in transmitter power results in a 6 dB increase?
- A) Multiply the original power by 2
- B) Multiply the original power by 1.5
- C) Multiply the original power by 4
- D) Multiply the original power by 3
C) Multiply the original power by 4 -- +6 dB = 4×.
B-005-008-004: If a signal transmitted with a power of 200 watts is received with an S-meter reading of "10 dB over S9," what would be the new reading if power was reduced to 20 watts?
- A) S9
- B) S9 plus 3 dB
- C) S9 minus 10 dB
- D) S9 plus 5 dB
A) S9 -- 200/20=10×=10dB reduction.
B-005-008-005: If a signal transmitted with a power of 150 watts is received with an S-meter reading of "20 dB over S9," what would be the new reading if power was reduced to 15 watts?
- A) S9
- B) S9 plus 10 dB
- C) S9 plus 5 dB
- D) S9 plus 3 dB
B) S9 plus 10 dB -- 150/15=10×=10dB; 20-10=10dB over S9.
B-005-008-006: What is the "decibel" used for?
- A) To describe a waveform on an oscilloscope
- B) To describe very high frequency radio waves
- C) To measure a single side band signal
- D) To measure the ratio of two signals
D) To measure the ratio of two signals
B-005-008-007: The power output from a transmitter increases from 1 watt to 2 watts. How many decibels does that increase represent?
- A) 3 dB
- B) 10 dB
- C) 6 dB
- D) 1 dB
A) 3 dB -- 2/1=2×=3dB.
B-005-008-008: The power of a transmitter is increased from 5 watts to 50 watts by a linear amplifier. The power gain, expressed in dB, is:
- A) 45 dB
- B) 20 dB
- C) 10 dB
- D) 30 dB
C) 10 dB -- 50/5=10×=10dB.
B-005-008-009: You add a 9 dB gain amplifier to your 2-watt hand-held. What is the power output of the combination?
- A) 11 watts
- B) 20 watts
- C) 18 watts
- D) 16 watts
D) 16 watts -- 9dB=8×; 2×8=16.
B-005-008-010: The power of your transmitter is 100 watts and your transmission line introduces a loss of 6 dB. How much power is delivered to the antenna?
- A) 33 watts
- B) 25 watts
- C) 50 watts
- D) 17 watts
B) 25 watts -- -6dB=÷4; 100/4=25.
B-005-008-011: A local amateur radio operator reports receiving your 100-watt 2-metre simplex transmission with an S-meter reading of "30 dB over S9." What power could you use to reduce that reading to S9?
- A) 0.1 W
- B) 1 W
- C) 10 W
- D) 33.3 W
A) 0.1 W -- -30dB=÷1000; 100/1000=0.1.
5.9 Inductors and Capacitors B-005-009
Resistors dissipate energy as heat. Inductors and capacitors, by contrast, store energy temporarily and release it. They are the building blocks of filters, tuned circuits, and power supplies.
Capacitor = Spring: A capacitor stores energy in an electric field between its plates, just as a compressed spring stores mechanical energy. Press it down (charge it) and it pushes back when released. It charges up quickly and can release energy in a burst.
Inductor = Flywheel: An inductor stores energy in a magnetic field around its coil, like a flywheel stores rotational kinetic energy. Once the flywheel is spinning (current is flowing), it resists any attempt to speed up or slow down. An inductor resists changes in current.
Inductors in Series and Parallel
Inductors combine exactly like resistors:
$$\text{Series: } L_{total} = L_1 + L_2 + \ldots \qquad \text{Parallel: } \frac{1}{L_{total}} = \frac{1}{L_1} + \frac{1}{L_2} + \ldots$$
For two equal inductors: in series the total is twice the value of one; in parallel it is half the value of one.
Capacitors in Series and Parallel
Capacitors combine opposite to resistors. This is the single most common source of confusion on this topic, so pay close attention:
$$\text{Parallel: } C_{total} = C_1 + C_2 + \ldots \qquad \text{Series: } \frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots$$
For two equal capacitors: in parallel the total is twice the value of one (they add); in series the total is half the value of one (they use the reciprocal formula).
Capacitors are the opposite of resistors and inductors! Capacitors in parallel add up directly. Capacitors in series use the reciprocal formula. Everything is flipped.
"Caps are contrary." Inductors follow the same rules as resistors. Capacitors do the opposite.
Component Construction
The exam asks what physical parameters determine inductance and capacitance:
- Air-core inductor: Inductance depends on coil diameter, coil length, and number of turns of wire.
- Parallel-plate capacitor: Capacitance depends on surface area of the plates and spacing between the plates.
Never apply a reverse voltage to a polarized electrolytic capacitor! Doing so can cause the capacitor to overheat, leak, or explode. Always observe correct polarity.
Worked Examples
Worked Example: Two 12 mH chokes in series (B-005-009-008)
Q: If you wire two 12-millihenry chokes in series, what is the inductance of the combination?
Step 1: Inductors in series add (just like resistors): \(L_{total} = L_1 + L_2\)
Step 2: \(L_{total} = 12 + 12 = 24\) mH
Answer: (D) 24 millihenries
Worked Example: Two 20 mH inductors in parallel (B-005-009-009)
Q: If you wire two 20-millihenry inductors in parallel, what is the inductance of the combination?
Step 1: Equal-value inductors in parallel: \(L_{total} = L / 2\)
Step 2: \(L_{total} = 20 / 2 = 10\) mH
Answer: (A) 10 millihenries
Worked Example: Two 20 μF capacitors in series (B-005-009-010)
Q: If you wire two 20-microfarad capacitors in series, what is the capacity of the combination?
Step 1: Caps in series use the reciprocal formula. Equal values: \(C_{total} = C / 2\)
Step 2: \(C_{total} = 20 / 2 = 10\) μF
Answer: (D) 10 microfarads
Worked Example: Two 24 μF capacitors in parallel (B-005-009-011)
Q: If you wire two 24-microfarad capacitors in parallel, what is the capacity of the combination?
Step 1: Caps in parallel ADD (opposite of resistors!): \(C_{total} = C_1 + C_2\)
Step 2: \(C_{total} = 24 + 24 = 48\) μF
Answer: (C) 48 microfarads
Two equal inductors in series = ____ the value; in parallel = ____ the value.
Twice; half (same rules as resistors)
Quick rule for equal-value components:
In series: Inductors double, capacitors halve.
In parallel: Inductors halve, capacitors double.
Practice Flash Cards -- B-005-009 (Inductors & Capacitors)
B-005-009-001: If two equal-value inductors are connected in series, what is their total inductance?
- A) Twice the value of one inductor
- B) Half the value of one inductor
- C) The value of one inductor times 4
- D) The value of one inductor divided by 4
A) Twice the value of one inductor
B-005-009-002: If two equal-value inductors are connected in parallel, what is their total inductance?
- A) The value of one inductor divided by 4
- B) Half the value of one inductor
- C) Twice the value of one inductor
- D) The value of one inductor times 4
B) Half the value of one inductor
B-005-009-003: If two equal-value capacitors are connected in series, what is their total capacitance?
- A) Half the value of either capacitor
- B) Twice the value of one capacitor
- C) The value of one capacitor times 4
- D) The value of one capacitor divided by 4
A) Half the value of either capacitor -- Caps are opposite to resistors.
B-005-009-004: If two equal-value capacitors are connected in parallel, what is their total capacitance?
- A) Half the value of either capacitor
- B) The value of one capacitor times 4
- C) The value of one capacitor divided by 4
- D) Twice the value of one capacitor
D) Twice the value of one capacitor
B-005-009-005: You are constructing an air-core inductor using a coil of wire. What parameters determine its inductance?
- A) Coil diameter, coil length and number of turns of wire
- B) Coil diameter, coil length and operating frequency
- C) Type of wire, coil length and number of turns of wire
- D) Coil diameter, coil orientation and number of turns of wire
A) Coil diameter, coil length and number of turns of wire
B-005-009-006: A capacitor is made of two identical metal plates separated by air. What parameters determine its capacitance?
- A) Surface area of the plates and applied voltage
- B) Surface area of the plates and spacing between the plates
- C) Operating frequency and spacing between the plates
- D) Type of metal and spacing between the plates
B) Surface area of the plates and spacing between the plates
B-005-009-007: What precaution must you take when using polarized electrolytic capacitors?
- A) Do not parallel with capacitors other than electrolytics
- B) Use them exclusively at radio frequencies
- C) Do not use them in series combinations
- D) Never apply a reverse voltage
D) Never apply a reverse voltage
B-005-009-008: If you wire two 12-millihenry chokes in series, what is the inductance of the combination?
- A) 6 millihenries
- B) 48 millihenries
- C) 3 millihenries
- D) 24 millihenries
D) 24 millihenries -- 12+12=24.
B-005-009-009: If you wire two 20-millihenry inductors in parallel, what is the inductance of the combination?
- A) 10 millihenries
- B) 40 millihenries
- C) 80 millihenries
- D) 5 millihenries
A) 10 millihenries -- 20/2=10.
B-005-009-010: If you wire two 20-microfarad capacitors in series, what is the capacity of the combination?
- A) 40 microfarads
- B) 80 microfarads
- C) 5 microfarads
- D) 10 microfarads
D) 10 microfarads -- Caps in series halve: 20/2=10.
B-005-009-011: If you wire two 24-microfarad capacitors in parallel, what is the capacity of the combination?
- A) 96 microfarads
- B) 4 microfarads
- C) 48 microfarads
- D) 12 microfarads
C) 48 microfarads -- Caps in parallel add: 24+24=48.
5.10 Reactance and Impedance B-005-010
In DC circuits, only resistance opposes current. In AC circuits, inductors and capacitors create an additional form of opposition called reactance. Unlike resistance, reactance depends on frequency -- the same inductor or capacitor offers different opposition at different frequencies.
Inductive Reactance
An inductor's opposition to AC current is called inductive reactance, calculated as \(X_L = 2\pi f L\). The key takeaway is straightforward: inductive reactance increases as frequency increases. Think of the flywheel analogy -- the faster things change, the more the flywheel resists. At low frequencies, an inductor is barely an obstacle. At high frequencies, it blocks strongly.
Capacitive Reactance
A capacitor's opposition to AC current is called capacitive reactance, calculated as \(X_C = \frac{1}{2\pi f C}\). Because frequency is in the denominator, the relationship is reversed: capacitive reactance decreases as frequency increases. Conversely, capacitive reactance increases as frequency decreases. At high frequencies, a capacitor lets current flow easily; at low frequencies, it blocks.
"Inductors Love Frequency, Capacitors Hate It": Inductive reactance goes UP with frequency (\(X_L\) gets larger). Capacitive reactance goes DOWN with frequency (\(X_C\) gets smaller).
Impedance
Impedance (Z) is the total opposition to AC current, combining both resistance and reactance. It is also described as the ratio of AC voltage to AC current in a circuit. The formula is:
$$Z = \sqrt{R^2 + (X_L - X_C)^2}$$
Practical Applications: Ferrite Chokes and Bypass Capacitors
The frequency-dependent nature of reactance is exploited in practical filtering components:
| Component | Property | Application |
| Ferrite choke coil | High reactance at radio frequencies | Blocks RF interference while allowing audio/DC through |
| Bypass capacitor (RF filter) | Low reactance at radio frequencies | Diverts interfering RF signals to ground in audio circuits |
| RF bypass capacitor | High reactance at audio frequencies | Passes RF but has little effect on audio signals |
| RF choke coil | Low reactance at low frequencies | Allows desired low-frequency signals to pass through |
Notice how the inductor (choke) and capacitor (bypass) work as a team: the choke blocks RF (high reactance at RF), while the bypass cap shunts RF to ground (low reactance at RF). Together they keep RF out of audio circuits.
What term equals the ratio of AC voltage to AC current in a system or circuit?
Impedance
Practice Flash Cards -- B-005-010 (Reactance & Impedance)
B-005-010-001: How does an inductor react to AC?
- A) As the amplitude of the applied AC increases, the reactance decreases
- B) As the amplitude of the applied AC increases, the reactance increases
- C) As the frequency of the applied AC increases, the reactance decreases
- D) As the frequency of the applied AC increases, the reactance increases
D) As the frequency of the applied AC increases, the reactance increases
B-005-010-002: How does a capacitor react to AC?
- A) As the frequency of the applied AC increases, the reactance decreases
- B) As the frequency of the applied AC increases, the reactance increases
- C) As the amplitude of the applied AC increases, the reactance increases
- D) As the amplitude of the applied AC increases, the reactance decreases
A) As the frequency of the applied AC increases, the reactance decreases
B-005-010-003: The reactance of capacitors increases as:
- A) Frequency decreases
- B) Applied voltage increases
- C) Applied voltage decreases
- D) Frequency increases
A) Frequency decreases
B-005-010-004: What is the term for the opposition to alternating current caused by the combined effect of reactance and resistance?
- A) Conductance
- B) Impedance
- C) Reluctance
- D) Admittance
B) Impedance
B-005-010-005: What term equals the ratio of AC voltage to AC current in a system or circuit?
- A) Conductance
- B) Impedance
- C) Resistance
- D) Reactance
B) Impedance
B-005-010-006: What circuit parameter change causes an inductor's reactance to increase?
- A) An increase in voltage
- B) An increase in frequency
- C) A decrease in frequency
- D) An increase in current
B) An increase in frequency
B-005-010-007: What property allows a coil wound on a ferrite core to mitigate the effects of an offending radio signal?
- A) Low reactance at radio frequencies
- B) Low reactance at audio frequencies
- C) High reactance at audio frequencies
- D) High reactance at radio frequencies
D) High reactance at radio frequencies
B-005-010-008: What property allows a bypass capacitor in an audio circuit to divert an interfering RF signal?
- A) Low reactance at audio frequencies
- B) High reactance at audio frequencies
- C) Low reactance at radio frequencies
- D) High reactance at radio frequencies
C) Low reactance at radio frequencies
B-005-010-009: What property allows an RF bypass capacitor to have little effect on an audio circuit?
- A) High reactance at radio frequencies
- B) Low reactance at audio frequencies
- C) High reactance at audio frequencies
- D) Low reactance at radio frequencies
C) High reactance at audio frequencies
B-005-010-010: What property allows an RF choke coil to have little effect on signals meant to flow through the coil?
- A) Low reactance at low frequencies
- B) High reactance at low frequencies
- C) Low reactance at high frequencies
- D) High reactance at high frequencies
A) Low reactance at low frequencies
B-005-010-011: In general, the reactance of inductors increases with:
- A) Increasing applied voltage
- B) Increasing AC frequency
- C) Decreasing AC frequency
- D) Decreasing applied voltage
B) Increasing AC frequency
A transformer transfers electrical energy from one circuit to another through electromagnetic induction, using two coils of wire (windings) wrapped around a shared core. Transformers can step voltage up or down, and they are essential for power supplies and impedance matching in radio equipment.
Turns ratio determines voltage: V_s/V_p = N_s/N_p
Transformer Equations
The voltage ratio follows the turns ratio, while the current ratio is its inverse (because power is conserved):
$$\frac{V_s}{V_p} = \frac{N_s}{N_p} \qquad \qquad \frac{I_p}{I_s} = \frac{N_s}{N_p}$$
In an ideal (lossless) transformer, power in equals power out: \(V_p \times I_p = V_s \times I_s\). This means if voltage goes up, current must come down by the same factor, and vice versa. Most exam problems assume an ideal transformer unless stated otherwise.
A transformer can also change impedance by carrying different voltages and currents in each winding. The impedance ratio follows the square of the turns ratio: \(\frac{Z_p}{Z_s} = \left(\frac{N_p}{N_s}\right)^2\).
Worked Examples
Worked Example: Primary current from power (B-005-011-001)
Q: A transformer with a 120-volt primary voltage supplies 250 watts to a transmitter. Neglecting losses, what is the approximate primary current?
Step 1: Use \(P = V \times I\), rearranged: \(I_p = \frac{P}{V_p}\)
Step 2: \(I_p = \frac{250}{120} \approx 2.1\) amperes
Answer: (D) 2.1 amperes
Worked Example: Step-up transformer voltage (B-005-011-004)
Q: The primary winding of a transformer has 250 turns, and the secondary has 500 turns. If the input voltage is 120 volts, what is the secondary voltage?
Step 1: \(V_s = V_p \times \frac{N_s}{N_p}\)
Step 2: \(V_s = 120 \times \frac{500}{250} = 120 \times 2 = 240\) V
Answer: (C) 240 V
Worked Example: Secondary current from power balance (B-005-011-003)
Q: A transformer with a single 12-volt secondary draws 0.5 amperes through its 120-volt primary. Assuming no losses, what current is drawn from the secondary?
Step 1: Power in = Power out: \(V_p \times I_p = V_s \times I_s\)
Step 2: \(120 \times 0.5 = 12 \times I_s\)
Step 3: \(I_s = \frac{60}{12} = 5\) amperes
Answer: (D) 5 amperes
Worked Example: Step-up 1:5 primary current (B-005-011-008)
Q: A step-up transformer with a primary to secondary turns ratio of 1:5 delivers 50 milliamperes to a load. Assuming 100% efficiency, what is the primary current?
Step 1: Current is inverse of turns ratio: \(I_p = I_s \times \frac{N_s}{N_p}\)
Step 2: \(I_p = 50 \times \frac{5}{1} = 250\) mA
Answer: (A) 250 mA
Other Transformer Facts
- The strength of the magnetic field around a conductor in air is directly proportional to the current in the conductor.
- Maximum induced voltage in a coil occurs when current is going through its greatest rate of change.
- Coupling (induction) between two wires is maximum when the wires are close and parallel.
- A permanent magnet is most likely made from steel.
- Warm iron laminations in a transformer confirm that energy transfer from primary to secondary is not perfect -- some energy is lost as heat.
A transformer primary consumes 10 watts. If the secondary voltage is 5 volts, what is the secondary current (neglecting losses)?
2 amperes (I = P/V = 10/5 = 2)
Exam shortcut for transformer problems: Always use \(P_{in} = P_{out}\) (assume ideal unless stated otherwise). Find the unknown current or voltage using \(P = V \times I\).
Practice Flash Cards -- B-005-011 (Transformers)
B-005-011-001: A transformer with a 120-volt primary voltage supplies 250 watts to a transmitter. Neglecting losses, what is the approximate primary current?
- A) 0.48 amperes
- B) 1.4 amperes
- C) 3.1 amperes
- D) 2.1 amperes
D) 2.1 amperes -- I=250/120=2.08.
B-005-011-002: How can a transformer with two windings change impedance?
- A) By allowing the difference to be dissipated in core losses
- B) By matching winding resistance to impedance
- C) By carrying different voltages and currents in each winding
- D) By using the correct magnetic coupling between windings
C) By carrying different voltages and currents in each winding
B-005-011-003: A transformer with a single 12-volt secondary draws 0.5 amperes through its 120-volt primary. Assuming no losses, what current is drawn from the secondary?
- A) 2.5 amperes
- B) 25 amperes
- C) 50 amperes
- D) 5 amperes
D) 5 amperes -- P=120×0.5=60W; I=60/12=5.
B-005-011-004: The primary winding of a transformer has 250 turns, and the secondary has 500 turns. If the input voltage is 120 volts, what is the secondary voltage?
- A) 620 V
- B) 60 V
- C) 240 V
- D) 480 V
C) 240 V -- 120 × (500/250) = 240.
B-005-011-005: The strength of the magnetic field around a conductor in air is:
- A) Directly proportional to the diameter of the conductor
- B) Inversely proportional to the voltage on the conductor
- C) Directly proportional to the current in the conductor
- D) Inversely proportional to the diameter of the conductor
C) Directly proportional to the current in the conductor
B-005-011-006: Maximum induced voltage in a coil occurs when:
- A) Current is going through its least rate of change
- B) The magnetic field around the coil is not changing
- C) Current is going through its greatest rate of change
- D) The current through the coil is DC
C) Current is going through its greatest rate of change
B-005-011-007: A transformer primary winding consumes 10 watts. Neglecting losses, if the secondary voltage is 5 volts, what is the secondary current?
- A) 1 ampere
- B) 5 amperes
- C) 2 amperes
- D) 0.5 amperes
C) 2 amperes -- I=10/5=2.
B-005-011-008: A step-up transformer with a primary to secondary turns ratio of 1:5 delivers 50 milliamperes to a load. Assuming 100% efficiency, what is the primary current?
- A) 250 mA
- B) 2500 mA
- C) 10 mA
- D) 0.25 mA
A) 250 mA -- Ip = Is × (Ns/Np) = 50×5 = 250.
B-005-011-009: When is coupling (induction) between two wires maximum?
- A) When the wires are close and parallel
- B) When the wires are close and at right angles
- C) When the wires are separated and parallel
- D) When the wires are separated and at right angles
A) When the wires are close and parallel
B-005-011-010: A permanent magnet would most likely be made from:
- A) Aluminum
- B) Brass
- C) Steel
- D) Copper
C) Steel
B-005-011-011: What confirms the fact that the transfer of energy from the primary to the secondary of a transformer is not perfect?
- A) Warm iron laminations
- B) Noisy operation
- C) Large secondary current
- D) High primary voltage
A) Warm iron laminations -- Heat = wasted energy.
5.12 Resonant Circuits B-005-012
When an inductor and capacitor are combined in a circuit, something remarkable happens: at one specific frequency, the inductive reactance and capacitive reactance become equal and cancel each other out. This is resonance, and it is the principle that allows your radio to tune in to a single station out of thousands.
The Resonance Condition
At resonance, \(X_L = X_C\). The two reactances are equal in magnitude and opposite in sign, so they cancel. The frequency at which this occurs is:
$$f_0 = \frac{1}{2\pi\sqrt{LC}}$$
Resonance occurs when inductive reactance equals capacitive reactance (\(X_L = X_C\)). A tuned circuit requires exactly two components: an inductor and a capacitor.
Series vs. Parallel Resonance
The behaviour at resonance depends on whether the inductor and capacitor are connected in series or parallel:
| Property | Series Resonant | Parallel Resonant |
| Impedance at resonance | Low (minimum) | High (maximum) |
| Current at resonance | Maximum | Minimum (from source) |
| Behaves like | Pure resistance (low) | Pure resistance (high) |
SERIES RESONANT CIRCUIT PARALLEL RESONANT CIRCUIT
+--[L]--[C]--+ +--[L]--+
| | | |
[V] [R] +---+ +---+
| | | +--[C]--+ |
+------------+ [V] [R]
| |
At resonance: +---------------+
XL = XC, they cancel
Only R remains At resonance:
Z is MINIMUM XL = XC
Current is MAXIMUM Z is MAXIMUM
Series vs. parallel resonant circuits
Bandwidth and Selectivity
Bandwidth is the range of frequencies around resonance where the effect is still significant. A narrow bandwidth means better selectivity -- the circuit can pick out one station more precisely. The Q factor (quality factor) determines bandwidth: higher Q means narrower bandwidth.
Adding a resistor to a resonant circuit affects Q and bandwidth but does not change the resonant frequency itself, because the resonant frequency depends only on L and C.
Resonant circuits in a receiver are used to select the desired signal frequencies -- this is how tuning works.
At resonance, a parallel tuned circuit exhibits what kind of impedance?
High impedance (maximum)
Practice Flash Cards -- B-005-012 (Resonant Circuits)
B-005-012-001: Resonance is the condition that exists when:
- A) Inductive reactance is the only opposition in the circuit
- B) The circuit contains no resistance
- C) Resistance is equal to the reactance
- D) Inductive reactance and capacitive reactance are equal
D) Inductive reactance and capacitive reactance are equal
B-005-012-002: At resonance, what impedance does a parallel tuned circuit exhibit?
- A) High impedance
- B) Low impedance
- C) Impedance equal to reactance of the circuit
- D) Impedance equal to resistance of the circuit
A) High impedance
B-005-012-003: While the resonant frequency of a tuned circuit is a single frequency, the effect of resonance is significant over a certain range of frequencies. What is this range called?
- A) Shape factor
- B) Response curve
- C) Bandwidth
- D) Quality factor
C) Bandwidth
B-005-012-004: What two components are required to form a tuned circuit?
- A) Resistor and transistor
- B) Capacitor and resistor
- C) Diode and transistor
- D) Inductor and capacitor
D) Inductor and capacitor
B-005-012-005: When a parallel coil-capacitor combination is supplied with AC of different frequencies, there will be one frequency where the impedance will be highest. This is the:
- A) Reactive frequency
- B) Resonant frequency
- C) Maximum frequency
- D) Inductive frequency
B) Resonant frequency
B-005-012-006: In a parallel-resonant circuit at resonance, the circuit has:
- A) High impedance
- B) Low impedance
- C) Low mutual inductance
- D) High mutual inductance
A) High impedance
B-005-012-007: In a series resonant circuit at resonance, the circuit has:
- A) Low mutual inductance
- B) High mutual inductance
- C) Low impedance
- D) High impedance
C) Low impedance
B-005-012-008: A coil and an air-spaced capacitor are arranged to form a resonant circuit. The resonant frequency will remain the same if we:
- A) Wind more turns on the coil
- B) Add a resistor to the circuit
- C) Increase the area of plates in the capacitor
- D) Insert Mylar sheets between the plates of the capacitor
B) Add a resistor to the circuit -- Only L and C affect resonant frequency.
B-005-012-009: Resonant circuits in a receiver are used to:
- A) Amplify audio signals
- B) Adjust voltage levels
- C) Select the desired signal frequencies
- D) Filter direct current
C) Select the desired signal frequencies
B-005-012-010: Resonance is the condition that exists when:
- A) Resistance is equal to the reactance
- B) Inductive reactance and capacitive reactance are equal and opposite in sign
- C) Inductive reactance is the only opposition in the circuit
- D) The circuit contains no resistance
B) Inductive reactance and capacitive reactance are equal and opposite in sign
B-005-012-011: What happens to current when a series RLC circuit is tuned to the frequency of the source?
- A) It is limited by capacitive reactance
- B) It reaches minimum
- C) It reaches maximum
- D) It is limited by inductive reactance
C) It reaches maximum -- Series resonance = minimum impedance = maximum current.
5.13 Test Instruments B-005-013
To troubleshoot and measure circuits, you need instruments. The exam tests your knowledge of which instrument to use, how to connect it, and what it looks like to the circuit.
Connecting Instruments
The two most fundamental instruments are the voltmeter and ammeter. They are connected differently because they measure different things:
Voltmeter in PARALLEL (high resistance), Ammeter in SERIES (low resistance)
"Voltage is across, Current is through." You measure voltage ACROSS a component (parallel). You measure current THROUGH the wire (series). Remember: "V-P, A-S" (Voltmeter-Parallel, Ammeter-Series).
Why does this matter? A voltmeter must have high resistance so it does not draw significant current from the circuit it is measuring. An ammeter must have low resistance so it does not significantly impede the current it is measuring. When measuring voltage across a circuit component, the voltmeter appears to be a high value resistance (essentially an open circuit). When measuring current, the ammeter appears as a low value resistance.
Types of Instruments
| Instrument | Measures | Connection |
| Voltmeter | Voltage (potential difference) | Parallel |
| Ammeter | Current | Series |
| Multimeter | Voltage, current, and resistance | Varies by function |
| RF Wattmeter | Power at transmitter output (direct measurement) | In-line |
| Ohmmeter | Resistance | Component removed from circuit |
An ammeter is also the correct instrument for measuring the final power amplifier current.
Measurement Quality
| Term | Definition | Example |
| Accuracy | Ability to display values that are true to reality | A thermometer reading 100°C when water is actually boiling |
| Precision | Ability to give consistent, repeatable results | A meter that always reads 5.02 V (even if actual is 5.00 V) |
| Resolution | Smallest change the instrument can detect | A meter reading to 0.01 V vs. one reading to 0.1 V |
What instrument provides a direct measurement of power at the output of a transmitter?
An RF wattmeter
Practice Flash Cards -- B-005-013 (Test Instruments)
B-005-013-001: How is a voltmeter usually connected to a circuit under test?
- A) In quadrature with the circuit
- B) In phase with the circuit
- C) In parallel with the circuit
- D) In series with the circuit
C) In parallel with the circuit
B-005-013-002: How is an ammeter usually connected to a circuit under test?
- A) In parallel with the circuit
- B) In series with the circuit
- C) In quadrature with the circuit
- D) In phase with the circuit
B) In series with the circuit
B-005-013-003: What does a multimeter measure?
- A) Resistance and reactance
- B) SWR and power
- C) Voltage, current and resistance
- D) Resistance, capacitance and inductance
C) Voltage, current and resistance
B-005-013-004: What is the correct instrument to measure the final power amplifier current?
- A) A wattmeter
- B) A voltmeter
- C) An ammeter
- D) An ohmmeter
C) An ammeter
B-005-013-005: When measuring the voltage across a circuit component, what does the voltmeter appear to be in the circuit?
- A) An open circuit
- B) A perfect conductor
- C) A high value resistance
- D) A low value resistance
C) A high value resistance
B-005-013-006: When measuring current drawn from a DC power supply, what does the ammeter placed in the circuit appear as?
- A) A perfect conductor
- B) An additional load
- C) A high value resistance
- D) A low value resistance
D) A low value resistance
B-005-013-007: What instrument can provide a direct measurement of power at the output of a transmitter?
- A) RF wattmeter
- B) Wavemeter
- C) Field-strength meter
- D) Ammeter
A) RF wattmeter
B-005-013-008: Potential difference is measured by means of:
- A) A voltmeter
- B) A wattmeter
- C) An ohmmeter
- D) An ammeter
A) A voltmeter
B-005-013-009: What instrument is used to measure electrical current?
- A) Wattmeter
- B) Voltmeter
- C) Ammeter
- D) Wavemeter
C) Ammeter
B-005-013-010: What term describes the ability of an instrument to display values that are true to reality?
- A) Stability
- B) Accuracy
- C) Precision
- D) Resolution
B) Accuracy
Quick Reference Summary
Unit Prefix Quick Reference
| Prefix | Symbol | Factor | Example |
| Giga | G | \(10^9\) | 1 GHz = 1,000 MHz |
| Mega | M | \(10^6\) | 1 MHz = 1,000 kHz |
| Kilo | k | \(10^3\) | 1 kΩ = 1,000 Ω |
| Milli | m | \(10^{-3}\) | 1 mA = 0.001 A |
| Micro | μ | \(10^{-6}\) | 1 μF = 0.000001 F |
| Pico | p | \(10^{-12}\) | 1 pF = 0.000001 μF |
Essential Formulas
| Formula | Use |
| \(V = I \times R\) | Ohm's Law -- voltage |
| \(I = V / R\) | Ohm's Law -- current |
| \(R = V / I\) | Ohm's Law -- resistance |
| \(P = V \times I\) | Power from voltage and current |
| \(P = I^2 \times R\) | Power from current and resistance |
| \(P = V^2 / R\) | Power from voltage and resistance |
| \(R_{series} = R_1 + R_2 + \ldots\) | Resistors / inductors in series |
| \(1/R_{par} = 1/R_1 + 1/R_2 + \ldots\) | Resistors / inductors in parallel |
| \(C_{parallel} = C_1 + C_2 + \ldots\) | Capacitors in parallel (opposite!) |
| \(1/C_{series} = 1/C_1 + 1/C_2 + \ldots\) | Capacitors in series (opposite!) |
| \(X_L = 2\pi f L\) | Inductive reactance |
| \(X_C = 1 / (2\pi f C)\) | Capacitive reactance |
| \(V_s/V_p = N_s/N_p\) | Transformer voltage ratio |
| \(f_0 = 1 / (2\pi\sqrt{LC})\) | Resonant frequency |
| \(\text{dB} = 10 \log_{10}(P_2/P_1)\) | Decibel calculation |
Decibel Quick Reference
| dB | Power Ratio | dB | Power Ratio |
| +3 dB | × 2 | -3 dB | ÷ 2 |
| +6 dB | × 4 | -6 dB | ÷ 4 |
| +9 dB | × 8 | -9 dB | ÷ 8 |
| +10 dB | × 10 | -10 dB | ÷ 10 |
| +20 dB | × 100 | -20 dB | ÷ 100 |
| +30 dB | × 1,000 | -30 dB | ÷ 1,000 |
Component Combination Quick Reference
| Component | Series | Parallel |
| Resistors | Add: \(R_1 + R_2\) | Reciprocal: \(\frac{R_1 R_2}{R_1+R_2}\) |
| Inductors | Add: \(L_1 + L_2\) | Reciprocal: \(\frac{L_1 L_2}{L_1+L_2}\) |
| Capacitors | Reciprocal: \(\frac{C_1 C_2}{C_1+C_2}\) | Add: \(C_1 + C_2\) |
Remember: Capacitors are the opposite of resistors/inductors!
Key Facts to Memorize
- Best conductor: Copper (among common exam choices)
- Good insulators: Glass, air, porcelain
- Power unit: Watt
- Open circuit: No current flows
- Short circuit: Excessive current, blows fuse
- Continuity: Circuit is closed/complete
- Audio range: 20 Hz to 20 kHz
- Household AC: Sine wave at 60 Hz
- Doubling voltage across a fixed resistance = 4× power
- Inductive reactance: Increases with frequency
- Capacitive reactance: Decreases with frequency
- Resonance: XL = XC
- Parallel resonant: High impedance | Series resonant: Low impedance
- Tuned circuit: Inductor + capacitor
- Voltmeter: Parallel, high resistance
- Ammeter: Series, low resistance
- Electrolytic caps: Never apply reverse voltage!
- Permanent magnets: Made from steel
- Accuracy: True to reality