This page is part of the N0NJY General Class self-study course for Technician operators upgrading to General.
Overview
Technician covered basic DC theory. General Class goes deeper into AC circuits: impedance, reactance,
resonance, transformers, power calculations, and decibels. These are the foundations for understanding
antenna systems, transmission lines, and transceiver design.
Impedance and Reactance
Resistance (R): Opposition to current that dissipates energy as heat. Measured in ohms.
Constant regardless of frequency.
Inductive Reactance (XL): XL = 2πfL — Increases with frequency. An inductor resists
changes in current.
Capacitive Reactance (XC): XC = 1/(2πfC) — Decreases as frequency increases.
A capacitor resists changes in voltage.
Impedance (Z): The total opposition to AC current flow, combining resistance and reactance.
Z = R + jX. The magnitude |Z| = √(R² + X²). Measured in ohms.
Series and Parallel Resonance
A resonant circuit contains an inductor and capacitor tuned so that XL = XC. At the resonant frequency:
f = 1 / (2π√LC)
- Series resonance: Impedance is at its minimum (equal to resistance only)
- Parallel resonance: Impedance is at its maximum
- Q factor: Q = XL/R. High Q = sharp, narrow resonance. Low Q = broad resonance.
Transformers and Baluns
A transformer uses mutual inductance to transfer energy between windings.
- Voltage turns ratio: Vs/Vp = Ns/Np
- Impedance transformation: Zs/Zp = (Ns/Np)²
- A 2:1 turns ratio transforms a 200-ohm load to 50 ohms
A balun (BALanced-UNbalanced) connects a balanced antenna (dipole) to unbalanced feedline (coax).
A 1:1 current choke balun is generally the best choice for a dipole-to-coax connection.
A 4:1 balun transforms 200 ohms to 50 ohms.
AC Power
- True power (watts): P = E × I × cos(θ)
- Power factor: PF = cos(θ) = P / S
- PEP (Peak Envelope Power): The average power during one RF cycle at the modulation peak.
The FCC measures transmitter output power as PEP.
- RMS voltage: VRMS = Vpeak × 0.707
Decibels
The decibel (dB) is a logarithmic ratio expressing power relationships:
dB = 10 × log10(P2/P1)
Values you must memorize:
- +3 dB = power doubled | −3 dB = power halved
- +10 dB = power × 10 | −10 dB = power ÷ 10
- +20 dB = power × 100 | +30 dB = power × 1000
- 0 dBd = 2.15 dBi (dipole vs. isotropic reference for antenna gain)
Practice Questions
Q1 (G5A01) — What is impedance?
- A. The rate of energy transfer in a resonant circuit
- B. The opposition to the flow of current in an AC circuit
- C. The opposition to the flow of current in a DC circuit
- D. The force that produces voltage in a circuit
Q2 (G5B01) — What dB change represents a two-times increase or decrease in power?
- A. Approximately 6 dB
- B. Approximately 3 dB
- C. Approximately 10 dB
- D. Approximately 1 dB
Q3 (G5A03) — What is the unit of impedance?
- A. Volt
- B. Ampere
- C. Ohm
- D. Watt
Q4 (G5C01) — What causes a voltage to appear across the secondary winding of a transformer?
- A. Capacitive coupling between the windings
- B. Mutual inductance
- C. Resistive coupling through the core
- D. Electrostatic charge buildup
Q5 (G5B03) — How many watts are consumed by a 12-VDC light bulb drawing 0.2 amperes?
- A. 2.4 watts
- B. 60 watts
- C. 0.024 watts
- D. 24 watts
Q6 (G5A07) — What happens when the impedance of a load equals the internal impedance of the power source?
- A. Maximum voltage is delivered to the load
- B. Maximum current is delivered to the load
- C. Maximum power transfer occurs
- D. No power is delivered to the load
Q7 (G5B09) — What is the RMS voltage of a sine wave with a peak value of 17 volts?
- A. 12 volts
- B. 17 volts
- C. 24 volts
- D. 8.5 volts
Q8 (G5A11) — What is inductive reactance?
- A. The opposition to DC current in an inductor
- B. The opposition to AC current due to the inductance of a coil
- C. The amount of energy stored in a capacitor
- D. The resistance of the wire in an inductor
Answer Key
- B — Impedance is the total opposition to AC current flow
- B — 3 dB represents a doubling or halving of power
- C — Ohms are the unit of impedance
- B — Mutual inductance causes the secondary voltage
- A — P = E × I = 12 × 0.2 = 2.4 watts
- C — Maximum power transfer occurs when source and load impedances are equal
- A — RMS = Peak × 0.707 = 17 × 0.707 ≈ 12 volts
- B — Inductive reactance is the opposition to AC current due to inductance