Electrical Fundamentals

Part II (i) Electrical units — volt, ampere, ohm, watt; wavelength, frequency & their relationship

Picture electricity as water in a pipe. Voltage is the pressure pushing the water; current is how much water flows per second; resistance is how narrow the pipe is; power is the useful work the flow does. The analogy is exact enough to carry you through every formula in this chapter.

Push harder (more volts) and more current flows; narrow the pipe (more ohms) and less flows. The three lock together:

A 28 V aircraft bus supplies a 14 Ω heater. Current = V/R = 28 ÷ 14 = 2 A.
If the resistance halves to 7 Ω, current doubles to 4 A (inverse relationship). If the bus sags to 24 V across the original 14 Ω, current = 24 ÷ 14 ≈ 1.71 A.

From 7.2: 28 V across 14 Ω drawing 2 A. P = V × I = 28 × 2 = 56 W.
Check with I²R = 2² × 14 = 4 × 14 = 56 W. Check with V²/R = 28² ÷ 14 = 784 ÷ 14 = 56 W. All agree.

1 kW = 1000 W. Efficiency = useful output power ÷ input power (always < 100% because some energy becomes heat). A transmitter rated 50 W output draws more than 50 W from the bus because the power amplifier is not perfectly efficient.

DC (Direct Current) flows in one direction at a steady value — e.g. a battery.
AC (Alternating Current) reverses direction periodically, tracing a sine wave; the number of complete cycles per second is the frequency (hertz), and the time for one cycle is the period (T = 1/f). Every radio-frequency signal is alternating current.

An AC wave is described by its peak value, its peak-to-peak (twice peak), and its RMS (root-mean-square) — the equivalent DC value that would do the same heating work. For a sine wave: RMS = 0.707 × peak (and peak = 1.414 × RMS). Mains and aircraft AC voltages are quoted as RMS.

A typical aircraft has a 28 V DC system (battery/bus feeding avionics) and a 115 V AC at 400 Hz system (for many instruments and as the form every radio signal takes on the way to the antenna). 400 Hz is used rather than 50/60 Hz because transformers and motors can be smaller and lighter at the higher frequency — a real weight saving.

Resistance increases with length, decreases with cross-sectional area (a thicker wire conducts better), depends on the material (its resistivity), and — for most metals — increases with temperature. Hence a long, thin, hot wire has the highest resistance.

Series: 10 Ω + 20 Ω = 30 Ω. Across 30 V the current = 30/30 = 1 A; the 10 Ω drops 10 V and the 20 Ω drops 20 V (voltage divides).
Parallel: two 10 Ω resistors → 1/R = 1/10 + 1/10 = 2/10, so R = 5 Ω (half of one). For two equal resistors, parallel total = half the value; the current splits equally.

Resistors oppose current the same at any frequency. Capacitors and inductors oppose AC by an amount that depends on frequency — and that frequency dependence is exactly what lets a radio tune to one station and reject the rest.

Capacitor — stores energy in an electric field; blocks DC, passes AC more easily as frequency rises (its reactance falls with frequency). Unit: farad (F).
Inductor — stores energy in a magnetic field; passes DC, opposes AC more as frequency rises (its reactance rises with frequency). Unit: henry (H).
Their frequency-dependent opposition is called reactance (X), measured in ohms.

Put an inductor and capacitor together and at one special resonant frequency their reactances cancel, giving a sharp peak in response. A tuned LC circuit selects that frequency and rejects others — this is how a receiver is "tuned" to a station and how a transmitter sets its operating frequency.

Every radio wave travels at the speed of light. Its frequency (cycles per second) and its wavelength (length of one cycle) are two views of the same wave — fix the speed and one decides the other.

1 kHz = 1,000 Hz · 1 MHz = 1,000,000 Hz · 1 GHz = 1,000 MHz. Period and frequency are reciprocals: a 1 MHz signal has a period of 1 microsecond.

A 120 MHz VHF signal: λ = c/f = (3 × 10&sup8;) ÷ (120 × 10&sup6;) = 2.5 m.
A 300 MHz signal: λ = (3 × 10&sup8;) ÷ (300 × 10&sup6;) = 1 m. A 3 MHz HF signal: λ = (3 × 10&sup8;) ÷ (3 × 10&sup6;) = 100 m. Higher frequency → shorter wavelength.