6.1 Electrical Elements, Units & Power Types

6.1 Electrical Elements, Units & Power Types

Quick Answer: Multiply an AC voltmeter reading by an AC ammeter reading and you get apparent power (VA). The basic unit of electrical power is the watt. Energy stored in an electrostatic field is measured in joules, and a capacitor is the device that stores it. Reactive power is the out-of-phase L/C power; true power is what resistors actually dissipate. On AC, true power = apparent power × power factor.

Element 3 begins with electrical principles because every later topic—components, circuits, transmitters, antennas—assumes you can name units, separate power types, and avoid treating every V×I product as “watts on the workbench.” Topic 3-A (key topics 001–008) is the foundation under all of general radiotelephone theory.

Core SI units GROL techs must own

Exam hook — basic unit of electrical power: the watt. Not the ohm (resistance), not the volt (potential), not the ampere (current). Power is the rate of doing electrical work; energy is the amount of work stored or delivered over time.

Exam hook — electrostatic energy: the amount of electrical energy stored in an electrostatic field is expressed in joules. The device that stores electrical energy in an electrostatic field is a capacitor (not a battery, transformer, or inductor). Batteries store chemical energy; inductors store energy in a magnetic field; transformers couple magnetic fields between windings but are not the classic “electrostatic store” answer.

Apparent power: voltmeter × ammeter

What is the product of the readings of an AC voltmeter and an AC ammeter called? Apparent power.

If a voltmeter across an AC circuit reads 120 V and an ammeter in series reads 2 A, then:

[ S = V \times I = 120,\mathrm{V} \times 2,\mathrm{A} = 240,\mathrm{VA} ]

That product is apparent power (often written S or |S|). It is not automatically true power, because on AC the voltage and current may not be in phase. Meters that display RMS voltage and RMS current give you the magnitudes; they do not by themselves tell you the phase angle between v and i.

Memory hook: VA is what the meters multiply; watts is what the resistor burns; VAR is what L and C bounce.

True power vs reactive power

Reactive power is the exam term for the out-of-phase power associated with inductors and capacitors. In a pure inductance or pure capacitance (ideal case), voltage and current are 90° apart. Instantaneous power still exists, but over a full cycle the average energy delivered to the ideal L or C is zero—the energy is stored and returned.

Pool-level fact you will reuse in later math topics: in a circuit with both inductors and capacitors, reactive power alternates between magnetic and electric fields and is not dissipated. It is not “burned up as heat in the reactive fields.” Heat is a true-power story (resistance, losses, radiation loads).

How do you compute true power when AC voltage and current are out of phase?

[ P_{\mathrm{true}} = S \times \mathrm{PF} = V_{\mathrm{RMS}} , I_{\mathrm{RMS}} , \cos\phi ]

In words: multiply apparent power by the power factor. Power factor is the cosine of the phase angle between voltage and current (for sinusoidal steady state). Multiplying RMS V by RMS I alone gives apparent power, not true power, when phase shift is present. Dividing or subtracting “power factor from apparent power” is not the rule.

Worked example — three power numbers from one circuit

A marine SSB PA supply rail is measured at 24.0 V RMS into a load that draws 5.0 A RMS. An instrument (or calculation) shows power factor 0.80 lagging because of inductive filtering and cabling.

1. Apparent power: (S = 24 \times 5 = 120,\mathrm{VA})
2. True power: (P = 120 \times 0.80 = 96,\mathrm{W})
3. Reactive power (magnitude): using (Q = \sqrt{S^2 - P^2}) gives (Q = \sqrt{120^2 - 96^2} = 72,\mathrm{VAR})

If you only memorize “power is V times I,” you would report 120 W and miss that only 96 W is true power. Element 3 rewards the distinction.

P = VI relationships without the traps

On DC (or purely resistive AC), true power collapses to the simple product:

[ P = VI = I^2 R = \frac{V^2}{R} ]

On AC with reactance, keep these rules of thumb:

1. V_RMS × I_RMS → apparent power (VA), always a valid magnitude product for sinusoids when both meters read RMS.
2. True power requires resistance or power factor: (P = I^2 R) in the resistive elements, or (P = S \times \mathrm{PF}).
3. Inductive reactance formula (preview of §6.3, already in 3-A-001): (X_L = 2\pi f L). Reactance is not resistance, but it still limits AC current and creates the phase shift that produces reactive power.
4. Do not call reactive power “peak envelope power.” PEP is an RF envelope concept used later for transmitters; reactive power is the L/C out-of-phase power.

Maximum power transfer

Assuming a source with fixed internal resistance (or fixed internal impedance), maximum power is transferred to the load when the load impedance equals the internal impedance of the source. That is the classic conjugate-match idea stated at Element 3 level: match the load to the source, do not deliberately run the load “much higher” or “much lower” if the goal is maximum delivered power.

In RF work you will revisit this as antenna and transmission-line matching. At Topic 3-A, remember the pure statement: load Z = source internal Z for maximum power transfer.

Putting §6.1 together for exam day

When a stem mentions meters, watts, joules, capacitors, reactive power, or power factor, run this checklist:

1. V_AC × I_AC → apparent power (VA).
2. Basic power unit → watt.
3. Electrostatic stored energy unit → joule; device → capacitor.
4. Out-of-phase L/C power → reactive power (not dissipated like true power).
5. True power with phase shift → apparent × power factor.
6. Max power transfer → load impedance matches source internal impedance.

Master these six hooks and you clear the electrical-elements slice of Topic 3-A before moving into magnetism, materials, RLC properties, waveforms, and semiconductors.