GEK-1 — Electrical Quantities and Calculation Setup

Build a calculation before reaching for the calculator

General Electrical Knowledge accounts for four questions in the current Tennessee LLE outline. The authorized references include the 2017 NEC Handbook, Ugly's Electrical References (2020), and OSHA 29 CFR Part 1926. The code handbook supplies definitions and installation context, while a formula reference helps with arithmetic; neither replaces a disciplined setup. Use this sequence on every numerical problem:

1. Identify the requested quantity. Write its symbol and required unit.
2. List known values with units. Do not copy bare numbers.
3. State assumptions. Examples include DC, purely resistive load, balanced three-phase load, or negligible conductor loss.
4. Convert prefixes before substitution. Keep one consistent unit system.
5. Select and rearrange the formula symbolically. Isolate the unknown first.
6. Substitute, calculate, label, and check. Round only at the end unless the question directs otherwise.

This method prevents the most common errors: mixing kilowatts with watts, using energy as though it were power, and solving for the wrong variable.

Know what each quantity means

Voltage (V) is electric potential difference: the electrical pressure available to move charge between two points. Its unit is the volt (V). Current (I) is the rate of charge flow, measured in amperes (A). Resistance (R) opposes current and is measured in ohms (Ω). These quantities are related by Ohm's law:

• V = I × R
• I = V ÷ R
• R = V ÷ I

Power (P) is the rate at which electrical energy is converted or transferred. Real power is measured in watts (W). For a DC circuit or a purely resistive AC load, P = V × I. Combining that relationship with Ohm's law gives P = I²R and P = V²/R. Those two derived forms are useful only when their listed variables are known.

Energy (E) is power accumulated over time. It is measured in watt-hours (Wh) or kilowatt-hours (kWh) for common billing and usage calculations: E = P × t. Power and energy are not interchangeable. A 2 kW heater is a load rating; operating it for 3 hours uses 6 kWh of energy. The unit is kWh, not “kW per hour.”

Control prefixes and units

Prefixes change magnitude. milli means 10⁻³, kilo means 10³, and mega means 10⁶. Therefore, 750 mA equals 0.750 A, 4.8 kW equals 4,800 W, 2.2 MΩ equals 2,200,000 Ω, and 9,000 VA equals 9 kVA. Convert before inserting values into a formula.

Units also expose bad setups. In I = P/V, watts divided by volts produce amperes. In R = V/I, volts divided by amperes produce ohms. If a proposed power answer is labeled in kWh, the calculation has confused a rate with an amount.

Worked example: current and resistance

A 240 V single-phase heater is rated 4.8 kW. Assume a purely resistive load, power factor 1.00, and rated voltage at the heater. Find current.

Convert power: 4.8 kW × 1,000 W/kW = 4,800 W. Rearrange P = VI to I = P/V. Then I = 4,800 W ÷ 240 V = 20 A. The answer is reasonable because a 2.4 kW load at the same voltage would draw half as much, or 10 A.

If a separate 120 V resistive load draws 8 A, its resistance is R = V/I = 120 V ÷ 8 A = 15 Ω. Its power is P = VI = 120 V × 8 A = 960 W. Cross-check with P = I²R = (8 A)² × 15 Ω = 960 W. Agreement between two valid methods is strong error detection.

Worked example: energy

A 1.5 kW load operates for 3.0 hours at constant output. Assume no cycling. E = Pt = 1.5 kW × 3.0 h = 4.5 kWh. In watt-hours, the same answer is 1,500 W × 3.0 h = 4,500 Wh. If the load cycles at 50 percent duty, the constant-output assumption would be false and the energy would be lower.

Rearrange instead of guessing

Treat the formula as an equation. From V = IR, divide both sides by R to obtain I = V/R; divide by I to obtain R = V/I. From P = VI, divide by I to obtain V = P/I. Writing the rearranged form before substitution reduces keypad mistakes and makes the work reviewable.

Reasonableness checks

Use three quick tests. First, check direction: at fixed voltage, greater resistance should mean less current. Second, check scale: 4,800 W at 240 V should produce tens of amperes, not thousands. Third, check units and precision: 19.9997 A is usually reported as 20.0 A or 20 A, depending on the given precision. Do not round intermediate values so aggressively that the final answer changes.

Finally, distinguish a calculated operating value from a code selection. Calculating 20 A does not by itself select a conductor or overcurrent device. Later chapters add continuous-load treatment, ampacity tables, terminal limitations, and equipment rules. Theory supplies the starting value; the 2017 NEC determines how that value is applied.

Calculator checkpoint

Keep the unrounded calculator value available until the final step. Before choosing an answer, compare its unit, sign, and order of magnitude with the stated conditions. A precise display cannot rescue an incorrect formula or unit conversion.