General Electrical Theory and Fundamentals
Domain 1 tests the electrical fundamentals that explain why a low-voltage circuit works or fails: Ohm's law, power, series and parallel loads, conductor resistance, voltage drop, and basic AC impedance. Questions often hide a simple calculation inside a field scenario, so the exam expects you to choose the right formula and keep units consistent.
Core relationships
- Ohm's law: V = I × R. Power: P = V × I, P = I²R, and P = V²/R.
- In series, current is common and voltages add. In parallel, voltage is common and currents add.
- Conductor resistance is published per 1,000 ft. A run has two conductors, so use 2 × one-way length.
- Voltage drop for a DC or resistive load is Vdrop = 2 × L × I × Rper1000 / 1000, with L in feet.
- AC loads have impedance, not just resistance; power factor and transformer regulation can change results.
Worked example
A 24 V DC access-control load draws 0.40 A at the end of a 120 ft run of 18 AWG copper. 18 AWG has about 6.51 Ω per 1,000 ft. The conductor length is 240 ft, so R = 6.51 × 240 / 1,000 = 1.56 Ω. Voltage drop = 0.40 × 1.56 = 0.63 V, about 2.6%. The device sees 23.37 V. If its minimum is 22 V, the design is acceptable. If the load were 0.80 A, drop would be 1.25 V, about 5.2%, leaving 22.75 V-still acceptable but with much less margin after terminations and temperature rise. Cable power loss would be I²R = 0.64 × 1.56 = 1.0 W.
Traps that cost marks
- Using the one-way distance instead of the loop length.
- Treating AWG ampacity as resistance data; they answer different questions.
- Assuming the source voltage appears at the device.
- Ignoring terminations, splices, and higher resistance at elevated temperature.
- Adding parallel load currents incorrectly or averaging them.
Exam-ready summary
Choose the relationship, convert units, calculate the loop, then compare with the device's published operating range. A result that is “close” is a design warning, not a pass.