2.1 Ohm's Law, Power, and Basic Circuits

The Arkansas Journeyman Electrician exam administered by Prov includes nine General Knowledge of the Electrical Trade and Calculations items. Several of those questions test pure electrical theory — Ohm's law, power relationships, and basic circuit math — rather than Code lookup. You still bring NEC 2023 into the room, but these problems are solved with formulas and careful unit handling, not Article hunting.

Ohm's Law Triangle

Ohm's law states that voltage across a resistive load equals current through the load times resistance:

E = I × R

Rearranged forms you must recall instantly:

Electricians often write V for voltage; exam questions may use E (electromotive force). Treat them as the same quantity in volts.

Worked Example — Find Current

A resistive heating element measures 24 Ω and is connected across a 120 V branch circuit. What is the current?

I = E / R = 120 / 24 = 5 A

That 5 A load on a 15 A or 20 A branch is within rating before continuous-load factors; the theory question usually stops at finding 5 A.

Worked Example — Find Resistance

A luminaire draws 0.83 A at 120 V. Approximate filament resistance equivalent:

R = E / I = 120 / 0.83 ≈ 144.6 Ω

On multiple-choice items, expect rounded options such as 145 Ω or 144 Ω.

The Power Family

Power in watts (W) is the rate of energy use. Three equivalent formulas cover nearly every JW theory item:

Kilowatts: 1 kW = 1,000 W. Always convert before mixing with amperes and volts unless the question already works in kW with a matching formula.

Worked Example — Power from Current and Voltage

A water heater element draws 18.8 A on a 240 V circuit:

P = I × E = 18.8 × 240 = 4,512 W ≈ 4.5 kW

Worked Example — Power from Voltage and Resistance

Same heater, resistance 12.8 Ω at operating temperature on 240 V:

P = E² / R = (240)² / 12.8 = 57,600 / 12.8 = 4,500 W

Slight differences come from rounding of measured current versus nameplate ohms — exam problems give clean numbers so answers match exactly.

Single-Phase kW to Amperes

For DC or single-phase AC with unity power factor (resistive loads):

I = (kW × 1,000) / E

Example: A 3 kW baseboard heater on 240 V:

I = (3 × 1,000) / 240 = 3,000 / 240 = 12.5 A

Continuous heating loads are later sized at 125% for branch-circuit and OCPD selection (Article 210/424 concepts), but the pure calculation question often asks only for the load current: 12.5 A.

Three-Phase Power (When PF Appears)

If the Prov item gives three-phase voltage and power factor:

P = E × I × 1.732 × PF (watts, line-to-line E)

I = P / (E × 1.732 × PF)

Example: A three-phase load of 15 kW at 208 V, PF 0.85:

I = 15,000 / (208 × 1.732 × 0.85) ≈ 15,000 / 306.3 ≈ 49 A

If PF is omitted and the load is described as resistive, treat PF as 1.0.

Horsepower Shortcuts (Nameplate Context)

Motors are covered heavily elsewhere on the outline, but theory items sometimes convert hp to watts: 1 hp ≈ 746 W. A 1 hp motor input is higher than 746 W because of efficiency and power factor — if the question says "output" use 746 W; if it gives efficiency, divide output by efficiency to get input watts, then find current.

Example: 1 hp motor, 85% efficient, 240 V single-phase, PF 0.9:

Input W = 746 / 0.85 ≈ 878 W

I = 878 / (240 × 0.9) ≈ 4.1 A

Unit Discipline

Common traps on open-book calculation days:

• Leaving current in mA (divide by 1,000 to get amperes before Ohm's law)
• Mixing kW with a formula that expects watts
• Using 208 V when the stem says 240 V (or vice versa)
• Forgetting that P = I²R uses amperes squared — a 10 A load on 2 Ω dissipates 200 W, not 20 W

Quick Reference Card for Exam Day

Relating Power Formulas Without a Calculator Panic

If you know any two of P, I, E, or R (with a consistent set), you can find the others. Suppose a question gives only P = 1,440 W and E = 120 V. Current is I = P / E = 1,440 / 120 = 12 A, then R = E / I = 120 / 12 = 10 Ω. Checking with P = I²R: 12² × 10 = 1,440 W. Building that habit catches wrong answer choices that swapped watts and amperes.

For multiwire branch circuits and shared neutrals, theory questions still treat each ungrounded conductor's load with Ohm's law separately unless the stem clearly combines loads. Do not invent three-phase math on a single-phase 120/240 V dwelling stem.

Practice these until you can pick the correct rearrangement in under ten seconds. On a 60-question, 3-hour Arkansas JW exam, theory items should be quick wins that bank time for Code lookups in grounding, services, and special occupancies.