Journeyman Electrician Practice Test 3: Calculations, Motors & Special Occupancies

Because the physical heating element doesn't change, its resistance stays the same. When voltage drops, power drops with the square of the voltage (P is proportional to E²), not in straight proportion. Method 1 (The Short Way): Square the voltage ratio. (208V / 240V)^2 = 0.751. Multiply by original power: 12 kW * 0.751 = 9.01 kW. Method 2 (The Long Way): First, find the element's fixed resistance using R = E^2 / P. (240V)^2 / 12,000W = 4.8 ohms. Now, apply the new voltage to that resistance using P = E^2 / R. (208V)^2 / 4.8 ohms = 9,013 Watts (9.0 kW).

Did you pick 10.4 kW? You fell for the trap! You can't just use a straight percentage because power is based on the SQUARE of the voltage. Master those two formulas: R = E^2 / P and P = E^2 / R.

VD = 2 × K × I × D / CM = 2 × 12.9 × 40 × 150 / 10,380 = 154,800 / 10,380 = 14.9V. That's 14.9 / 240 = 6.2%, which exceeds the 3% branch circuit recommendation.

Always plug into VD = 2KID/CM first, then check if the percentage is under 3% for branch circuits!

Max VD = 208 × 0.03 = 6.24V. D = VD × CM / (2 × K × I) = 6.24 × 16,510 / (2 × 12.9 × 30) = 133.1 ft.

208V gives more distance than 120V for the same wire and current.

Three-phase voltage drop = 1.732 × I × R × D / 1000 = 1.732 × 60 A × 0.197 Ω × 150 ft / 1000 = 3.07 V line-to-line.

For three-phase, use 1.732 (√3) instead of the single-phase factor of 2 — it is the most common mix-up on voltage drop problems.

Using the power formula R = E² / P. R = 240² / 4,800. R = 57,600 / 4,800 = 12 ohms.

When you only have Volts and Watts, square the Volts first! Think: 'Volts times Volts divided by Watts'.

Because the raceway exceeds 24 inches, it is subject to the standard 40% fill limit (Chapter 9, Table 1) and not the 60% nipple exception. When all conductors are the exact same size and type, use Annex C. Looking at Annex C, Table C.1 for EMT: a 1/2" conduit holds a maximum of two #6 THHN conductors, while a 3/4" conduit holds up to four. Therefore, 3/4" is the minimum size.

The '50-foot run' detail is the test writer's way of telling you: 'Do NOT use the 60% nipple rule!' Once you know you're at 40% fill, and the wires match, skip the calculator and jump straight to Annex C.

Per Informative Annex C, Table C.1 (EMT), 1/2-inch EMT can accommodate 9 #12 AWG THHN conductors at the 40% fill limit of Chapter 9, Table 1.

Annex C tables give you the answer directly! C.1 is for EMT - memorize common combinations.

According to Table 310.16, #12 AWG THHN has a 90°C base ampacity of 30 A. For an ambient temperature of 10°C (the '10°C or less' row), the correction factor in Table 310.15(B)(1)(1) for 90°C conductors is 1.15. Calculation: 30 A × 1.15 = 34.5 A.

Temperature correction works both ways! In cold ambient temperatures, the conductor can dissipate heat more efficiently, allowing for a higher ampacity. Just remember: 26-30°C (78-86°F) is your baseline 1.00 multiplier!

Chapter 9, Table 1, Note 3 requires equipment grounding and bonding conductors to be counted toward conduit fill, using the real cross-sectional area of the conductor — whether it is bare or insulated. Bare conductor areas come from Chapter 9, Table 8.

Don't forget: If it's a bare conductor, you won't find it in Table 5. You have to flip to Table 8 (Conductor Properties) to get the area for your calculation!

Step 1: Navigate to Chapter 9, Table 4 and find the specific page for Electrical Metallic Tubing (EMT). Step 2: Locate Trade Size 2. Step 3: Look under the '100% Total Area' column. The value is 3.356 square inches.

Did you pick 1.342 or 2.013? You fell for the column trap! 1.342 is the 40% fill area (for pulling 3 or more wires). 2.013 is the 60% fill area (for short nipples). The question asked for the TOTAL area (100%), which is 3.356 sq in.

Per NEC 430.22, branch circuit conductors must have ampacity not less than 125% of the motor FLC from Table 430.250 — NOT the nameplate FLA. 34A × 1.25 = 42.5A minimum ampacity required.

Never use nameplate FLA for conductor sizing — always use Table 430.250 FLC × 125%!

First, find the FLC from Table 430.250 for a 25 HP, 460V motor, which is 34A. According to Table 430.52(C)(1), the maximum percentage for an inverse-time breaker is 250% of the FLC. So, 34A × 2.50 = 85A. However, 85A is not a standard circuit breaker size listed in 240.6(A). Per 430.52(C)(1) Exception No. 1, if the calculated value does not correspond to a standard size, you are permitted to use the next higher standard size. The next standard size up from 85A is 90A.

The 'next size up' rule for motor branch-circuit protection is a crucial exception. Don't just stop at the calculated value; always check if you need to round up to the next standard breaker or fuse size.

Per NEC 430.24, feeder conductors for several motors are sized at 125% of the largest motor FLC plus 100% of the others: (44 A × 1.25) + 26 A + 14 A = 55 + 40 = 95 A minimum ampacity.

Largest motor at 125%, all others at 100%.

Per NEC 430.32(A)(1), a separate overload device is sized from the motor NAMEPLATE full-load current: 125% for motors with a marked service factor of 1.15 or greater OR a marked temperature rise of 40°C or less; 115% for all other motors. This motor is SF 1.0 and has no 40°C rise marking, so it is an 'other motor': 22 A × 1.15 = 25.3 A maximum. Use NAMEPLATE amps for overloads, not Table FLC.

No 1.15 SF and no 40°C rise marking = 115% overload max. And always use NAMEPLATE amps for overload sizing — one of the few times you don't use Table 430.250!

NEC 450.21(B) requires indoor dry-type transformers rated over 112.5 kVA to be installed in a transformer room of fire-resistant construction. The fire resistance rating must be not less than 1 hour, unless one of the NEC exceptions applies.

Indoor dry-type over 112.5 kVA: 1-hour fire-resistant room is the default. Class 155+ insulation with proper separation can exempt it.

Per 500.5(B)(2), a Class I, Division 2 location is one where flammable liquids/vapors are normally confined in closed containers and would only be released by accidental rupture or failure. Sealed drums of solvent that release vapor only on accidental spill fit this definition exactly. It is Class I (flammable liquid), Division 2 (accidental condition only).

Apply the rule of thumb! Division 1 = everyday operations. Division 2 = accidental conditions. Since the flammable vapors are safely sealed in drums and only escape if a drum breaks (an accident), it is a Division 2 location.

NEC 680.26(B)(1) requires that the equipotential bonding grid use a minimum 8 AWG solid copper conductor. This conductor connects all metal parts and creates a grid that equalizes the voltage potential around the pool.

Remember: #8 solid copper for pool bonding grid. It must be SOLID, not stranded — solid copper holds up better when buried in concrete or soil.

Per 695.4(B)(1), only a single disconnecting means and associated overcurrent protective device are permitted in the supply path between the source and the listed fire pump controller. Multiple disconnects increase the chance of accidental power loss during a fire — the rule prioritizes reliability over selectivity.

Fire pumps are about reliability. Fewer shutdown points means less chance of accidentally killing power to the pump.

For indoor locations, NEC 625.50 requires the coupling means of the EVSE (or WPTE) to be stored or located not less than 18 in. (450 mm) and not more than 4 ft above the floor level. For outdoor locations the minimum is 24 in. (600 mm) above the parking surface, with the same 4 ft maximum.

Indoor: 18 inches minimum. Outdoor: 24 inches minimum. Memorize the pair — examiners love this one.

Per NEC 550.16(A), the grounded (neutral) conductor and the equipment grounding conductor must be kept separate in a mobile home. The neutral-to-ground bond is made at the service disconnecting means (the outdoor pedestal), not inside the mobile home.

A mobile home panelboard works like a sub-panel — neutral and ground stay separate! The bond happens at the outdoor disconnect.