Applied Electrical Theory and Calculations

Applied Electrical Theory and Calculations (Minnesota Class A Journeyworker)

Learning Objectives

1.1 Ohm's Law and Watt's Law

Ohm's law fixes the relationship among voltage, current, and resistance: E = I × R, so I = E / R and R = E / I. Watt's law ties power to voltage and current in a dc or purely resistive ac circuit: P = E × I, with the companion forms P = I² × R and P = E² / R. In an ac circuit the product of voltage and current gives apparent power in volt-amperes, equal to real power in watts only when current and voltage are in phase. Dividing resistance by voltage yields conductance, not current, and that reversed form is a standard distractor.

Example — Ohm's law. Statement: E = 240 V, R = 12 Ω, find I. Formula: I = E ÷ R. Substitution: 240 ÷ 12. Result: I = 20 amperes. At unity power factor the same load draws P = 240 × 20 = 4,800 watts, or 4.8 kW.

1.2 Transposing Equations

Transposition isolates the quantity you want: whatever is applied to one side is applied to the other, and a quantity that multiplies on one side divides on the other. The single-phase voltage drop formula carries a factor of 2, a length, a resistance, a current, and a divisor of 1000:

VD = (2 × L × R × I) ÷ 1000

where L is the one-way circuit length in feet, R is the conductor resistance in ohms per 1000 ft, and I is the current in amperes. The factor of 2 exists because current travels out on the ungrounded conductor and back on the grounded conductor.

Example — transposing for current. Statement: VD = (2 × L × R × I) ÷ 1000, solve for I. Formula: multiply both sides by 1000, then divide both sides by (2 × L × R). Substitution: 1000 × VD = 2 × L × R × I, so I = (1000 × VD) ÷ (2 × L × R). Result: I = (1000 × VD) ÷ (2 × L × R). The companion forms solve for L and R, and a three-phase circuit uses VD = (1.732 × L × R × I) ÷ 1000.

1.3 Percentages, Area, and Volume

A percentage is a ratio over 100. To find what percentage one quantity is of another, divide the part by the whole and multiply by 100; to apply a percentage, multiply the base by the percentage written as a decimal. Load-versus-rating questions and sizing questions are the same arithmetic read in opposite directions.

Example — load as a percentage of rating. Statement: 40 A load on a 50 A branch circuit. Formula: percent = (load ÷ rating) × 100. Substitution: (40 ÷ 50) × 100 = 0.80 × 100. Result: 80 percent. The 50 A rating comes from 210.19(A)(1): the noncontinuous load plus 125 percent of the continuous load, or 40 × 1.25 = 50 A, so 125 percent is the sizing multiplier, not the utilization of the circuit.

Example — available fill area in 1 in. EMT. Statement: total internal area = 0.864 in², over two conductors. Formula: available area = total area × permitted percentage. Substitution: 0.864 × 0.40 = 0.346 in², the value listed in the 40 percent column of Table 4. Result: 0.346 in² available, with 53 percent permitted for one conductor or cable and 31 percent for two (Chapter 9 Table 1).

Example — box volume. Statement: 4 × 2⅛ square box = 30.3 in³, 4 × 1¼ square box = 18.0 in³. Formula: difference = larger volume − smaller volume. Substitution: 30.3 − 18.0. Result: 12.3 in³ more capacity in the deeper box, which takes 15 conductors of 14 AWG against 9 in the shallower box.

1.4 Voltage Drop

Voltage drop depends on current, conductor resistance, and circuit length. Resistance comes from Chapter 9 Table 8 in ohms per 1000 ft: uncoated copper is 3.07 at 14 AWG, 1.93 at 12 AWG, 1.21 at 10 AWG, and 0.764 at 8 AWG. The Informational Note to 210.19 states that conductors sized to prevent a voltage drop exceeding 3 percent at the farthest outlet of power, heating, and lighting loads, where the total voltage drop on both feeders and branch circuits to the farthest outlet does not exceed 5 percent, provide reasonable efficiency of operation.

Example — voltage drop from total conductor resistance. Statement: I = 20 A, total conductor resistance = 0.60 Ω, E = 120 V. Formula: VD = I × R, then percent = (VD ÷ E) × 100. Substitution: 20 × 0.60 = 12 V; (12 ÷ 120) × 100 = 10 percent. Result: 12 volts, or 10 percent — well beyond the 3 percent branch-circuit recommendation and the 5 percent total.

Example — voltage drop using Table 8. Statement: single-phase 120 V circuit, L = 50 ft, 12 AWG uncoated copper (R = 1.93 Ω per 1000 ft), I = 20 A. Formula: VD = (2 × L × R × I) ÷ 1000, then percent = (VD ÷ E) × 100. Substitution: (2 × 50 × 1.93 × 20) ÷ 1000 = 3,860 ÷ 1000 = 3.86 V; (3.86 ÷ 120) × 100 = 3.2 percent. Result: 3.86 V, about 3.2 percent — hence the practice of upsizing long runs.

1.5 Series and Parallel Circuits

In a series circuit the current is the same through every element, voltages add, and resistances add: R = R₁ + R₂ + R₃. In a parallel circuit the voltage is the same across every branch, currents add, and the reciprocals of the resistances add: 1/R = 1/R₁ + 1/R₂ + 1/R₃. Two resistors in parallel reduce to the product-over-sum shortcut, R = (R₁ × R₂) ÷ (R₁ + R₂), and n identical resistors give one resistance divided by n.

Example — two unequal parallel resistors. Statement: 20 Ω and 30 Ω in parallel across 120 V. Formula: R = (R₁ × R₂) ÷ (R₁ + R₂). Substitution: (20 × 30) ÷ (20 + 30) = 600 ÷ 50. Result: 12 ohms — 120 ÷ 20 = 6 A and 120 ÷ 30 = 4 A give 10 A total; adding the resistances to get 50 Ω is the series rule misapplied.

Example — three identical parallel loads. Statement: three 60 Ω resistors in parallel across 120 V. Formula: I(branch) = E ÷ R, and I(total) = 3 × I(branch), or R(eq) = 60 ÷ 3. Substitution: 120 ÷ 60 = 2 A per branch, so 2 + 2 + 2 = 6 A; equivalently 120 ÷ 20 = 6 A. Result: 6 amperes; treating them as a 180 Ω series string gives 0.67 A, the classic distractor.

1.6 Single-Phase and Three-Phase Power, and Power Factor

Single-phase apparent power is S = E × I; three-phase apparent power is S = 1.732 × E × I, where E is the line-to-line voltage and 1.732 is the square root of 3. Real power is P = E × I × PF single-phase and P = 1.732 × E × I × PF three-phase, so power factor is PF = P ÷ S. In a balanced wye system the line-to-neutral voltage is the line-to-line voltage divided by 1.732; in a balanced delta system the line voltage equals the phase voltage and the line current is 1.732 times the phase current.

Example — single-phase apparent power and power factor. Statement: E = 240 V, I = 25 A, wattmeter reads 5,400 W. Formula: S = E × I, and PF = P ÷ S. Substitution: S = 240 × 25 = 6,000 VA; PF = 5,400 ÷ 6,000 = 0.90. Result: 6,000 volt-amperes at 90 percent power factor — equipment is sized on volt-amperes, not watts.

Example — three-phase apparent power. Statement: balanced three-phase 480 V load, 30 A per phase, unity power factor. Formula: S = 1.732 × E × I. Substitution: 480 × 30 = 14,400 VA taken single-phase, and the 1.732 factor carries the same load to about 25 kVA three-phase. Result: about 25 kVA; omitting 1.732 understates the load by roughly 42 percent.

Example — line-to-neutral voltage of a 480-volt wye system. Statement: E(line-to-line) = 480 V, balanced wye. Formula: E(line-to-neutral) = E(line-to-line) ÷ 1.732. Substitution: 480 ÷ 1.732 ≈ 277 V. Result: approximately 277 volts, the standard utilization voltage of a 480Y/277-volt system.

1.7 Motor Calculations

Motor calculations start from the table full-load current rather than the nameplate, as 430.6(A)(1) requires. Table 430.248 covers single-phase ac motors and Table 430.250 covers three-phase ac motors; 430.250 states that the listed voltages are rated motor voltages and the currents are permitted for system voltage ranges of 110 volts to 120 volts, 220 volts to 240 volts, 440 volts to 480 volts, and 550 volts to 600 volts. Both tables are read at the lower voltage of the range: a 1 hp single-phase motor shows 16 amperes at 115 volts and 8.0 amperes at 230 volts, and the halving holds throughout, with 9.8 and 4.9 amperes for ½ hp and 80 and 40 amperes for 7½ hp, because a dual-voltage motor connects two run windings in parallel for the lower voltage and in series for the higher voltage. Table 430.250 gives 1 hp at 460 volts as 2.1 amperes, 25 hp as 34 amperes, and 50 hp as 65 amperes, and the footnote states that for 90 and 80 percent power factor the figures shall be multiplied by 1.1 and 1.25, respectively.

Example — branch-circuit conductors for a three-phase motor. Statement: 25 hp, 460 V, three-phase, FLC = 34 A from Table 430.250. Formula: 430.22 requires ampacity ≥ full-load current × 1.25. Substitution: 34 × 1.25 = 42.5 A. Result: an ampacity of at least 42.5 A.

Example — reduced power factor correction. Statement: the same 34 A motor at 80 percent power factor, footnote factor 1.25. Formula: adjusted current = FLC × footnote factor. Substitution: 34 × 1.25 = 42.5 A at 80 percent; 34 × 1.1 = 37.4 A at 90 percent. Result: 42.5 A at 80 percent and 37.4 A at 90 percent — the footnote 1.25 and the 430.22 1.25 are separate requirements.

Example — motor control circuit conductor protection. Statement: 18 AWG copper control conductors within the enclosure with separate protection provided (Column A of Table 430.72(B)(2)). Formula: read the table at the conductor size and applicable column. Substitution: 18 AWG copper — Column A = 7 A, Column B = 25 A, Column C = 7 A. Result: 7 amperes, the same limit 240.4(D)(1) sets for 18 AWG copper; 25 A applies only in Column B.

Under 430.74, a three-wire start-stop control circuit with one grounded conductor must be arranged so that a ground fault remote from the controller will not start the motor and will not bypass manually operated or automatic safety shutdown devices. Where a capacitor sits on the load side of the motor overload device, 460.11 requires the overload device to be based on the improved power factor, and the capacitor is disregarded for the conductor rating under 430.22.

Overload protection, short-circuit and ground-fault protection, and the disconnecting means are three separate motor calculations. 430.32(A)(1) sizes the overload device from the motor nameplate full-load current: 125 percent of that current where the motor has a marked service factor of 1.15 or greater or a marked temperature rise of 40 °C or less, and 115 percent for all other motors, with modification permitted by 430.32(C). The disconnecting means is rated under 430.110(A) at not less than 115 percent of the full-load current rating of the motor, except that a listed unfused motor-circuit switch having a horsepower rating not less than the motor horsepower is permitted to have a lower current rating. Motor control circuits have a separate rule: 430.75 requires them to be arranged so that they will be disconnected from all sources of supply when the disconnecting means is in the open position, with the disconnecting devices located immediately adjacent to each other.

1.8 Transformer, Overcurrent, and Conductor Sizing

Conductor ampacity for these sizing steps is determined under 310.15 and Table 310.16, and overcurrent devices must also have an interrupting rating not less than the available fault current at their terminals, as 110.9 requires.

Transformer overcurrent protection follows 450.5, with the maximum rating or setting taken from Table 450.5(B) as a percentage of the transformer-rated current: with primary-only protection, 125 percent where that value corresponds to a standard rating and, where it does not, up to 167 percent for primary currents less than 9 amperes or 300 percent for primary currents under 2 amperes; with both primary and secondary protection, 250 percent on the primary side. Capacitor banks are protected separately: 460.9 requires an OCPD in each ungrounded conductor for each capacitor bank, with the rating or setting as low as practicable, and 460.10 requires a disconnecting means in each ungrounded conductor that opens all ungrounded conductors simultaneously and is rated not less than 135 percent of the rated current of the capacitor. Neither device is required for a capacitor connected on the load side of a motor overload protective device.

Example — autotransformer overcurrent protection. Statement: autotransformer rated 1000 V, nominal, or less, rated full-load input current 20 A. Formula: 450.6(A)(2) — OCPD ≤ input current × 1.25. Substitution: 20 × 1.25 = 25 A, a standard rating in Table 240.6(A). Result: 25 amperes, with an individual OCPD in each ungrounded input conductor (450.6(A)(1)); if the calculation misses a standard rating and the input current is 9 A or more, the next higher rating of 240.6 is permitted, and below 9 A up to 167 percent is permitted.

Example — available short-circuit current at a transformer. Statement: three-phase transformer, secondary full-load current 200 A, nameplate impedance 5 percent. Formula: with an infinite primary source, I(sc) = FLC ÷ per-unit impedance. Substitution: 200 ÷ 0.05 = 4,000 A. Result: 4,000 amperes; a 10 percent impedance transformer gives 200 ÷ 0.10 = 2,000 A, and 10,000 A comes from multiplying instead of dividing.

Example — small conductor overcurrent limit. Statement: 20 A breaker on a branch circuit wired with 14 AWG copper. Formula: 240.4(D) — protection shall not exceed the listed value for the size, after correction factors for ambient temperature and number of conductors. Substitution: 14 AWG copper limit = 15 A (240.4(D)(4)); device = 20 A. Result: not permitted — the list gives 18 AWG copper 7 A, 12 AWG copper 20 A, and 12 AWG aluminium 15 A; Table 240.6(A) supplies standard ratings from 10 through 6000 amperes.

Example — continuity of a shared grounded conductor. Statement: 120/240 V multiwire branch circuit whose shared grounded conductor passes through a receptacle's terminals. Formula: apply 300.15(B). Substitution: continuity of a grounded conductor shall not depend on device connections where removing the device would interrupt it. Result: the shared grounded conductor may not depend on device terminals — removing the device opens the neutral, the legs act in series across 240 V, and the other leg sees about twice rated voltage.

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Summary

Applied calculations on the Minnesota Class A journeyworker exam rest on a small set of laws and tables, and each problem is a two-step operation: choose the governing formula, then substitute the values given. Ohm's law and Watt's law produce current, voltage, resistance, and power from any two known quantities, and their transposed forms — including the single-phase voltage drop equation VD = (2 × L × R × I) ÷ 1000 and its 1.732 three-phase counterpart — solve for whatever is asked. Percentages convert a load into a comparison, so a 40-ampere load on a 50-ampere circuit is 80 percent, and that rating comes from applying 125 percent to the continuous load under 210.19(A)(1). Voltage drop is computed from current and the resistance data of Chapter 9 Table 8 and compared with the 3 percent and 5 percent values of the Informational Note to 210.19. Series resistances add while parallel resistances combine as reciprocals or by product over sum, and apparent power is the direct product of voltage and current single-phase or that product times 1.732 three-phase, with 480 volts divided by 1.732 giving 277 volts line-to-neutral.