3.6 Electrical Calculations, Units & Conversions

What the DCO Actually Asks

Domain II subdomain B lists three knowledge statements: perform mathematical calculations using basic algebra, geometry, and trigonometry; apply fundamental electrical laws to simple circuits (Ohm's law, Kirchhoff's law); and apply conversions between units (microhms to milliohms, horsepower to kilowatts). The exam is closed-book but provides an on-screen calculator and the NETA Formulae Sheet, so the failure mode is never "I forgot the formula." It is a decimal-place error in a unit conversion or picking the wrong current in a three-phase problem.

Unit Conversions That Show Up Constantly

Worked conversion. A Ductor reads 187 microhms across a breaker pole. The manufacturer's limit is printed as 0.25 milliohms. Convert one side: 0.25 mohm = 250 microhms. 187 < 250, so the pole passes. Candidates who compare 187 to 0.25 without converting conclude the pole is 748 times over limit and pick the wrong answer.

Worked conversion. A 150 hp motor is 94% efficient at full load. Input power = 150 x 0.746 / 0.94 = 119 kW. If someone reports 112 kW at the switchgear, the motor is running below full load — not evidence of a fault.

Ohm's Law and Power

E = I x R        I = E / R        R = E / I
P = E x I        P = I^2 x R      P = E^2 / R

The I squared R form is the one that matters for testing work, because it explains why a slightly high contact resistance becomes a thermography finding. A 4000 A bus joint at 20 microhms dissipates 0.000020 x 4000^2 = 320 W at one joint. Let that joint degrade to 100 microhms and it dissipates 1,600 W — a five-fold jump from a resistance change nobody would notice on a handheld meter. This is the single most useful calculation on a NETA job.

Kirchhoff's Laws

• Kirchhoff's Current Law (KCL): the sum of currents entering a node equals the sum leaving. This is the arithmetic behind differential (87) protection — currents into a zone must equal currents out, and any imbalance is either a fault or a CT error.
• Kirchhoff's Voltage Law (KVL): around any closed loop, the sum of voltage rises equals the sum of voltage drops. This is the arithmetic behind voltage-drop troubleshooting on control circuits — if 125 V DC leaves the panel and 118 V arrives at the trip coil, 7 V is being dropped somewhere in the loop, and that somewhere is a resistance you can find.

Worked KCL. A bus has 800 A and 350 A flowing in, and two feeders drawing 600 A and 400 A out. In = 1150 A, out = 1000 A. The 150 A discrepancy is not allowed by KCL, so either a measurement is wrong, a CT ratio is misapplied, or there is a ground-return path carrying current you did not account for.

Series and Parallel Circuits

Series:    R_total = R1 + R2 + R3 ...
Parallel:  1/R_total = 1/R1 + 1/R2 + 1/R3 ...
Two only:  R_total = (R1 x R2) / (R1 + R2)

The qualitative rules that answer most exam items without arithmetic:

• In series, current is the same everywhere and voltages divide in proportion to resistance.
• In parallel, voltage is the same across every branch and currents divide inversely with resistance.
• Total parallel resistance is always less than the smallest branch. If an answer choice for two parallel resistors is larger than either one, it is wrong before you calculate.

Worked parallel. Three ground rods measure 15, 20, and 30 ohms and are bonded together. Ideal parallel combination: 1/R = 1/15 + 1/20 + 1/30 = 0.0667 + 0.0500 + 0.0333 = 0.15, so R = 6.67 ohms. In the field the measured value will be higher because the rods' resistance shells overlap, which is exactly why spacing rods at least one rod-length apart matters.

Series and Parallel Capacitance and Inductance

Capacitance behaves opposite to resistance, and this reversal is a reliable exam trap:

This is why a long cable run has high capacitance: the conductor-to-shield capacitances of every foot are effectively in parallel, so they add. That accumulated capacitance is what drives the charging current you see in a DC hipot and why VLF testing exists.

Single-Phase vs. Three-Phase Power

Single-phase:  P = E x I x PF
Three-phase:   P = 1.732 x E_line x I_line x PF
Three-phase:   S (kVA) = 1.732 x E_line x I_line / 1000

Worked three-phase. A 480 V three-phase motor draws 65 A at 0.85 power factor. P = 1.732 x 480 x 65 x 0.85 = 45,940 W = 45.9 kW. In horsepower output terms, at 93% efficiency: 45.9 x 0.93 / 0.746 = 57 hp, consistent with a 60 hp nameplate running slightly under full load.

Worked transformer FLA. A 1500 kVA, 480 V secondary transformer has a full-load secondary current of 1,500,000 / (1.732 x 480) = 1,804 A. This calculation shows up whenever a question asks whether a measured load is reasonable for the transformer or whether a CT ratio was chosen sensibly.

Geometry and Trigonometry You Actually Need

The DCO mentions geometry and trigonometry, and the applications are narrow:

• Right triangles for the power triangle. Real power (kW) on the base, reactive power (kVAR) vertical, apparent power (kVA) the hypotenuse. kVA = square root of (kW squared + kVAR squared), and power factor = kW / kVA = cosine of the phase angle. A load at 0.80 PF has a phase angle of about 36.9 degrees.
• Circle area for conductor cross-section when comparing conductor sizes, and the fact that doubling circular-mil area halves resistance.
• Vector addition for phasor problems — adding two currents that are 120 degrees apart is not adding their magnitudes, which is why the neutral current of a balanced three-phase four-wire system with linear loads is zero rather than three times the phase current.

Time Discipline

With 100 items in 120 minutes you have about 72 seconds per question. Calculation items are the ones that blow that budget. The technique that works: estimate first, then compute. If a 480 V three-phase 100 A load is asked for in kW, you know 1.732 x 480 x 100 = about 83 kVA before power factor, so any answer choice above 83 kW is impossible regardless of what your arithmetic says. Eliminating impossible answers by magnitude is faster than solving.