2.5 Cost of Power in a Circuit, Efficiency & Troubleshooting/Test Instruments

Key Takeaways

  • Operating cost is energy, not power: kWh = (watts ÷ 1000) × hours, and cost = kWh × the utility energy rate, so a load's cost depends as much on run hours as on its wattage.
  • Commercial bills add a demand charge in dollars per kW billed on the highest 15- or 30-minute demand interval of the month, which is why a short peak can cost more than steady consumption.
  • Apparent power in kVA equals kW ÷ power factor, so correcting power factor from 0.72 to 0.95 on a 100 kW load drops apparent power from 139 kVA to 105 kVA and cuts I²R losses proportionally to the square of the current reduction.
  • A megohmmeter measures insulation resistance in megohms at an applied DC test voltage (commonly 500 V or 1000 V) and must never be used on an energized circuit or on solid-state equipment left connected.
  • A three-point fall-of-potential ground resistance tester is the instrument that verifies the 25-ohm threshold in NEC 250.53(A)(2) Exception, which is why a single rod electrode is normally supplemented rather than tested.
Last updated: August 2026

2.5 Cost of Power in a Circuit, Efficiency & Troubleshooting/Test Instruments

Two sub-topics inside General Electrical Knowledge are neither Ohm's law nor code lookup: cost of power used in a circuit and troubleshooting and test systems. They are the questions where the exam asks what the electricity is worth and how you would find the fault.


1. From Watts to Dollars

Power is a rate; energy is the accumulation. The utility bills you for energy:

kWh=Pwatts1000×thoursCost=kWh×rate\text{kWh} = \frac{P_{\text{watts}}}{1000} \times t_{\text{hours}} \qquad \text{Cost} = \text{kWh} \times \text{rate}

Worked example 1 — resistive load. A 4,500 W electric water heater element runs an average of 3 hours per day at $0.13/kWh.

  • Daily energy: (4,500 ÷ 1000) × 3 = 13.5 kWh
  • Daily cost: 13.5 × $0.13 = $1.76
  • 30-day cost: $52.65

Worked example 2 — motor load, where efficiency enters. A 10 HP, 230 V, three-phase motor runs 8 hours a day, 22 days a month, at 92% efficiency and $0.11/kWh. Horsepower is output, so input power is output divided by efficiency:

  • Output: 10 HP × 746 W/HP = 7,460 W
  • Input: 7,460 ÷ 0.92 = 8,109 W = 8.11 kW
  • Monthly energy: 8.11 × 8 × 22 = 1,427 kWh
  • Monthly energy cost: 1,427 × $0.11 = $157

The 8% the motor wastes is not free — at 92% efficiency the customer pays for 649 W of heat every hour the motor runs.

Energy charge versus demand charge

Billing componentUnitWhat drives it
Energy charge$/kWhTotal consumption over the billing period
Demand charge$/kWThe single highest 15- or 30-minute average demand in the month
Power-factor penalty / kVA billing$/kVA or a multiplierReactive current the utility must carry but cannot bill as kWh

This is why a plant that starts three large compressors simultaneously can pay more in demand charges than a plant that runs longer but staggers its starts. Staggered starting, soft starters, and load-shedding controls exist to shave the demand peak.

Power factor as money

kVA=kWPFPF=cosθ=kWkVA\text{kVA} = \frac{\text{kW}}{PF} \qquad PF = \cos\theta = \frac{\text{kW}}{\text{kVA}}

A 100 kW industrial load at 0.72 PF draws 100 ÷ 0.72 = 139 kVA. Corrected to 0.95 PF, the same real work needs 100 ÷ 0.95 = 105 kVA — a 24% reduction in current. Because conductor and transformer losses go as I², cutting current by 24% cuts those losses by roughly 42%. Capacitor banks and synchronous condensers pay for themselves out of that difference plus the avoided penalty.

Conversion constants worth memorizing

  • 1 HP = 746 W
  • 1 kW = 3,412 BTU/hr
  • 1 ton of cooling = 12,000 BTU/hr = 3.517 kW
  • 1 kWh = 3.6 MJ

2. Troubleshooting and Test Systems

Troubleshooting is a discipline, not a guess. The sequence that survives an exam item and a service call is the same:

  1. Define the symptom precisely. "The lights flicker when the A/C starts" is a different fault from "the lights are dim all the time."
  2. Verify the supply before the load. Confirm voltage at the source; a tripped upstream device or an open neutral explains most "dead equipment" calls.
  3. Half-split the circuit. Test at the electrical midpoint, then discard the half that tests good. Each measurement should cut the suspect length in half rather than walk it end to end.
  4. Prove the instrument. Test-before-touch and test-after on a known live source — a non-contact tester that failed silently is worse than no tester.
  5. Repair, then verify under load. A high-resistance connection only shows itself when current flows.

The instrument set

InstrumentMeasuresFault it findsCritical caution
Digital multimeter (DMM)Volts, amps, ohms, continuityOpen circuits, voltage drop, blown fusesMust carry a CAT rating for the point of measurement — CAT III for distribution panels, CAT IV at the service and outdoors
Clamp-on ammeter (true RMS)Current without breaking the circuitOverloads, unbalanced phases, neutral current from harmonicsAn averaging meter reads low on nonlinear loads; clamping all conductors at once should read near zero unless there is a ground fault
Megohmmeter (insulation tester)Insulation resistance in megohms at 500 V or 1000 V DCDegraded, wet, or damaged insulation before it failsNever apply to an energized circuit; disconnect solid-state equipment and electronic ballasts or the test voltage will destroy them; discharge the conductor after testing
Low-resistance ohmmeter (ductor)Micro- and milliohmsHigh-resistance bolted joints, corroded busbar splicesRequires a de-energized, isolated joint
Non-contact voltage testerPresence of an AC fieldQuick live/dead screening onlyNot a verification instrument — it can false-negative on shielded cable and false-positive on induced voltage
Phase rotation testerABC versus ACB sequenceMotors that run backwards after a service changeVerify rotation before coupling any pump, fan, or conveyor
Ground resistance testerEarth electrode resistance (three-point fall-of-potential)Electrode systems that cannot meet 25 ohmsAuxiliary probes must be spaced along a straight line clear of buried metal
Infrared thermal imagerSurface temperatureLoose terminations, unbalanced phases, failing breakersRequires load on the circuit; an unloaded connection looks fine
Receptacle testerBasic wiring patternReversed hot/neutral, open groundCannot detect a bootleg ground (neutral jumpered to the ground terminal), which reads "correct"

Fault signatures the exam likes

  • Open neutral on a multiwire branch circuit. Line-to-neutral voltages swing in opposite directions — one leg reads high (say 190 V) and the other low (say 40 V) while line-to-line stays near 240 V. The 120 V loads are in series across 240 V, and the lighter-loaded leg gets the higher voltage.
  • High-resistance connection. Voltage measured across the connection rises as current increases, and the joint runs hot under thermography. A resistance measurement on a de-energized, unloaded joint often reads acceptably low and hides the problem.
  • Shared neutral overload. A clamp on the neutral of a three-phase, four-wire circuit feeding nonlinear loads can read higher than any phase conductor because triplen harmonics add arithmetically in the neutral.
  • Bootleg ground. A receptacle tester shows correct wiring, but the equipment grounding conductor carries normal load current — found by clamping the EGC and seeing current where there should be none.
  • Voltage present but no current. An open in the load path; a meter reading source voltage across an open switch or an open element is normal, not a fault indication in itself.
Test Your Knowledge

A 15 kW resistive process heater operates 6 hours per day, 26 days per month. Energy is billed at $0.12 per kWh and the utility also applies a $14.00 per kW demand charge to the monthly peak. If this heater sets the facility peak, what is its approximate total monthly cost?

A
B
C
D
Test Your Knowledge

An industrial customer has a 100 kW real load operating at 0.72 power factor. What apparent power does the service carry, and what changes if the power factor is corrected to 0.95?

A
B
C
D
Test Your Knowledge

A journeyman must verify the insulation condition of a 480-volt feeder that supplies a variable-frequency drive. What is the correct instrument and procedure?

A
B
C
D
Test Your Knowledge

On a 120/240-volt multiwire branch circuit, one leg measures 191 volts to neutral and the other measures 49 volts to neutral, while line-to-line measures 240 volts. What is the most probable fault?

A
B
C
D