30.1 Current Draw, Voltage, Power & Motor Terminals

30.1 Current Draw, Voltage, Power & Motor Terminals

Trade Area E (Maintenance Analysis) is 9 percent of Class A and 10 percent of Class B. Chapter 29 diagnosed the refrigeration circuit with gauges and a P–h chart. This chapter diagnoses the electrical circuit and, in 30.3, the air and water instruments that prove the machine is actually moving what the nameplate claims. The numbered Trade E items that land here are reading and analyzing electrical circuits; testing current draw, voltages, and power consumption; and testing motor terminals. Installing the same contactors, capacitors, and starters was Chapter 27.3–27.4 and Chapter 28. Calibration is Chapter 31. The 2026 Air Conditioning CBT books that carry this cluster are Refrigeration & Air Conditioning Technology, 9th Edition (2021), and Air Conditioning and Refrigeration Troubleshooting Handbook, 2nd Edition (2003). OSHA 29 CFR 1926 (July 1, 2025 edition on the reference list) still wants lockout/tagout before a terminal cover comes off for an ohm check. The CILB card is not a live-work permit.

Quick Answer: Ohm's law is (E = IR). Power is (P = IE) (watts). True power on AC motors is (P = E \times I \times \text{PF}) (single-phase) or (P = \sqrt{3} \times E \times I \times \text{PF}) (three-phase). Compare clamp-on amps to nameplate RLA/FLA. Identify hermetic terminals C, R, S with an ohmmeter after LOTO: R–S highest, C–R lowest; infinite is open; case continuity is grounded.

Why this is analysis, not a parts catalog

A dirty condenser, a failed run capacitor, a 208-volt supply on a 230-volt nameplate, a shorted start winding, and a single-phased three-phase compressor can all present as “the compressor is bad” until you read the schematic and measure. Trade E scores that sequence. F.S. 489.105(3)(f) and (g) still fence the line-voltage work: replace/disconnect/reconnect on the line or load side of a dedicated existing single-phase disconnect; repair or replace power wiring, disconnects, breakers, or fuses on dedicated HVAC circuits with a circuit-breaker lock; and run low-voltage HVAC control wiring. New building three-phase feeders are still an electrical contractor. Diagnosis of a 30-ton motor is on both exams. Installing that 30-ton system as one unit is Class A — Class B stops at 25 tons cooling and 500,000 Btu heating in any one system.

Ohm's law, (P = IE), and power factor

Voltage (E) in volts, current (I) in amperes, resistance (R) in ohms:

(E = I \times R \qquad I = E / R \qquad R = E / I)

Power (P) in watts for DC and for purely resistive AC is the product the outline wants memorized:

(P = I \times E)

Two other forms of the same resistive law: (P = I^{2}R) and (P = E^{2}/R). A 5 kW heat strip on 240 V draws

(I = P / E = 5000 / 240 = 20.8\ \text{A})

and the true power is (P = IE) because the load is essentially resistive (power factor ≈ 1.0).

Motors are not purely resistive. The winding is an inductor. Current lags voltage. Power factor (PF) is the qualitative ratio of true power (watts — what the shaft and the refrigerant actually get) to apparent power (volt-amperes — what the supply and the conductors must be sized for):

(\text{PF} = \dfrac{\text{true power (W)}}{\text{apparent power (VA)}})

PF sits between 0 and 1. A loaded HVAC motor often runs around 0.80–0.95 — use the nameplate when it prints PF or kW. An unloaded motor (belt off, compressor pumping almost no gas) can show a worse (lower) PF even as amps fall. Run capacitors on single-phase motors supply leading current and improve PF. Do not quote (P = IE) as true watts on a running compressor unless the item says the load is resistive or PF is 1.0.

Single-phase true power: (P = E \times I \times \text{PF})

Three-phase true power: (P = \sqrt{3} \times E \times I \times \text{PF} \approx 1.732 \times E \times I \times \text{PF})

Apparent power is (S = E \times I) (single-phase, VA) or (S = \sqrt{3} \times E \times I) (three-phase). kW is true power / 1000; kVA is apparent / 1000. A motor that draws more kVA than kW is not “wasting heat in the meter” in a way you fix with a bigger breaker — it is telling you the current is higher than the useful work, which is why low PF and high amps show up together on a weak capacitor or a failing winding.

Nameplates, a worked amp-draw example, and voltage

FLA (full-load amps) is the current a motor is designed to draw at rated load, rated voltage, and rated frequency. RLA (rated-load amps) is the hermetic-compressor cousin printed on condensing-unit nameplates. LRA (locked-rotor amps) is inrush with the rotor stalled — commonly on the order of 4–7 times RLA. A clamp-on that catches the first cycle of a start will look like a short until the motor is up. MCA (minimum circuit ampacity) and MOCP (maximum overcurrent protection) size the circuit and the breaker; they are not the running-load test. Exam questions that ask whether the compressor is overloaded want running amps versus RLA/FLA, not versus LRA or MOCP.

Worked amp draw versus FLA. A 3-ton single-phase condensing unit in Hillsborough County is nameplated 230 V, RLA 14.1 A, LRA 73 A. Outdoor air is 95°F. You measure 228 V on the load side of the dedicated disconnect and 19.8 A on the common with a clamp-on around one conductor.

(19.8 / 14.1 = 1.40 \rightarrow 40\ \text{percent over RLA})

Voltage is essentially nameplate, so this is not a 208-volt-on-a-230-motor story. The compressor is overloaded: dirty condenser, recirculated discharge air, overcharge, restricted airflow on the evaporator, or a weak run capacitor until you prove otherwise. Do not compare 19.8 A to LRA 73 A and call the motor healthy because “it is under locked-rotor.” Locked-rotor is a start number.

Power at that running point, assuming PF 0.90:

(P = 228 \times 19.8 \times 0.90 = 4{,}063\ \text{W} \approx 4.06\ \text{kW})

Apparent power is (228 \times 19.8 = 4{,}514) VA. If you quote (P = IE) without PF you are stating VA, not watts.

Low voltage raises current on a loaded motor. A 230 V nameplate sitting on 208 V (about a 10 percent sag) must draw more amps for the same watts, and starting torque falls — hard starts, extra heat, tripped overloads. Measure voltage at the motor terminals under load, not only at the panel with the unit off. Three-phase voltage unbalance of only a couple of percent can multiply into a much larger current unbalance (the textbook qualitative flag is on the order of current unbalance ≈ 8 × voltage unbalance). Measure all three legs at the motor. Single-phasing (one lost leg) drives the other two toward overload and is how a magnetic starter with oversized heaters burns a legal 15-ton semi-hermetic.

A three-phase 15-ton compressor nameplated 460 V, FLA 42 A, PF 0.85:

(P = 1.732 \times 460 \times 42 \times 0.85 \approx 28{,}400\ \text{W} \approx 28.4\ \text{kW})

Class B may install that 15-ton machine. Class B still has to compute the same (\sqrt{3}) power on a 30-ton exam item; Class B does not contract the 30-ton system.

Motor terminals — C, R, S and the ohmmeter

A single-phase hermetic compressor presents three pins: Common (C), Run (R), Start (S). The run winding is heavier wire and the lowest resistance. The start winding is finer wire and higher resistance. R to S is the sum of the other two and is therefore the highest.

Worked identification (after LOTO, capacitor discharged, leads lifted):

• C–R = 1.6 Ω
• C–S = 2.9 Ω
• R–S = 4.5 Ω
• any pin to shell = infinite

R–S equals C–R plus C–S. The lowest pair (1.6 Ω) is run. The windings are not open, not shorted, and not grounded. Live amp and volt tests come next, with the cover strategy the listing allows.

Lockout/tagout before testing internals. An ohmmeter on C, R, and S is a de-energized test. OSHA 1926.417 (and the 1910.147 concepts the contractor exam still uses) require the dedicated HVAC disconnect open, locked, and tagged, plus the circuit-breaker lock F.S. 489.105 already wants on that dedicated circuit. Verify zero voltage with a rated meter before the probes go on the pins. A clamp-on amp test is the opposite: it is a live reading on one insulated conductor. It is not an invitation to ohm a live winding.

Reading the circuit, then a Florida scenario

Analyze the schematic before you throw parts. Safeties (HP, LP, oil failure, freeze, overflow, high-limit) sit in series with the contactor coil. Loads (compressor, condenser fan, strips) sit across the line through the contactor or starter. A 24-volt call at Y with no contactor pull-in is a coil-circuit problem (open safety, open transformer, open coil). A contactor pulled in with no compressor amps is a load-circuit problem (open winding, open internal overload, open overload heater, burned load wire). Jumping the coil “to see if it runs” bypasses every safety at once — that is not analysis, and it is how you restart into a closed discharge valve.

Tampa Class B scenario. Gulfstream Cooling is on a legal 4-ton single-phase heat pump. The helper condemns the compressor because it “hums and trips.” No LOTO, a screwdriver across the capacitor, and no ohm chart. The correct sequence: lock the pad disconnect and the HVAC breaker; discharge the capacitor through a resistor; ohm C–R–S and case; if the windings identify and are ungrounded, restore power and clamp common. If running amps sit at 18 A on a 14.1 RLA with 230 V and a 45 µF can that measures 35 µF, you have a failed capacitor and a dirty coil, not a shorted compressor. The same campus has a 30-ton RTU. Class B still has to read its three-phase FLA, voltage unbalance, and (P = \sqrt{3}EIPF) on the exam. Class B does not contract that 30-ton system.

Traps: (1) Comparing running amps to LRA. (2) Quoting (P = IE) as kW on an inductive motor with no PF. (3) Ohm tests on live terminals. (4) Clamping the whole power cable (hot and neutral cancel). (5) Calling an open-hot internal overload a burned compressor. (6) Class B installing the 30-ton motor because “we only tested it.”