Millivolt Drop on a Hot Connection: What the Numbers Mean

Correct answer: C) 30 mΩ, dissipating 1.92 W

Ohm's law applied to the connection itself, not to the string. The measurement is the voltage across that one joint, so the resistance is that voltage divided by the current flowing through it: 0.240 V ÷ 8.0 A = 0.030 Ω, or 30 mΩ. The heat is the same two numbers multiplied: 0.240 V × 8.0 A = 1.92 W. You can check it the other way round with I²R, since 8.0² × 0.030 also gives 1.92 W, and getting the same answer by two routes is worth the extra ten seconds on a roof.

Now the part the arithmetic is actually for. That string is producing somewhere around 600 V × 8.0 A = 4,800 W, so the 1.92 W lost in this joint is about four hundredths of one percent of it. You will never find this fault by looking at the production data. It is invisible on the monitoring portal, it will not move the string against its neighbours, and a performance ratio calculation will not blink.

And yet it is worth a truck roll, because 1.92 W is being released inside an object the size of a thumbnail. A connector is not designed to shed that continuously, so the temperature climbs until something conducts it away or something gives, and a degrading connection heats more as it degrades, which heats it further. This is the mechanism behind a large share of PV fires. The thermal camera found it because heat is the symptom; the millivolt drop quantified it because resistance is the cause.

The O&M point: the same calculation means two opposite things depending on who is doing it. A designer uses Ohm's law on conductor runs to keep losses within a voltage drop budget, where a fraction of a percent is genuinely negligible and gets rounded away. In service you are applying it to a single joint, where the same fraction of a percent is the thing that burns the building down. Reading "0.04% of production" as "not worth attending to" is the specific error this question is built to catch.

Why the other options are wrong

A) 30 Ω, dissipating 1.92 kW

This is the millivolt slip, and it is carried through both halves consistently, which is exactly what makes it look right. Dividing 240 by 8 gives 30, and multiplying 240 by 8 gives 1,920, so the arithmetic is sound and only the units are wrong. Keep hold of the magnitudes as a sanity check: a 30 Ω connection in a string carrying 8 A would drop 240 V across a single joint and release nearly 2 kW inside a connector, which is not a hot spot on a thermal image, it is a fire already in progress.

B) 33 Ω, dissipating 1.92 W

The ratio is upside down. 8.0 ÷ 0.240 gives 33, which is amps per volt rather than volts per amp, so it is a conductance dressed up as a resistance. The power figure beside it is right, which is the trap: getting one of the two numbers correct makes the pair feel checked. Resistance is always the voltage you measured divided by the current that produced it.

D) 30 mΩ, dissipating 4.8 kW

The resistance is correct and the power comes from the wrong voltage. 600 V is what the string develops across its whole length; the joint only has 240 mV across it, and power dissipated in a component is calculated from the voltage across that component. Multiplying the full string voltage by the current gives you the power the string is producing, which is 4.8 kW, and that is the number this joint is stealing from, not the number it is dissipating.

References

• NABCEP OMAT JTA v.2026.5 — Domain 2, Task 3.B: "Ohm’s law"
• No external standard is cited for the correct answer, because none is needed: the result follows from Ohm’s law and the definition of electrical power. The interpretation that follows — that a loss too small to see in production data can still be a fire risk — is the reason the JTA places this in Core Knowledge rather than in a calculation exercise.