What Voltage Drop Actually Costs the Equipment

Why this matters

Voltage drop gets taught as a compliance number and remembered as a percentage, which is why so many techs can compute it and so few can say what it does. The cost is not paid in the wire. It is paid at the equipment, and the currency depends entirely on what kind of load is standing there: one load quietly does less work and never complains, another holds its output exactly and sends the bill somewhere you are not looking. Two pieces of equipment on the same feeder, seeing the same drop, can produce a performance complaint and a nuisance trip six months apart, and a tech who does not know which response belongs to which load will diagnose both as separate faults.

The setup this card runs everything against

One feeder, one number, two very different pieces of equipment.

A 240 V single-phase feeder is long enough that at its design current of 50.0 A it drops 4 percent. Four percent of 240 V is 9.6 V, and 9.6 V at 50.0 A means the round-trip resistance of the run is 9.6 divided by 50.0, or 0.192 ohms. That resistance is a property of the conductor and its length, not of the load, so it stays fixed for the rest of this card while everything else moves.

Two things follow immediately and both get missed in the field. First, drop only exists while current flows, so the same run measures full nominal voltage at the equipment with the load off. A run that "tested good" at no load has not been tested. Second, the 9.6 V does not disappear. It becomes heat in the conductor at a rate of current squared times 0.192 ohms, inside a raceway that already has a thermal problem of its own (see the derating and bundling cards for what that heat costs the conductor's rating).

Any voltage reading taken at equipment terminals under load is taken on energized conductors, so 29 CFR 1910.333(a)(1) applies: de-energize before working on or near exposed live parts unless the employer can demonstrate that de-energizing introduces additional or increased hazards or is infeasible due to equipment design or operational limitations. A loaded voltage reading does not exist on a dead circuit, which is the ordinary case for that second clause; take it under the employer's energized-work program with boundaries and PPE established under NFPA 70E-2021 130.5 and 130.7, in the edition your employer's program or your authority having jurisdiction has adopted, and prove dead live-dead-live per NFPA 70E-2021 120.5 whenever the same job turns into work on the conductors.

Three responses a load can have, and they are not variations of each other

Constant impedance. A resistance heating element, an incandescent lamp, a resistive load bank. Its resistance is fixed, so current follows voltage down (Ohm's law) and power follows voltage down squared. Lower voltage means less current AND less output, which makes this family self-limiting: the drop partially corrects itself because the load stops asking for as much.

Constant power. A drive's rectifier front end, a switch-mode power supply, a regulated battery charger. Within its normal operating range and above its own undervoltage threshold, it delivers the output its load demands regardless of input voltage, so lower input voltage means HIGHER input current. That runs the feedback the wrong way: more current means more drop, which means still more current. Below the undervoltage threshold the behaviour changes entirely and the unit trips rather than continuing to compensate, so this response describes the operating range and not the failure.

An induction motor across the line. In between, and it moves during the event. Torque produced varies roughly with the square of applied voltage, so at reduced voltage the motor's torque capability falls, but the load's torque demand does not. The motor settles at higher slip and draws more current to hold the same shaft power, so it behaves like a constant-power load in the direction of current while paying an additional penalty in rotor heating.

Those three have different signs. Nothing you learned about one transfers to another.

Outcome one: the resistance heater

The heater's nameplate is 240 V and its design current is the feeder's 50.0 A, so its element resistance is 240 divided by 50.0, or 4.80 ohms. Put it on the end of the run and solve the whole circuit rather than assuming the design current still flows.

Total resistance is 4.80 plus 0.192, or 4.992 ohms. Current is 240 divided by 4.992, or 48.08 A. Drop across the run is 48.08 times 0.192, or 9.23 V, which is 3.85 percent rather than the 4 percent the design current predicted. Terminal voltage is 230.77 V.

Output is voltage squared over resistance. At the terminals: 230.77 squared is 53,255, divided by 4.80 is 11,095 W. At full nominal: 240 squared is 57,600, divided by 4.80 is 12,000 W. The heater delivers 11,095 of 12,000, or 92.5 percent, so about 7.5 percent less heat.

Read what happened. The drop came in slightly UNDER the design prediction, because the load backed off and less current means less drop. Nothing gets hot that should not, nothing trips, nothing is damaged, and no protective device has an opinion. The entire cost is 7.5 percent less output, which shows up as a tank that takes longer to recover or a space that holds temperature on a mild day and falls behind on a design day. Nobody calls it a wiring problem. They call it undersized equipment, and the next quote adds capacity that the building did not need.

Conductor loss in this case is 48.08 squared times 0.192, or about 444 W of heat in the raceway.

Outcome two: the drive on the same feeder

Now put a drive on the identical run, sized so that at full nominal voltage it draws the same 50.0 A, which at 240 V is 12,000 VA. Drive front ends run at close to unity displacement power factor, so for this illustration treat that as 12,000 W of input power and hold it constant, which is the constant-power assumption stated above and valid only inside the drive's normal operating range.

Solve for where it actually settles. Terminal voltage times current equals 12,000, and terminal voltage is 240 minus current times 0.192. Substituting gives 240 I minus 0.192 I squared equals 12,000, which rearranges to I squared minus 1250 I plus 62,500 equals zero. The operating root is 52.18 A.

Terminal voltage is 240 minus 52.18 times 0.192, which is 240 minus 10.02, or 229.98 V. The drop is 10.02 V, or 4.18 percent - higher than the 4 percent design figure, not lower, and higher for exactly the reason the heater's came in low.

Conductor loss is 52.18 squared times 0.192, or about 523 W. Against the heater's 444 W in the same wire, that is 18 percent more heat for a load that was nominally identical on paper.

And here is the part that matters most: the driven machine did not slow down. The drive held its output the whole time, so there is no performance complaint, no service call, and nothing to diagnose. What the installation lost was margin. The drive's DC bus is now sitting lower, its undervoltage threshold is unchanged, and the headroom between them has shrunk by the drop. The cost gets collected later, when a large motor starts across the building and adds a momentary sag on top of the standing one, and the drive faults on undervoltage while every other load rides through. That fault gets written up as a power quality event, and the standing drop that used up the margin is never mentioned on the ticket.

Why the same gate produced opposite readings

Same feeder, same 0.192 ohms, same 50.0 A design current. The resistive load settled at 3.85 percent drop and 48.08 A; the constant-power load settled at 4.18 percent drop and 52.18 A. The difference is entirely the sign of the load's response, and it produces three inversions worth keeping:

• The drop you calculate at design current is a ceiling for a constant-impedance load and a floor for a constant-power load.
• The load that complains is the one that is not being damaged. The load that is quietly heating its own feeder says nothing.
• Adding load to a feeder that already has drop hurts a constant-power load more than proportionally, because every existing constant-power load on that feeder responds to the new drop by drawing more.

What this does to a motor started across the line

The third response is the one that turns a tolerable running drop into a starting problem. A motor's torque varies roughly with the square of applied voltage, so a motor seeing 96 percent of rated voltage has about 92 percent of its rated torque available (0.96 squared is 0.9216). That is usually fine while running, because a motor running at steady state is using a fraction of the torque it can make.

At start it is not fine, because starting is where the motor uses nearly all of it, and starting is also where the current is several times running current, so the drop is several times larger at the exact moment the margin is smallest. Working out whether a given run survives that moment is its own procedure and the how-to card on run length owns it; do not answer it from a running-current calculation.

What changes the answer

• Duration. A drop present only during a start is a starting-torque and contactor-dropout question. A drop present continuously is a heating and margin question. They are not the same defect and they are not fixed the same way.
• The equipment's own listed range. Utilization equipment carries an operating voltage range on its nameplate or in its literature, and that range, not a percentage rule, is what the equipment is entitled to. Where the nameplate range and a calculated percentage disagree about whether an installation is acceptable, the nameplate governs for the equipment and the adopted code governs for the installation, and they answer different questions.
• How much of the drop is the run and how much is upstream. A 4 percent drop that is half source impedance and half conductor cannot be fixed by upsizing the conductor alone, and a feeder upgrade quoted on the assumption that it can will underdeliver by exactly the source's share.

How to verify you got this right

Take voltage at the equipment terminals twice under the energized-work gate above, once with the load off and once at its normal duty, and record both with the current at the same instant. The difference between them is the drop your run is actually producing, and it is the only version of the number that is a measurement rather than a calculation. Then check the direction against the load type: if the measured drop came in below the figure you calculated at design current, you are on a constant-impedance load and the calculation was conservative. If it came in above, you are on a constant-power load and every calculation done at nameplate current on that feeder has been understating it. If the two disagree with what the load is, one of the two numbers is wrong and it is usually the assumed load type, not the meter.

References

• 29 CFR 1910.333(a)(1), general industry electrical safety-related work practices and the de-energizing gate
• NFPA 70E-2021, 120.5, 130.5 and 130.7, as adopted by your employer's electrical safety program or your authority having jurisdiction
• Equipment nameplate and manufacturer literature for listed operating voltage range and drive undervoltage thresholds
• See related: Voltage Drop Calculations Reference; Why Voltage Sags Under Load; How to Work Out Whether a Run Is Too Long