Ohm's Law and Voltage Drop

7 min · difficulty 2/10

A single-phase 120 V service feeds a house 500 ft down a small conductor, and the load is pulling 40 A. You meter the transformer terminal and read a full 120 V. You walk to the house, meter the service, and it reads 112 V. The transformer is fine and the meters agree. Before any equation, predict it: which way does the voltage go along that span, and what physically moves it?

It sags, and the conductor itself is the cause. Wire is not a perfect path. Every foot has a little resistance, and when you push current through resistance you spend voltage doing it. That spent voltage is the drop, and Ohm’s law sizes it: V = I times R. The harder the load pulls, the bigger the drop, which is why the sag only shows up under heavy current.

Voltage drop around a loaded single-phase loop A single-phase circuit drawn as a loop. A 120 V source on the left feeds current out along a top conductor to a load drawing 40 A on the right, and the current returns along a bottom conductor. The conductor loop has 0.2 ohm of total resistance counting both conductors. The voltage drop is 40 A times 0.2 ohm equals 8 V, so the 120 V source arrives at the load as 112 V. A callout explains the current flows out one conductor and back the other, so the 8 V is dropped across the whole round-trip loop. Voltage drop around a loaded loop single-phase service, 40 A load 120 V source I = 40 A R(loop) = 0.2 Ω total of both conductors, out and back load 40 A 112 V The drop V_drop = 40 A × 0.2 Ω = 8 V 120 V − 8 V = 112 V at load Why the whole loop Current flows out one conductor and back the other, so the 8 V is dropped across both, the round-trip resistance.
Voltage drop is current times the conductor resistance. With 40 A through a 0.2 ohm loop (both conductors, out and back), 8 V is lost, so a 120 V source reaches the load at 112 V.

Put numbers on our span. Treat the 0.2 ohm as the total loop resistance, the round trip out and back through both conductors, since current has to travel the full circuit. With 40 A through 0.2 ohm, the drop is 40 times 0.2, or 8 V. Subtract that from the 120 V source and the load sees 112 V, exactly what the house meter read.

So what do you do when a span sags too far? The load sets the current, so the lever you actually control is resistance. A larger conductor lowers resistance, and lower resistance means less drop for the same load. A shorter run or a higher serving voltage do the same job. Adding length or pushing more current only makes it worse.

Power factor penalties, reactive charges, and %Z mismatches show up as real line items on a utility bill or a bid rejection. DistroForge Insider applies these fundamentals to pricing and equipment-selection intelligence.

Question 1 of 4

A house at the end of a long 120 V service drop measures 112 V at the meter while the transformer terminal still reads 120 V, and only under heavy load. What is the cause?

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