How do you calculate voltage drop along a conductor and what typical limits apply?

• A. ΔV = R × I; limit 10–15%
• B. ΔV = Z × I; limit 0–1%
• C. ΔV = I × Z (or I × R for purely resistive paths); keep ΔV within a prescribed percentage of supply (commonly 3–5%)
• D. ΔV = I ÷ Z; limit 50%

Answer: (C)

Voltage drop along a conductor comes from the current flowing through the conductor’s impedance. In AC circuits you use the impedance Z, which includes both the resistance and the reactance; if the path is purely resistive, you can use the resistance R alone. So the drop is ΔV = I × Z (and for purely resistive paths, ΔV = I × R).

Keep in mind why this matters: a voltage drop reduces the voltage available at the load, which can cause devices to underperform, overheat, or draw more current to compensate. That’s why designers limit the drop to a small percentage of the supply, typically about 3–5% for feeders and branch circuits, with the total system sometimes targeted around 5% or a bit less depending on the code and application. Long runs, smaller conductors, or higher load amplify the drop, while larger conductors or shorter runs reduce it.

The best fit among the options is the one that uses ΔV = I × Z (or ΔV = I × R for purely resistive paths) and acknowledges keeping ΔV within a prescribed limit around 3–5%. Using ΔV = R × I is valid only for purely resistive cases and misses the general AC impedance. An expression like ΔV = I ÷ Z is incorrect dimensionally, and the limits 0–1%, 10–15%, or 50% are not typical practice for most electrical designs.