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Voltage Drop Calculations 3/26/2003
By Mike Holt for EC&M Magazine

Do you know how and why to calculate voltage drop?

Contrary to common belief, the NEC generally does not require you to size conductors to accommodate voltage drop. It merely recommends that you adjust for voltage drop when sizing conductors. The recommendations are in the Fine Print Notes to Sections 210.19(A), 215.2(A)(4), 230.31(C) and 310.15(A)(1). Fine Print Notes are recommendations, not requirements [90.5(C)].

The NEC recommends that the maximum combined voltage drop for both the feeder and branch circuit should not exceed five percent, and the maximum on the feeder or branch circuit should not exceed three percent (see Figure 8-10). This recommendation is a performance issue, not a safety issue.

Note: Graphic are not included with this newsletter.

If the NEC doesn't require you to size for voltage drop, why even think about it? Here are some reasons:

System efficiency. If a circuit supports much of a load, a larger conductor will pay for itself many times over in energy savings alone.
System performance. Lighting loads perform best when voltage drop is minimal. You get the light of a higher-watt system simply by running larger wires.
Troubleshooting. If you follow the NEC voltage drop recommendations, you don't have to guess whether your field measurements indicate a problem or if the voltage is low due to not accommodating voltage drop in the design.
Load protection. Undervoltage for inductive loads can cause overheating, inefficiency, and shorter life span. When a conductor resistance causes the voltage to drop below an acceptable point, increase the conductor size (see Figure 8-9).
What is it?

The voltage drop of a circuit is in direct proportion to the resistance of the conductor and the magnitude of the current. If you increase the length of a conductor, you increase its resistance-and you thus increase its voltage drop. If you increase the current, you increase the conductor voltage drop. Thus long runs often produce voltage drops that exceed NEC recommendations.

Here's a pop quiz. What is the minimum NEC-recommended operating voltage for a 115V load connected to a 120V source (see Figure 8-11)?

(a) 120V (b) 115V (c) 114V (d) 116V

Answer: (c) 114V

The maximum conductor voltage-drop recommended for both the feeder and branch circuit is five percent of the voltage source (120V). The total conductor voltage drop (feeder and branch circuit) should not exceed 120V x 0.05 = 6V. Calculate the operating voltage at the load by subtracting the conductor voltage drop from the voltage source: 120V - 6V = 114V.

Doing the calculations

You can determine conductor voltage drop by the Ohm's Law method or by the formula method. You can use the Ohm's Law method (I x R) for single-phase only. Regardless of which method you use, observe the following:

For conductors 1/0 AWG and smaller, the difference in resistance between dc and ac circuits is so little that it can be ignored. In addition, you can ignore the small difference in resistance between stranded and solid wires.
VD = Voltage Drop
I = The load in amperes at 100%, not at 125%, for motors or continuous loads.
R = Conductor Resistance, Chapter 9, Table 8 for dc or Chapter 9, Table 9 for ac.
Let's do an Ohm's Law Method sample calculation. What is the voltage drop of two 12 AWG THHN conductors that supply a 16A, 120V load located 100 ft from the power supply (see Figure 8-12)?

(a) 3.2V (b) 6.4V (c) 9.6V (d) 12.8V

Answer: (b) 6.4V

The math is straightforward:

VD = I x R
I = 16A; R = 2V per 1,000 ft, Chapter 9, Table 9: (2V/1,000 ft) x 200 ft = 0.4V
VD = 16A x 0.4V = 6.4V
Formula Method

This method is a bit more involved that the Ohms Law method, but the big advantage is you can use it for three-phase or single-phase. Here are some additional items to observe:

Single-phase VD = 2 x K x I x D/CM.
Three-phase VD = 1.732 x K x I x D/CM. The difference between this and the single phase formula is you replace the 2 with 1.732.
K = Direct-Current Constant. K represents the dc resistance for a 1,000-circular mils conductor that is 1,000 ft long, at an operating temperature of 75C. K is 12.9 ohms for copper and 21.2 ohms for aluminum.
Q = Alternating-Current Adjustment Factor: For ac circuits with conductors 2/0 AWG and larger, you must adjust the dc resistance constant K for the effects of self-induction (eddy currents). Calculate the "Q" Adjustment Factor by dividing the ac ohms-to-neutral impedance listed in Chapter 9, Table 9 by the dc resistance listed in Chapter 9, Table 8.
I = Amperes: The load in amperes at 100% (not at 125% for motors or continuous loads).
D = Distance: The distance the load is from the power supply. When calculating conductor distance, use the length of the conductor-not the distance between the equipment connected by the conductor. To arrive at this length, add distance along the raceway route to the amount of wire sticking out at each end. An approximation is good enough. Where we specify distances here, we are referring to the conductor length.
CM = Circular-Mils: The circular mils of the circuit conductor as listed in NEC Chapter 9, Table 8.
Let's do a three-phase example. A 3, 36 kVA load rated 208V is wired to the panelboard with 80 ft lengths of 1 AWG THHN aluminum. What is the approximate voltage drop of the feeder circuit conductors (see Figure 8-15)?

(a) 3.5V (b) 7V (c) 3% (d) 5%

Answer: (a) 3.5V

How did we do that? Applying the three-phase formula, where:

K = 21.2 ohms, aluminum

I = 100A (36,000/(208 x 1.732))

D = 80 ft

CM = 83,690 (Chapter 9, Table 8)

VD = 1.732 x 21.2 x 100A x 80/83,690 = 3.51 VD

Algebraic variations

Using basic algebra, you can apply the same basic formula to find one of the other variables if you already know the voltage drop. For example, suppose you want to know what size conductor you need to reduce the voltage drop to the desired level. Simply rearrange the formula. For three-phase, it would look like this: CM (3) = 1.732 x K x I x D/VD. Of course, for single-phase calculations, you would use 2 instead of 1.732.

Suppose you have a 3, 15 kVA load rated 480V and 390 ft of conductor. What size conductor will prevent the voltage drop from exceeding three percent (see Figure 8-17)?

K = 12.9 ohms, copper
I = 18A (15,000/(480 x 1.732))
D = 390 ft
VD = 480V x 0.03 = 14.4V
CM = 1.732 x 12.9 x 18 x 390/14.4V = 10,892, 8 AWG, Chapter 9, Table 8
You could also rearrange the formula to solve a problem like this one: What is the maximum length of 6 AWG THHN you can use to wire a 480V, 3, 37.5 kVA transformer to a panelboard so voltage drop does not exceed three percent (see Figure 8-19)?

D (3) = CM x VD/1.732 x K x I

CM = 26,240, 6 AWG Chapter 9, Table 8
VD = 480V x 0.03 = 14.4V
K = 12.9 ohms, copper
I = 45A (37,500/480 x 1.732))
D = 26,240 x 14.4/1.732 x 12.9 x 45 = 376 ft

Sometimes, the only method of limiting voltage drop is to limit the load. Again, we can rearrange the basic formula algebraically: I = CM x VD/1.732 x K x D. Suppose an installation contains 1 AWG THHN conductors, 300 ft long in an aluminum raceway, to a 3, 460/230V power source. What is the maximum load the conductors can carry, without exceeding the NEC recommendation for voltage drop (see Figure 8-21)? Let's walk it through:

I = CM x VD/1.732 x K x D

CM = 83,690 (1 AWG), Chapter 9, Table 8
VD = 460V x 0.03 = 13.8V
K = 12.9 ohms, copper
D = 300 ft
I = 83,690 x 13.8/1.732 x 12.9 x 300 = 172A

Note: The maximum load permitted on 1 AWG THHN at 75C is 130A [110.14(C) and Table 310.16].

As you can see, voltage drop is something you calculate. And calculating it is easy. The NEC doesn't require you to do these calculations, so voltage drop isn't a safety issue. However, a system that meets NEC requirements may not be efficient in terms of power consumption or optimum in terms of equipment longevity and performance. As the old saying goes, "Wire is cheap, but performance loss is costly." First ensure your system meets NEC requirements. Then, look at it for voltage drop so you get the right cost trade-offs between performance and cost.