2.4 Voltage Drop Calculations & Chapter 9 Conductor Properties

2.4 Voltage Drop Calculations & Chapter 9 Conductor Properties

Electrical conductors are manufactured from materials with low, but finite, electrical resistance. As load current flows through feeder and branch conductors, energy is converted into heat, creating a potential loss along the length of the run known as voltage drop. If unmitigated over long distances, voltage drop deprives utilization equipment of rated operating potential, resulting in motor overheating, nuisance tripping of electronic drives, diminished lighting lumen output, and excessive energy consumption.

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NEC Voltage Drop Recommendations vs. Mandatory Code

Understanding the legal status of voltage drop in the National Electrical Code is critical for the licensing examination:

1. Informational Note Status: The primary NEC guidelines on voltage drop appear in Informational Notes:
   • Informational Note to NEC 210.19: Recommends branch-circuit conductors sized to limit voltage drop to 3% at the farthest outlet of power, heating, and lighting loads.
   • Informational Note to NEC 215.2: Recommends feeder conductors sized to limit voltage drop to 3%.
   • Total Overall System Limit: Combined feeder and branch circuit voltage drop should not exceed 5% from the service equipment to the final utilization point.
   • Code Rule: Under NEC 90.5(C), Informational Notes are non-mandatory and unenforceable unless specifically adopted into state/local municipal amendments or specified in project engineering criteria.
2. Mandatory NEC Voltage Drop Requirements: In select critical applications, the NEC makes voltage drop compliance mandatory:
   • Fire Pumps (NEC 695.7): Voltage drop must not exceed 15% under motor starting conditions, and must not exceed 5% under normal full-load operating conditions.
   • Sensitive Electronic Equipment (NEC 647.4(D)): Limits total voltage drop to 1.5% for branch circuits and 2.5% total.

Voltage Drop Limits for Standard Nominal Voltages

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The Circular Mil Method and Formulas

The standard method for calculating voltage drop on the Minnesota Journeyman examination utilizes the circular mil formula.

Definitions of Variables

• $V_{\text{drop}}$: Total circuit voltage drop in volts ($\text{V}$).
• $K$ (Specific Conductor Resistivity): The resistance of a conductor $1\text{ circular mil}$ in cross-sectional area and $1\text{ foot}$ long, often called ohms per mil-foot:
  • Copper: $K \approx 12.9$ (DLI's examples use 12.8)
  • Aluminum: $K \approx 21.2$ (DLI's examples use 21.1)
  • On the exam, use the K value the question gives. If none is given, results with 12.8 or 12.9 differ by less than 1%, so the answer choice is rarely affected.
• $I$: Circuit design load current in amperes ($\text{A}$).
• $L$: One-way length of the circuit run in feet ($\text{ft}$).
• $\text{CM}$: Cross-sectional conductor area in circular mils (from NEC Chapter 9, Table 8).

Single-Phase Circuits

In a single-phase circuit, current flows out along the ungrounded conductor and returns along the neutral (or second ungrounded conductor), traversing twice the one-way circuit length. The factor of $2$ accounts for both conductors:

$V_{\text{drop}(1\phi)} = \frac{2 \times K \times I \times L}{\text{CM}}$

Three-Phase Balanced Circuits

In a balanced three-phase system, current returns through the remaining two phase conductors with a $120^\circ$ vector displacement. The single-phase factor of $2$ is replaced by $\sqrt{3} \approx 1.732$:

$V_{\text{drop}(3\phi)} = \frac{\sqrt{3} \times K \times I \times L}{\text{CM}} \approx \frac{1.732 \times K \times I \times L}{\text{CM}}$

Sizing Conductors to Comply with Voltage Drop

To find the minimum conductor size required to stay within an allowable voltage drop limit ($V_{\text{drop}}$), rearrange the formulas to solve for circular mils ($\text{CM}$):

$\text{CM}_{(1\phi)} = \frac{2 \times K \times I \times L}{V_{\text{drop}}}$ $\text{CM}_{(3\phi)} = \frac{\sqrt{3} \times K \times I \times L}{V_{\text{drop}}}$

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NEC Chapter 9 Table 8 Conductor Properties

Once the minimum required circular mil area is calculated, electricians refer to NEC Chapter 9 Table 8 (Conductor Properties) to select the next standard American Wire Gauge (AWG) or kcmil size with an area equal to or greater than the calculated value.

Table 8 gives direct-current resistance. For long feeders or inductive loads, NEC Chapter 9 Table 9 gives alternating-current resistance and reactance, but DLI's guide notes that its simplified formulas ignore skin effect, power factor, and harmonics. Each set of parallel conductors is treated as one conductor when calculating voltage drop.

DLI's worked voltage-drop examples

• Single-phase, K method: 240 V, 150 ft, 28 A, No. 8 copper: $VD = (2 \times 12.8 \times 28 \times 150) \div 16{,}510 = 6.5$ V.
• Three-phase, R method: 208 V, 205 ft, 33 A, conductors of 0.510 Ω per 1000 ft: $VD = (1.732 \times 0.510 \times 33 \times 205) \div 1000 = 5.98$ V.
• Percent drop: 480 V three-phase feeder, 280 ft, 135 A, 250 kcmil aluminum: $VD = (1.732 \times 21.1 \times 135 \times 280) \div 250{,}000 = 5.53$ V, which is $5.53 \div 480 = 1.15\%$.

The Ohm's Law Table 8 Alternative Method

Instead of the circular mil formula, voltage drop can be verified directly using conductor resistance from Table 8:

$R_{\text{conductor}} = R_{\text{table}} \times \left(\frac{L}{1000\text{ ft}}\right)$ $V_{\text{drop}(1\phi)} = 2 \times I \times R_{\text{conductor}}$ $V_{\text{drop}(3\phi)} = \sqrt{3} \times I \times R_{\text{conductor}}$

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Mandatory Proportional Upsizing of EGC (NEC 250.122(B))

One of the most frequently tested provisions on the Minnesota Journeyman examination is NEC 250.122(B):

"Where ungrounded conductors are increased in size from the minimum size that has sufficient ampacity for the intended installation, wire-type equipment grounding conductors, where installed, shall be increased in size proportionately according to the circular mil area of the ungrounded conductors."

Rationale for the Rule

When phase conductors are upsized to counteract voltage drop over long runs, conductor impedance decreases. If a phase-to-ground fault occurs at the end of that run, the circuit must maintain an equivalently low-impedance ground path to ensure sufficient fault current flows to rapidly trip the upstream overcurrent protective device. If the equipment grounding conductor (EGC) remained at its standard minimum size from Table 250.122, its relatively high resistance could limit ground-fault current or cause dangerous touch potential along metallic conduit enclosures.

Proportional Upsizing Formula

$\text{CM}_{\text{new EGC}} = \text{CM}_{\text{orig EGC}} \times \left(\frac{\text{CM}_{\text{upsized phase}}}{\text{CM}_{\text{orig phase}}}\right)$

Step-by-Step EGC Sizing Walkthrough

A $100\text{ A}$ single-phase feeder protected by a $100\text{ A}$ circuit breaker runs $350\text{ ft}$ from a main service to a subpanel. The calculated load current is $80\text{ A}$ at $240\text{ V}$. Copper conductors with $75^\circ\text{C}$ THHN insulation are specified.

1. Determine Original Minimum Sizes:
   • Per NEC Table 310.16 ($75^\circ\text{C}$), a $100\text{ A}$ breaker requires a minimum 3 AWG copper ungrounded conductor ($52,620\text{ cmil}$, rated at $100\text{ A}$).
   • Per NEC Table 250.122, a $100\text{ A}$ overcurrent device requires a minimum 8 AWG copper equipment grounding conductor ($16,510\text{ cmil}$).
2. Calculate Maximum Allowable Voltage Drop (3% Feeder Limit): $V_{\text{drop(max)}} = 240\text{ V} \times 0.03 = 7.20\text{ V}$
3. Calculate Required Ungrounded Conductor Area: $\text{CM} = \frac{2 \times K \times I \times L}{V_{\text{drop}}} = \frac{2 \times 12.9 \times 80\text{ A} \times 350\text{ ft}}{7.20\text{ V}} = \frac{722,400}{7.20} = 100,333\text{ cmil}$
4. Select Upsized Ungrounded Conductor:
   • Consulting NEC Table 8: 1 AWG is $83,690\text{ cmil}$ (insufficient).
   • Select 1/0 AWG copper ($105,600\text{ cmil}$).
5. Apply NEC 250.122(B) Proportional EGC Upsizing: $\text{Ratio} = \frac{\text{CM}_{\text{upsized}}}{\text{CM}_{\text{original}}} = \frac{105,600\text{ cmil}}{52,620\text{ cmil}} \approx 2.00684$ $\text{CM}_{\text{new EGC}} = 16,510\text{ cmil} \times 2.00684 \approx 33,133\text{ cmil}$
6. Select Upsized EGC from Table 8:
   • 6 AWG is $26,240\text{ cmil}$ (insufficient, less than $33,133\text{ cmil}$).
   • 4 AWG copper ($41,740\text{ cmil}$) is the next standard size.
   • Result: The ungrounded conductors must be upsized to 1/0 AWG, and the equipment grounding conductor must be upsized from 8 AWG to 4 AWG.

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Conductor Sizing Summary Workflow