CHAPTER 2
EXAMPLES AND TABLES
COMMENTARY AT 210.20(A) EXCEPTION
An overcurrent device that supplies continuous and noncontinuous loads must have a rating that is not
less than the sum of 100 percent of the noncontinuous load plus 125 percent of the continuous load.
Because grounded/neutral conductors are generally not connected to the terminals of an overcurrent
protective device, this requirement for sizing conductors subject to continuous loads does not apply.
Grounded/neutral conductors are typically connected to a neutral bus or neutral terminal bar located in
distribution equipment.
In addition, 210.19(A)(1) requires that the circuit conductors, chosen from the ampacity tables, have an
ampacity that is:

Not less than the sum of 100 percent of the noncontinuous load plus 125 percent of the
continuous load

Not less than the maximum load to be served after the application of any adjustment or
correction factors

The rating of the overcurrent device cannot exceed whichever calculation yields the greater
result.
Calculation Example
Determine the minimum-size overcurrent protective device and the minimum conductor size for the
following circuit:





30 amperes of continuous load
60°C overcurrent device terminal rating
40°C ambient temperature
Type THWN conductors
Four current-carrying copper conductors in a raceway
Solution
STEP 1. Determine the size of the overcurrent protective device (OCPD). 125 percent of 30 amperes =
37.5 amperes. Thus, the minimum standard-size overcurrent device, according to 240.6(A), is 40
amperes.
STEP 2. Determine the minimum conductor size based on the noncontinuous load plus 125 percent of
the continuous load. The ampacity of the conductor, with no adjustment or correction factors applied.
There is only a continuous load, so 125 percent of the 30-ampere continuous load results in 37.5
amperes.
Because of the 60°C rating of the overcurrent device terminal, a conductor chosen based on the
ampacities in the 60°C column of Table 310.15(B)(16) is necessary. The calculated load must not exceed
the conductor ampacity. Therefore, an 8 AWG conductor with a 60°C allowable ampacity of 40 amperes
is the minimum size permitted. Conductors with a higher allowable ampacity based on their insulation
temperature rating may be used but only at a 60°C allowable ampacity.
STEP 3. Because four current-carrying conductors are in the raceway, Table 310.15(B)(3)(a) applies. The
adjustment factor for four conductors is 80 percent. Because the ambient temperature is 40°C, Table
310.15(B)(2)(a) applies. The correction factor for the ambient temperature is 82 percent. Calculate the
ampacity of the conductor using these adjustment and correction factors:
𝐶𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟 𝑎𝑚𝑝𝑎𝑐𝑖𝑡𝑦 =
𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑙𝑜𝑎𝑑
[𝐹𝑎𝑐𝑡𝑜𝑟𝑠 𝑓𝑟𝑜𝑚 𝑇𝑎𝑏𝑙𝑒 310.15(𝐵)(3)(𝑎)𝑋 𝑇𝑎𝑏𝑙𝑒 310.15(𝐵)(2)(𝑎)]
30 𝑎𝑚𝑝𝑒𝑟𝑒𝑠
= 45.7 𝑎𝑚𝑝𝑒𝑟𝑒𝑠
0.80 𝑋 0.82
The conductor ampacity determined in Step 2 is not the calculated load that is used in this step. It is a
calculation that is used to determine a minimum conductor ampacity and circular mil area based only on
the load with a continuous load factor applied. It ensures sufficient dissipation of heat at the terminals
of overcurrent protective devices supplying continuous loads. The calculated load is 30 amperes. The
phrase “before the application of any adjustment or correction factors” has the effect of not requiring
that the conductor be subjected to a “double-derating.” This occurs when continuous loads are supplied
by a conductor that is also subject to ampacity adjustment because more than three current-carrying
conductors are in a cable or raceway or is subject to ampacity correction due to the installation being
made where the ambient temperature exceeds 30°C (86°F). In this instance, the application of the
adjustment and correction factors indicates that the conductor must have an ampacity of 45.7 amperes.
This would require a 6 AWG conductor at 60°C. However, an 8 AWG conductor with a 75°C insulation
has an ampacity of 50 amperes. The higher temperature-rated insulation can be used, for derating
purposes, as long as the 60°C conductor has an ampacity sufficient to carry the load without overheating
the terminations.
COMMENTARY AT 215.2(A)(B) INFORMATIONAL NOTE NO. 3
Reasonable operating efficiency is achieved if the voltage drop of a feeder or a branch circuit is limited
to 3 percent. However, the total voltage drop of a branch circuit plus a feeder can reach 5 percent and
still achieve reasonable operating efficiency. See Article 100 for the definitions of feeder and branch
circuit.
The 5 percent voltage-drop value is explanatory material and, as such, appears as an informational note.
The informational notes covering voltage drop are not mandatory (see 90.5). Where circuit conductors
are increased due to voltage drop, 250.122(B) requires an increase in circular mil area for the associated
equipment grounding conductors.
The resistance or impedance of conductors may cause a substantial difference between voltage at
service equipment and voltage at the point-of-utilization equipment. Excessive voltage drop impairs the
starting and the operation of electrical equipment. Undervoltage can result in inefficient operation of
heating, lighting, and motor loads. An applied voltage of 10 percent below rating can result in a decrease
in efficiency of substantially more than 10 percent — for example, fluorescent light output would be
reduced by 15 percent, and incandescent light output would be reduced by 30 percent. Induction
motors would run hotter and produce less torque. With an applied voltage of 10 percent below rating,
the running current would increase 11 percent, and the operating temperature would increase 12
percent. At the same time, torque would be reduced 19 percent.
In addition to resistance or impedance, the type of raceway or cable enclosure, the type of circuit (ac,
dc, single-phase, 3-phase), and the power factor should be considered to determine voltage drop.
This basic formula can be used to determine the voltage drop in a 2-wire dc circuit, a 2-wire ac circuit, or
a 3-wire ac single-phase circuit, all with a balanced load at 100 percent power factor and where
reactance can be neglected:
𝑉𝐷 =
2×𝐿×𝑅×𝐼
1000
where:
VD = voltage drop (based on conductor temperature of 75°C)
L = one-way length of circuit (ft)
R = conductor resistance in ohms (Ω) per 1000 ft (from Chapter 9, Table 8)
I = load current (amperes)
For 3-phase circuits (at 100 percent power factor), the voltage drop between any two phase conductors
is 0.866 times the voltage drop calculated by the preceding formula.
Calculation Example
Determine the voltage drop in a 240-volt, 2-wire heating circuit with a load of 50 amperes. The circuit
size is 6 AWG, Type THHN copper, and the one-way circuit length is 100 ft.
Solution
STEP 1. Find the conductor resistance in Chapter 9, Table 8.
STEP 2. Substitute values into the voltage-drop formula:
𝑉𝐷 =
2 × 𝐿 × 𝑅 × 𝐼 2 × 100 × 0.491 × 50
=
= 4.91 𝑣𝑜𝑙𝑡𝑠
1000
1000
STEP 3. Determine the percentage of the voltage drop:
%𝑉𝐷 =
4.91 𝑉
= 0.02 𝑜𝑟 2%
240 𝑉
A 12-volt drop on a 240-volt circuit is a 5 percent drop. A 4.91-volt drop falls within this percentage. If
the total voltage drop exceeds 5 percent, or 12 volts, larger-size conductors should be used, the circuit
length should be shortened, or the circuit load should be reduced.
See the commentary following Chapter 9, Table 9, for an example of voltage-drop calculation using ac
reactance and resistance. Voltage-drop tables and calculations are also available from various
manufacturers.
COMMENTARY AT 215.2(A)(1) EXCEPTION NO. 2
Example
Feeder grounded/neutral conductors that do not connect to the terminals of an overcurrent protective
device are not required to be sized based on 125 percent of the continuous load. For example, if the
maximum unbalanced load on a feeder neutral is calculated per 220.61 to be 200 amperes and the load
is considered to be continuous, the use of a 3/0 AWG, Type THW conductor is permitted as long as the
conductor terminates at a neutral bus or terminal bar within the electrical distribution equipment.
COMMENTARY AT 220.40 INFORMATIONAL NOTE
In the example shown in Exhibit CE1 below, each panelboard supplies a calculated load of 80 amperes.
The main set of service conductors is sized to carry the total calculated load of 240 amperes (3 × 80
amperes). The service conductors from the meter enclosure to each panelboard [2 AWG Cu = 95
amperes per 60°C column of Table 310.15(B)(16)] are sized to supply a calculated load of 80 amperes
and to meet the requirement of 230.90 relative to overcurrent (overload) protection of service
conductors terminating in a single service overcurrent protective device. The main set of service
conductors [250 kcmil THWN Cu = 255 amperes per 75°C column of Table 310.15(B)(16)] is not required
to be sized to carry 300 amperes based on the combined rating of the panelboards. The individual
service-entrance conductors to each panelboard (2 AWG THWN) meet the requirement of 230.90.
EXHIBIT CE1 Service conductors sized in accordance with 220.40.
See Exhibit 230.26 in the handbook for a similar example. In that example, the ungrounded service
conductors are not required to be sized for the sum of the main overcurrent device ratings of 350
amperes. Service conductors are required to have sufficient ampacity to carry the loads calculated in
accordance with Article 220, with the appropriate demand factors applied. See 230.23, 230.31, and
230.42 for specifics on size and rating of service conductors.
Part III of Article 220 contains the requirements for calculating feeder and service loads. Part IV provides
optional methods for calculating feeder and service loads in dwelling units and multifamily dwellings.
Except as permitted in 240.4 and 240.6, the rating of the overcurrent device cannot exceed the final
ampacity of the circuit conductors after all the correction and adjustment factors are applied (such as
where the ambient temperature, the number of current-carrying conductors, or both exceed the
parameters on which the allowable ampacity table values are based).
Calculation Example
Determine the minimum-size overcurrent protective device (OCPD) and the minimum conductor size for
a feeder circuit with the following characteristics:








30 degrees C ambient
3-phase, 4-wire feeder (full-size neutral)
125-A noncontinuous load
200-A continuous load
75°C overcurrent device terminal rating
Type THWN insulated conductors
Major portion of the load is nonlinear
Four current-carrying conductors (neutral conductor is a current-carrying conductor) in the
raceway
Solution
Based on 215.2(A), there are two application conditions that have to be looked at individually and then
compared in determining the minimum size of the feeder conductors. The condition that results in the
larger conductor size must be chosen.
STEP 1. Determine the minimum feeder conductor ampacity by first totaling the continuous and
noncontinuous loads according to 215.2(A)(1)(a).
Total load = 125% of continuous load + noncontinuous load
= (200 A × 1.25) + 125 A
= 250 A + 125 A
= 375 A
In accordance with 215.3, the same calculation is used to determine the minimum rating for the feeder
overcurrent protective device (OCPD)
Using 240.4(B) and 240.6(A), adjust the minimum standard-size OCPD to 400 A.
STEP1(A). Select the feeder conductor size before ampacity adjustment in accordance with
215.2(A)(1)(a).
Using the 75°C column of Table 310.15(B)(16) (because of the overcurrent device terminal), the
minimum-size Type THWN copper conductor that can be used to meet the minimum conductor
ampacity determined in accordance with 215.2(A)(1)(a) (375 amperes) is 500-kcmil copper, which has
an ampacity of 380 A. This conductor is permitted to be protected with a standard-size OCPD rated at
400 A.
If the OCPD used is listed for operation at 100% of its rating, the minimum conductor ampacity is
permitted to be 325 amperes (200A + 125A) and the feeder overcurrent protective device is permitted
to be 350 amperes (next standard size above 325 amperes). See 215.2(A)(1) Exception No. 1 and 215.3
Exception No. 1.
STEP 2. Determine the minimum feeder conductor ampacity needed to supply the calculated load and
then apply any required ampacity adjustment or correction factors according to 215.2(A)(1)(b).
Feeder ampacity (before applying ampacity adjustment or correction factors)
= continuous load + noncontinuous load
= 200 A + 125 A
= 325 A
STEP2(A). Apply adjustment or correction factors. Section 310.15(B)(5)(c) requires that the neutral
conductor be counted as a current-carrying conductor because a major portion of the load consists of
fluorescent and high-intensity discharge (HID) luminaires. Therefore, this feeder circuit consists of four
current-carrying conductors in the same raceway. Section 310.15(B)(3) requires an 80-percent
adjustment factor for four current-carrying conductors in the same raceway. To determine the minimum
ampacity for the conductor use the following formula:
Minimum conductor ampacity = Load
current/adjustment factor
= 325 A / 0.80
= 406.25 A
Because of the requirement to adjust the ampacity of the current-carrying conductors using an 80%
factor, the minimum conductor ampacity for this installation is 406.25 amperes.
STEP 2(B). Determine the minimum conductor size. Using Table 310.15(B)(16) the conductor can be
selected from either the 75 degree C or 90 degree C column. If the conductor insulation temperature
rating is 75 degrees C, the minimum size copper conductor is 600 kcmil with an ampacity of 420
amperes. If the conductor insulation temperature rating is 90 degrees C, the minimum size copper
conductor is 500 kcmil with an ampacity of 430 amperes. Both conductors, after ampacity adjustment,
have an adequate ampacity to supply this load.
600 kcmil @ 420A x 0.80 = 336 amperes
500 kcmil @ 430 x 0.80 = 344 amperes
STEP 2(C). Determine OCPD rating. According to 240.4(B) and 240.6(A), a conductor with a calculated
ampacity of 336 amperes or 344 amperes is permitted to be protected by a 350-A OCPD. Therefore,
either the 600 kcmil THWN (75 degree C insulation rating) copper conductor or the 500-kcmil, Type
THHN (90 degree C insulation rating) copper conductor can be used for this installation. The next
standard size overcurrent device would be 350 amperes. However, 215.3 requires that the OCPD be
rated for the noncontinuous load plus 125% of the continuous load. Exception No. 1 would allow the
OCPD to protect this load if the OCPD is listed for operation at 100% of its rating. If the OCPD is not
listed for 100% of it rating, then a 400 ampere rated device would be necessary, as indicated in Step 1. A
400 ampere rated device would be insufficient to protect conductors having an ampacity of 344
amperes because it is larger than the next standard size. If a 400 ampere OCPD is necessary, it is also
necessary to use 600 kcmil 90 degree C conductors, which would have an ampacity (after adjustment
factors) of 475 x 0.80 = 380 amperes
STEP 3. Compare the minimum conductor ampacity determined in accordance with Step 1 and Step 2.
The calculation in Step 1 results in the need for four 500-kcmil 75 degree C Type THWN copper
conductors in one raceway, each with an ampacity of 380 A, to supply a 375-A load (that consists of
continuous and noncontinuous loads) which is protected by a 400-A OCPD.
The result in Step 2 will depend on the OCPD available. If the OCPD is listed for operation at 100% of its
rating, four 500-kcmil 90 degree C Type THHN copper conductors in one raceway with an ampacity of
344 A can be used with an OCPD rating of 350 amperes (the next standard size in accordance with
240.6). If the OCPD is not listed for operation at 100% of it rating, a 400 ampere rated device is required.
Therefore, it is also necessary to use 600 kcmil 90 degree C conductors, with an ampacity (after applying
adjustment factor) of 380 amperes (475 x 0.80)
Section 215.2(A) requires the minimum conductor to be the larger result determined using Step 1 or
Step 2. Step 2 results in either a larger conductor size or a conductor with a higher insulation
temperature rating. Therefore a 500 kcmil copper conductor, with a 90 degree C insulation temperature
rating and ampacity of 430 amperes can be used to meet the minimum conductor requirement
determined in accordance with 215.2(A)(1)(b). Interestingly, the size of the conductor determined by
both conditions (215.2(A)(1)(a) and 215.2(A)(1)(b) is 500 kcmil. However the higher ampacity achieved
by using a 500 kcmil conductor with a 90 degree C insulation temperature rating is required in order to
meet the minimum conductor ampacity determined by 215.2(A)(1)(b). If a conductor with a 75 degree C
insulation temperature rating is used, the minimum copper feeder conductor is 600 kcmil with an
ampacity of 420 amperes.
COMMENTARY AT 250.30(A)(6)
A common grounding electrode conductor serving several separately derived systems is an alternative
to installing individual grounding electrode conductors from each separately derived system to the
grounding electrode system. In such an arrangement, a tapped grounding electrode conductor is
installed from the common grounding electrode conductor to the point of connection to the individual
separately derived system grounded conductor. This tap is sized from Table 250.66 based on the size of
the ungrounded conductors for that individual separately derived system.
The minimum size for this conductor is 3/0 AWG copper or 250-kcmil aluminum so that the grounding
electrode conductor always is of sufficient size to accommodate the multiple separately derived systems
it serves. This minimum size for the common grounding electrode conductor correlates with the
maximum size grounding electrode conductor required by Table 250.66. Therefore, the 3/0 AWG copper
or 250-kcmil aluminum becomes the maximum size required for the common grounding electrode
conductor. The sizing requirement for the common grounding electrode conductor is specified in
250.30(A)(6)(a), and the sizing requirement for the individual taps to the common grounding electrode
conductor is specified in 250.30(A)(6)(b).
The methods of connecting the individual tap conductor(s) to the common grounding electrode
conductor are specified in 250.30(A)(6)(c). The permitted methods include the use of busbar as a point
of connection between the taps and the common grounding electrode conductor. The connections to
the busbar must be made using a listed means. Exhibit CE2 shows a copper busbar used as a connection
point for the individual taps from multiple separately derived systems to be connected to the common
grounding electrode.
The following example, together with Exhibit CE3, illustrates the use of a common grounding electrode
conductor for grounding multiple separately derived systems (transformers in this case).
Calculation Example
A building is being renovated for use as an office building. The building is being furnished with four 45kVA, 480 to 120/208-volt, 3-phase, 4-wire, wye-connected transformers. Each transformer secondary
supplies an adjacent 150-ampere main circuit breaker panelboard using 1/0 AWG, Type THHN copper
conductors. The transformers are strategically placed throughout the building to facilitate efficient
distribution. Because there is no accessible effectively grounded structural steel in the same areas as
where the transformers are installed and the metal water piping is concealed after it leaves the lower
level, each transformer secondary (separately derived system) must be grounded to the water service
electrode within the first 5 ft of entry into the building. A common grounding electrode conductor has
been selected as the method to connect all the transformers to the grounding electrode system.
What is the minimum-size common grounding electrode conductor that must be used to connect the
four transformers to the grounding electrode system? What is the minimum-size grounding electrode
conductor to connect each of the four transformers to the common grounding electrode conductor?
Solution
STEP 1. Determine the minimum size for the common grounding electrode conductor.
Minimum size required [per 250.30(A)(6)(a)]: 3/0 copper or 250-kcmil aluminum
The common grounding electrode conductor does not have to be sized larger than specified by this
requirement. Additional transformers installed in the building can be connected to this common
grounding electrode conductor, and no increase in its size is required.
STEP 2. Determine the size of each individual grounding electrode tap conductor for each of the
separately derived systems. According to Table 250.66, a 1/0 AWG copper derived phase conductor
requires a conductor not smaller than 6 AWG copper for each transformer grounding electrode tap
conductor. This individual grounding electrode conductor will be used as the permitted tap conductor
and will run from where it connects to the grounded conductor in the transformer to a connection point
located on the common grounding electrode conductor. This conductor is labeled “Conductor B” in
Exhibit CE3.
EXHIBIT CE2 Listed connectors used to connect the common grounding electrode
conductor and individual taps to a centrally located copper busbar with a minimum
dimension of 1/4 in. thick by 2 in. wide.
EXHIBIT CE3 The grounding arrangement for multiple separately derived systems using
taps from a common grounding electrode conductor, according to 250.30(A)(6)(a) and
(A)(6)(b).