**Service Neutrals And Balancing**

**Article 220.61**covers requirements for determining the required amp rating of the neutral feeder conductor. It states that the neutral feeder load shall be the “maximum unbalance" of the feeder load. The maximum unbalanced load shall be the maximum connected load between the neutral and any one ungrounded conductor. In a 3-wire, 120/240 volt, single-phase feeder, the neutral must have current- carrying capacity at least equal to the current drawn by the total 120 volt load (connected between the more heavily loaded hot leg and the neutral).

It should be noted that straight 240 volt loads, connected between the two ungrounded conductors, do not place a load on the neutral because they are balanced.

Balancing

When calculating the neutral for a single-family dwelling, we must distribute all of the 120 volt loads as evenly as possible, between the two “hot” feeder conductors of a typical 240/120 volt, single-phase panel.

A typical house panel consists of two bus bars that circuit breakers snap into. A single 120 volt breaker connects to one phase

Say we have a 2,400 square foot house, our general lighting and receptacle loads would equal..

**(“A or “B”)**, a 240 volt breaker would connect to both phases**(“A and “B”)**.

Say we have a 2,400 square foot house, our general lighting and receptacle loads would equal..

2,400 sq. ft. x 3 vA =

120 volts x 15 amps (per ckt.) =

7,200 vA ÷ 1,800 vA =

**7,200 vA**120 volts x 15 amps (per ckt.) =

**1,800 vA**7,200 vA ÷ 1,800 vA =

**4 @ 15a each**We would need a total of four 15 amp circuits (#1 - #4) at 1,800 vA each for general lighting and receptacle loads. If we were to distribute these circuits as evenly as possible, on our 240/120 volt panel, it would look something like this...

We call this technique balancing.

Also, notice that the general lighting circuits can be less than 1,800 vA each. If the square foot area of a house was 1,600 square feet:

1,600 sq. ft. x 3 vA = 4,800 vA

4,800 vA ÷ 120 volts = 40 amps

40 amps ÷ 15 amps = 2.66 circuits (round up to 3)

4,800 vA ÷ 3 =

**Article 210.11(B)**of the Code encourages us to balance our loads as evenly as possible. We were lucky enough to balance our circuits evenly in the above panel. Sometimes it’s a little more difficult and we have to settle for a close approximation. Remember, we size our neutral conductor based on the maximum unbalanced load.Also, notice that the general lighting circuits can be less than 1,800 vA each. If the square foot area of a house was 1,600 square feet:

1,600 sq. ft. x 3 vA = 4,800 vA

4,800 vA ÷ 120 volts = 40 amps

40 amps ÷ 15 amps = 2.66 circuits (round up to 3)

4,800 vA ÷ 3 =

**1,600 vA per circuit**Cooking Appliance And Dryer Neutrals

**Article 220.61(B)(1)**also states that for a neutral feeder supplying household electric ranges, wall-mounted ovens, counter-mounted cooking units, and electric dryers, the maximum unbalanced load shall be considered as

**70%**of the load on the ungrounded conductors (as determined in accordance with

**Table 220.55**for ranges, and

**Table 220.54**for dryers).

It might seem a little strange to reduce cooking appliances and dryers twice. But actually we’re only reducing them once. Most of the load of a range or dryer is primarily heating elements (240 volts - balanced). A smaller neutral conductor is there only to cover the 120 volt components like clocks, timers, lights, etc.

Let’s calculate the neutral for the following appliances...

Neutral Size

Now all we have to do to size our neutral is add it all together:

situation because we’re at two voltage levels (240/120v). If we add together all of the

wattages and divide by either 240 or 120 volts we would get an incorrect answer.

For example...

**(1)**The maximum of the unbalanced 120 volt loads in amps**(Feeder“A” or “B”)**.**(2)**Cooking appliance and dryer amp loads at**70%**.We must convert everything to amps in thissituation because we’re at two voltage levels (240/120v). If we add together all of the

wattages and divide by either 240 or 120 volts we would get an incorrect answer.

For example...

Neutrals And Balancing Sample Problem...

Calculate the system feeders and neutral conductor for an 1,800 square foot house fed by a 240/120 volt service containing the following loads...

Neutral

Largest 120 volt load (6,635/120) = 55.3 amps

Total =

**(220.61)**To the Larger of the balanced 120 volt loads add 70% of the Cooking Appliances and Dryer...Largest 120 volt load (6,635/120) = 55.3 amps

__Range, Oven, Dryer (12,800/240) x .7 or 70%____= 37.3 amps__Total =

**92.6 amps**

Neutral Size (T310.15(B)(16)) = #3 THW