At first glance the math involved in calculating copper losses may appear awkward, but don't worry. A common calculator with basic functions makes it easy. A basic understanding of Ohm's Laws for electricity is helpful, but not necessary. An copy of the National Electrical Code, even a copy a few years old, will go a long way in helping you find information on the conductors you wish to calculate losses for.

### Things You'll Need

National Electrical Code book or a NEC pocket reference

Calculator

Measured length of a conductor in feet

Size of conductor in American wire gauge or kcm

Amount of current in amps that will be applied to the conductors

## Copper Losses

### Step 1

Open the NEC book to chapter 9, table 8, called conductor properties. Locate the size and type (solid or stranded) of wire and scan across to the ohm/kFt column. This is one of the values you will need. For example, 12 gauge stranded wire has 2.05 ohms of resistance per every 1000 feet.

### Step 2

Insert the value from Step 1 into the following equation. Insert the measured length of the circuit into the equation.

L2/1000 * R = RT = Total DC resistance of circuit conductors L= length of circuit R = conductor resistance per 1000 Ft (from step one)

examples : L = 250 Ft R = 2.05 (# 12 stranded wire)

### Step 3

With a calculator, solve for above equation from left to right. The following example is based on a 250 foot circuit using #12 AWG. EXAMPLE: (250)2/1000 * 2.05 = RT 500/1000 * 2.05 = RT .5 * 2.05 = RT 1.025 ohms = RT

### Step 4

Wire losses are generally given as I²R which is also an Ohm's law for power. The unit used for these losses will therefore be watts or W for short. This is used because losses are basically electrical power given off in the form of heat. Insert the value you got for RT in Step 3 and the expected full load current (given in amps or A) for the circuit into the loss formula below.

I²RT = Loss in watts

EXAMPLE: 10 amps ² * 1.025 ohms = Loss in watts

### Step 5

With a calculator, solve for above equation from left to right. Below an example is given for a ten amp load using a 250 Ft. circuit with #12 gauge wire.

EXAMPLE: 10 amps ² * 1.025 ohms = Loss in watts 100 * 1.025 = Loss in watts 102.5 W = Loss in watts

### Tip

Look over the article at least once before starting any math. Use the practice examples to check your math.

Another way of looking at electrical losses is pressure losses known as "voltage drop."This can be found in NEC article 215.2A Fine Print Note No. 4.

### Warning

Due to the differences in local building codes, this article is primarily for academic use. Before using this information for actual application consult your local and federal building codes and laws.