Divide the voltage flowing through the cable by **the desired current**. For example, if the wire will be subjected to 120 volts and you want 30 amps to pass through it, 120/30 equals 4. This is your desired resistance in ohms. 30,000 multiplied by 1.724 multiplied by 10-8 equals **0.0005172 ohm square**. This is the required cross-sectional area in inches squared.

There are 12 inches in a foot, so you need.00583 feet-inch or 612 cubic inches of space. One cubic inch holds **about 0.25 square inches**, so you need **about 20 linear inches** of space per run of cable.

You can also use voltage and length. If you know the voltage and the length of the cable, you can calculate the current. For example, if you know the voltage is 120 volts and the length is 100 feet, then the current is 120/100 or 12 amps. You can now divide 12 by 0.01 to find out how many milliohm meters the cable needs to have: 12000/0.01 = 1200 ohm meter. Or you can say that you need one thousandth of an ohm per foot of cable.

Finally, you can count the number of conductors in the cable and multiply their average diameter by the number of strands they contain.

Engineers frequently inquire about how to determine the resistance of a cable. With copper cable, there is a fairly simple formula that works well within an ohm or so. Resistance per meter is 7.6 ohms per km/by 1000 for 19.2/by 2.5mm2. 4.75 ohms per kilometre is obtained by multiplying 19/20 by 4mm2. If the conductor size is smaller than 4mm2, then use **the smaller value**. If it's larger, then multiply by 4 instead of 20.

Resistance of coaxial cable depends on its diameter: resistance = 120ohms/diameter. Coaxial cable is usually between 19 and 100 mm (7.5 and 39 inches) in diameter. The resistance varies with length as ln(r)/r, where r is the radius of the cable. For example, if the cable is 10 times as long but only half as thick, its resistance will be approximately equal to that of the shorter one. Or, if the longer cable has twice the resistance of the shorter one, then they are equivalent in value.

The resistance of **fiber-optic cable** is much less than 1ohm. But since it is very small compared to what we normally deal with, we can assume it is zero for **our purposes**.

The resistance of **metal wiring** in buildings is mainly due to the amount of metal involved. Wires are usually either aluminum or steel.

Case #3: Cable Purchasing

- Current capacity of 95 Sq.mm cable is 200 Amp, Resistance = 0.41 Ω/Km and. Reactance = 0.074 mho/Km.
- Total derating current of 70 Sq.mm Cable = 200 · 0.93 = 187 Amp.
- Voltage drop of cable = 1.732 · 139 · (0.41 · 0.8 + 0.074 · 0.6) · 200 · 100 / (415 · 1 · 1000) = 2.2%

(100 * 1000) / (1.732 * 415) = 139 Amps. Full Load Current = (KVA * 1000) / (1.732 * Voltage): (100 * 1000) / (1.732 * 415) = 139 Amps. For a single run, let's go with a 3.5-core 70 sq.mm cable.

- Current capacity of 70 Sq.mm cable is: 170 Amp,
- Total derating current of 70 Sq.mm cable = 170 · 0.93 = 159 Amp.

To determine the percentage voltage drop for a circuit, multiply the current (amps) by the cable length (metres), then divide the result by the value in the table. A 30m stretch of 6mm2 cable carrying 3 phase 32A, for example, will result in a 1.5 percent drop: 96 Amps / 615 = 1.5 percent. The more conductors in a cable, the less voltage drop across it.

Cable resistance is a major factor in determining **voltage drop** across a cable run. The more conductors in a cable, the lower the resistance per unit length, so fewer volts drop across the cable. For example, a 4-wire 120V subpanel cable can be used for branch circuits if it is specified that it carry no more than 15 amps, which means it should be no more than 2.5 mm2 in cross section. If the cable is either 4 or 4-wire, it will reduce the voltage drop across it compared to a 2-wire cable of **the same size wire gauge**. Voltage drops due to resistance can be reduced by using **larger cables** or multiple runs of smaller cables in parallel.

The second major factor affecting voltage drop across a cable run is the distance between electrical outlets (or fuse boxes). The farther apart these are, the greater the potential difference between them will be.

3 x (I2R)/1000 = power losses Where P is the power loss in kW units, I is the current (in amps), and R is the average conductor resistance (in ohms). How can you reduce the cable's resistance? The amount of power lost in a cable is determined by the cable length, cable size, and current flowing through the cable. If you can reduce any one of these factors, you can reduce the power loss in **your cable**.

The power loss in an electrical cable is the rate at which energy is lost to heat in **the conductor material**. This is different from actual power loss which is the rate at which energy is lost to the environment. Power loss depends on **several factors** such as the type of conductor used, its diameter, the number of strands it contains, its length, and the current it carries.

The power loss in a cable is given by this formula: 3 x (I2R)/1000 where I is the current carried by the cable, R is the average resistance of the cable and 1000 is the factor by which power loss in watts is multiplied to give power loss in kilowatts.

So, if we take **an example case** of a cable with I = 10 A, R = 0.5 OA and length l = 1 km then power loss will be 30 kW. In reality though, power loss does not depend only on the current carrying capacity of the cable but also on its length. As we know, voltage drops along a conductor length.