CHOOSING THE CORRECT **POWER CABLE** FOR DIRECT CURRENT

In any project where you need to handle electricity, there will be a time when you ask yourself, what **dc power cable** do I need? I will give you some clues but remember that we are working with direct and not alternating current.

I have preferred to tell the whole story a bit so that those interested know better were all these formulas and concepts come from. There is nothing worse than formulas without knowledge.

First, a basic notion. We need to know that in electricity, the three most important concepts are voltage, current, and resistance, or what is the same, Ohm’s law:

**V = I R**

*V, potential difference in volts (V)*

* I, current in amps (A)*

* R, resistance in ohms (Ω)*

Therefore we have a potential difference normally generated by batteries or a power supply and which is measured in volts (V). The power cable or any component that we have in our circuit supposes a resistance (ohms – Ω) to the passage of current (amps – A). For simplicity, the larger the cable section, the lower its resistance, and therefore more current can pass through it without heating (or heating as little as possible).

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## CABLE RESISTANCE

The resistance depends on the power cable material; copper has a resistivity coefficient ρ = 0.0172 Ω · mm2 / m (at 20ºC). This is ohms for each mm2 of cable section for each linear meter. If you have aluminum electrical cables, this value is higher, and therefore its resistance ρ = 0.028 Ω · mm2 / m. Iron has a much higher resistivity ρ = 0.089 Ω · mm2 / m. This is how you understand the reason behind the choice of copper as the preferred material for conductors.

It is not strictly necessary, but for information purposes, I leave you how the real resistance of the elctrical cable in question is calculated, using the following equation:

**R = ρ · L / s**

*R, resistance of the wire in Ω*

* ρ, the coefficient of resistivity of the material in Ω. mm² / m, at 20ºC*

* L, cable length in meters*

* s, cable section in mm2 (for electrical cables with circular section A = pi · (diam / 2) ^ 2)*

## TENSION FALL

One idea that has to be clear is that there is always a voltage drop in any power cable (at the end of the electric cable, we will have fewer volts than supplied). We have to define how many volts we need to make our appliance, bulb, motor, etc., work correctly.

Normally the voltage drop is defined as a percentage of the supplied voltage; that is, if our power supply is 12V and we determine a maximum voltage drop of 2%, we are saying that at least we want to have 11.76V available for our device (12V – 2% of 12V). If this voltage were not enough, we would only have two options or increase the power supply voltage or, which is normally done, increase the cable section so that its resistance is lower and, therefore, the voltage drop.

## CALCULATION OF INTENSITY

The current intensity is defined almost by its name; it is the flow of charge per second (or unit of time) that passes through the cable. Normally we know the intensity because it is already specified in the lights, motors, or any other device that we need. If we are given only the power, we also have it very easy:

**P = V I** , or what is the same:

**I = P / V**

P, power in watts (W)

* V, potential difference in volts (V)*

* I, current in amps (A)*

We divide the power by the voltage, and we have the intensity. For example, we have our 12V circuit with an LED strip with a maximum power of 24W; our calculation intensity will be 24W / 12V = 2A. Sometimes the maximum amperage is already specified, so we would already have it.

## FINALLY … THE CALCULATION OF THE CABLE SECTION

Well, now that we have these clear concepts, to the mess. Ohm’s law tells us that the voltage drop is ΔV = I · R (careful, because here we have to count the forward and backward cable’s resistance since logically the return also opposes resistance). We substitute R for the formula for calculating the resistance of the power cable that we saw before, and we obtain:

**ΔV = 2 I ρ L / s**

*ΔV, voltage drop in V*

* I, current intensity in A*

* ρ, the resistivity coefficient of the material in Ω. mm² / m, at 20ºC*

* L, cable length in meters in one direction.*

* s, cable section in mm2 (for cables with circular section A = pi · (diam / 2) ^ 2)*

If we clear the s to calculate the section of the electrical cable we have:

**s = 2 I ρ L / ΔV** (mm2)

We already have everything, the volts that we allow to fall, the cable’s length ( **in one direction,** that’s why it is multiplied by 2 in the formula), what material the electrical cable is made of, and the intensity that will pass through it. We substitute, and it gives us the necessary section that we will have to approximate to the next available commercial section.

The most common cables have the following sections in mm2:

0.14 – 0.25 – 0.5 – 0.75 – 1 – 1.5 – 2.5 – 4 – 6 -10 – 16 – 25 – 35 – 50 – 70 – 95 – 120

*Be careful not to confuse the section (mm2) with the diameter (mm)!*

A small example. We have a 48W 24V DC LED spotlight. The electrical cable is copper wire and has a length of 3m (in one direction) and a voltage drop of 1% (ΔV = 0.24V). We have that the maximum intensity will be 48W / 24V = 2A. We substitute in the calculation formula of the section and we have: s = 2 · 2A · 0.0172 Ω · mm2 / m · 3m / 0.24 V = 0.86mm2. We look at the power cables that we can buy, and we would choose the 1mm2—that easy.

## CONCLUSION

Realize how important both the length of the power cable and the voltage are in direct current(DC). If instead of having a 24V bulb we had a 12V bulb with the same power, the intensity that would pass through the cable would be exactly double (48W / 12V = 4A) and therefore, the result would be: s = 2 · 4A · 0.0172 Ω · mm2 / m · 3m / 0.12 V = 3.44mm2, and we would have to put a 4mm2 electric cable !! All this for the same power or, in other words, for the same job.

So, in general, the higher the voltage, the lower the intensity and, therefore the less copper wire.

## BONUS

To put the icing on the cake, we can calculate how much power the cables dissipate (in the form of heat).

**P = I ^ 2 R**

*P, power in watts (W)*

* I, current in amps (A)*

* R, resistance in ohms (Ω)*

In the previous example R = 2 · ρ · L / s = 0.0172 Ω · mm2 / m · 3m / 1mm2 = 0.1Ω (we multiply by 2 considering the boundary electrical wire as well).

P = 2A ^ 2 · 0.1Ω = 0.4W as heat over the entire length of the electrical wire.

Joule’s Law. **E = 0.24 Pt** (calories)

*E, energy in calories*

* P, power dissipated in watts (W)*

* t, time in seconds*

* 0.24 converts joules to calories*

E = 0.24 · 0.4W · 60s = 5.7 cal/min, to be able to calculate the temperature that the power cable reaches is somewhat more complicated and laborious, so we should look at it another time.