**When we work with 12V DC installations it is advisable to do some technical considerations.**

Before you begin we will indicate a series of definitions that we have to take into account when analyzing the problem:

**Electrical resistance (R): **

Is called **resistance electric** to the opposition or difficulty that is a current to the travel a circuit electric closed, and that allows stop or mitigate the free flow of electrons.

The unit of resistance is the **ohm** (W or Ω): ohm is the resistance that provides a driver when it circulates an Ampere (intensity) and between their extremes, there is a difference of one Volt potential (voltage).

The electrical resistance of a conductor depends on the nature of the material, its length and its section, as well as the temperature. The resistance that offers a material specific, with long and thickness defined, is applied to formula:

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Where **L** is the length of cord in feet, **S** section or thickness of wire mm2 and constant **ρ** of resistivity of the material that is made of the cable.

**With this conclude that:** The value of a resistance is directly proportional to the length of the conductor and inversely proportional to the same section.

At greater length, greater resistance.

A shorter, less resistance.

At most section, less resistance.

Lower section, greater resistance.

**Law of Ohm**

**Ohm’s law** States that the intensity of the electrical current (I) circulating by an electrical conductor is directly proportional to the applied potential difference (V) and inversely proportional to the resistance of the same (R).

Put another way, the difference of potential (V) appearing between the ends of a given conductor is proportional to the intensity of the current (I) that circulates through the aforementioned driver.

- I = intensity in
**amp**(to) - V = potential difference in
**volts**(V) - R = resistance in
**ohm**(Ω).

**Power electric**

When is is of current continuous (CC, DC) the power electric developed in a certain instant by a device of two terminals, is the product of the difference of potential between such terminals and the intensity of current that passes through the device. For this reason the power is proportional to the current and the voltage.

Where **I** is the value instant of the intensity of current and **V** is the value instant of the voltage. If **I** is expressed in amperes and **V** in Volt, **P** will be expressed in watts (Watts).

**We started with our experiment:**

In an installation of low voltage, direct current (12V DC) and to keep the power consumption similar to work with 220V compels us to dramatically increase the current which passes to compensate for the lack of tension. As explanation to it previous us help from the law of Ohm and of the power consumed:

As we have said before, all electrical cables have a resistance to the flow of current (stated in ohms per meter) that depends on the diameter, length and the material with which they are made (usually copper). This resistance, to high current, creates some inconveniences as for example, the loss of voltage and as consequence loss of power, (that is dissipates in form of heat along them cables). We can divide our analysis into three parts:

- Increase of current
- Voltage drop
- Short circuit
**increase of current**

When have a voltage low, to get keep the power is necessary an increase of current, as see in the formula of the power:

Keeping the power (P) constant and lowering the voltage (V), current (I) increases (every time we are dividing by a number more small)

**Example handy:** Have a light bulbs of 60W that feed to 220V and to 12V, are going to see that intensity of current need in both cases

**Case 1: 60W power and voltage 220V**

I = 60/220 = 0, 2727A

**Case 2: 60W power and voltage 12V**

I = 60/12 = 5A

Can see that to keep the power of 60W with a voltage of 12V is required a current electric (I) 18,33 times top that if work to a voltage of 220V.

**This affects us?**

We have seen a greater electric current is necessary to lower tension and to withstand this current, much larger, is necessary to work with greater section cables (cables more thick). The cables more fine have greater resistance to the flow of current and great part of the power delivered by the source of power is would lose in avoid this resistance increasing the risk of warming of the cable. By Ohm’s law know that if I have a constant voltage (V) and increases the intensity of current (I) resistance (R) should be lower, and for resistance is below or lower is necessary that the section of the cable is higher.

**fall of voltage**

The problem of **voltage drop** on a cord, the potential difference of an electrical circuit, according to the Ohm’s law (V = I * R) is directly proportional to the current and the resistance of the cable. If we take the previous example we can see very clearly, we have:

- 60w bulb
- Intensity of current 5A
- Copper cable
- The cable of 1mm 2 section
- Resistivity (ρ) 0,0172 (is the value for the copper)
- Cord length: 25m (as they are two wires one of ida and another back, we have 50 meters in total)

Resistance in this case is, by applying the formula of electrical resistance we have:

**R = 0,0172 * 50/1 = 0.86 Ohm**

Thus, by applying Ohm’s law to the copper cable, V = I * R

**V = 0.86 * 5 = 4.3 volt**, which uses copper cables

We see that the initial 12V we have, 4.3 volt consumes them the cable, we still have 7.7 volts, the difference is enormous, losing 4.3 volts to 12 is 35.8% of losses. In the case of working at 220V, cable is the same, we will have the same losses, the 4.3 volts, and opposite the 220V hardly it is significant, only lose 1.95%.

**How do we solve the problem of voltage drop?**

- Lowering consumption
- Reducing the length of the cables
- Increasing the section of the conductors of the cable
- Increasing the voltage of the source to offset losses

**Example:** instead of a cable of 1 mm 2 I install a cable 2, 5 mm 2 voltage drop would be 1.72 Volt.

**3 prevent short circuits**

Working with facilities of 12V can be dangerous. Currents that travel can be very high and there is a serious risk of fire if things are not done well. The facilities must be made with care and attention, using cables thick and well isolated through gutters and boxes of distribution. Connections must be robust to using terminals and connectors of good quality. And fundamentally, always put protection fuses.

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