When working with 12V DC installations it is appropriate to make some technical considerations. We explain you all the data to take into account
voltage drop,ohm,ohm,electrical power,electrical resistance,resistivity,installation
installation at 12v dc some technical aspects to take into account
When we work with 12V DC installations, it is appropriate to make some technical considerations.
Before starting, we are going to indicate a series of definitions that we have to take into account when analyzing the problem:
Electrical resistance (R):
It is called electrical resistance to the opposition or difficulty encountered by a current when traveling through a closed electrical circuit, making it possible to slow down or attenuate the free flow of electrons.
The unit of resistance is the ohm (W or Ω): and ohm is the resistance offered by a conductor when an ampere (intensity) circulates through it, and between its ends, there is a potential difference (voltage) of one volt.
The electrical resistance of a conductor depends on the nature of the material, length and section, in addition to the temperature. The resistance offered by a specific material, with defined length and thickness, is applied to the formula:
Where, L is the length of the cable in meters, S the section or thickness of the cable in mm2, and constant ρ of resistivity of the material that the cable is made of.
With this, we deduce that: The value of resistance is directly proportional to the length of the conductor and inversely proportional to its section.
The longer the length, the greater the resistance.
The shorter the length, the less resistance.
The larger the section, the less resistance.
The smaller the section, the higher the resistance.
Ohm’s law states that the intensity of the electric current (I) flowing in an electrical conductor is directly proportional to the applied potential difference (V) and inversely proportional to the resistance of the same (R).
In other words, the potential difference (V) that appears between the ends of a given conductor is proportional to the intensity of the current (I) that circulates through the said conductor.
- I = intensity in amps(A)
- V = Potential difference in volts(V)
- R = Resistance in ohms(Ω).
When it comes to direct current (DC, DC), the electrical power developed at a certain instant by a device with two terminals is the product of the potential difference between said terminals and the intensity of current that passes through the device. For this reason, power is proportional to current and voltage.
P = V*I
Where I is the instantaneous value of the current intensity, and V is the instantaneous value of the voltage. If I is expressed in amperes and V in volts, then P will be expressed in watts (watts).
We start with our experiment:
In a low-voltage installation, in direct current (12V DC) and that maintains the power consumed similar to working with 220V, it forces us to significantly increase the current that passes to compensate for the lack of voltage. As an explanation to the above, we use Ohm’s Law and the power consumed:
As we have commented before, all electric cables have a resistance to the flow of current (indicated in ohms per meter) that depends on the diameter, the length, and the material with which they are made (generally copper). This resistance, at high currents, creates some inconveniences such as loss of voltage and consequently loss of power (which is dissipated as heat along with the cables). We can divide our analysis into three parts:
- Current surge
- Tension fall
- Short circuit
1. Increase current
When we have a low voltage, in order to maintain power, an increase in current is necessary, as we see in the power formula:
I = P/V
Keeping the power (P) constant and lowering the voltage (V), the current intensity (I) increases (we are dividing each time by a smaller number)
Practical example: We have 60W bulbs that we feed at 220V and 12V; let’s see what current intensity we need in both cases
Case 1: Power 60W and voltage 220V
I = 60/220 = 0.2727A
Case 2: Power 60W and voltage 12V
I = 60/12 = 5A
We can see that to maintain the power of 60W with a voltage of 12V, an electric current (I) 18.33 times higher is necessary than if we work at a voltage of 220V.
How does this affect us?
We have seen that at a lower voltage, a greater electrical current is necessary, and to withstand this much higher current, it is necessary to work with cables with a larger section (thicker cables). The thinner cables have greater resistance to the flow of current, and much of the power delivered by the power supply would be lost in avoiding this resistance, increasing the risk of heating the cable. By ohm’s law, we know that if I have a constant voltage (V) and the current intensity (I) increases, the resistance (R) has to be lower, and for the resistance to be lower or lower, it is necessary that the cable section be older.
The problem of the voltage drop in an electrical cable, the potential difference of an electrical circuit, according to ohm’s law (V = I * R), is directly proportional to the current intensity and the resistance of the cable. If we take the previous example, we can see very; clearly, we have:
- 60w bulb
- Current intensity 5A
- Copper wire
- 1mm2 cable section
- Resistivity (ρ) 0.0172 (it is the value for copper)
- Cable length: 25m (as there are two cables, one going and the other back, we have 50 meters in total)
The resistance, in this case, is, applying the formula of electrical resistance we have:
R = 0.0172 * 50/1 = 0.86 Ohm
Thus, applying ohm’s law to the copper wire, V = I * R
V = 0.86 * 5 = 4.3 volts , which consumes the copper wire itself
We see that of the initial 12V we have, 4.3 volts are consumed by the cable, we have 7.7 volts left, the difference is enormous; losing 4.3 volts compared to 12 is a 35.8% loss. In the case of working at 220V, as the cable is the same, we will have the same losses, 4.3 volts, and compared to 220V, it is hardly significant; we only lose 1.95%.
How do we solve the problem of voltage drop?
- Lowering consumption
- Reducing the length of cables
- Increasing the section of the conductors, of the cable
- Increasing the supply voltage to compensate for losses
Example: if instead of having a 1mm2 electrical cable, I install a 2.5mm2 electrical cable, the voltage drop would be 1.72 volts.
3.Avoid a Short Circuit
Working with 12V installations can be dangerous. The currents that flow can be very high, and there is a serious risk of fire if things are not done well. Installations must be done with care and attention, using thick, well insulated cables through conduits and distribution boxes. Connections must be robust using good quality terminals and connectors. And fundamentally, always put protection fuses.