Distance plays in part as the resistance of the wire increases with it’s length. I will assume being a kilometer there may be other homes also connected to this line. So the total current is what’s consumed by all. Try to picture a wire being analogy to a water pipe. Only so much water can pass through the pipe. The amount of water coming out depends on the other branches usage, and what finally reaches your measuring point.
In a analogy, Voltage is the same as water pressure, Current is the amount of water, and Resistance is the size of the pipe. What you're experiencing is called a IR Drop or Voltage Drop.
The following defined E= Voltage, I=Current and R=Resistance. The formula “IR Drop” is E = IR. (Note this is DC Theory and not truly what’s happening over the length - * see below)
By transposing this, two other formulae tools help understand the issues. I= E/R and R= E/I. In your example 220 less 185 gives you a Voltage Drop of 35 Volts by time it reaches your point of measurement. In any case this is a considerable loss of about 16%. A 10% loss is significantly large but 16% well let's say it needs to be addressed.
In any case a good solution would be increasing the wire size so the resistance over the length is reduced, reducing the resistance and resulting voltage drop.
To look at some reasonable numbers to plug in the above formulae. (The following assumes you’re the only LOAD on the supply line from the transformer.)
Most older homes in the US have 200 Amp services (I really don’t know Europe’s standards). So let's find the wire resistance R=E/I. You’re getting a Voltage drop of 35V and the load is max’ed out at 200 Amps. R = E/I = 35/200 = 0.175 Ohm of wire resistance. If you increase the wire thickness that results in 0.09 ohms in turn your IR Drop would be reduced to 1/2 the 35 volt loss or (E=IR) E= 200 * 0.09 = 18 V or about 8% loss, which is acceptable. Let look at it this way, a 16% voltage loss is large. This is saying there is insignificant supply to the load. Over loaded wires heat up, and could fail. Under supplied devices are starved of power, and will be damaged over time, including lack of performance over their shorten life cycle.
As I said there could be other issues involved. One mentioned is being down stream from other user’s LOAD(s). The company controls the supply maybe reducing their voltages due to high demand. Others major factors that come quickly to mind is poor connections such as loose or corroded connections.
If you are familiar with Tesla vs Edison regarding AC vs DC in transmission of electricity over great distance. The wire length doesn’t affect the AC as it does with DC. Please keep in mind I used the DC formulae as a SIMPLE WAY to paint a picture without the very complicated math involved in the AC theory, bring reactance into play. Still there is loss. In any case a 35 V drop should be addressed by a professional to remedy it.
The following defined E= Voltage, I=Current and R=Resistance. The formula “IR Drop” is E = IR. (Note this is DC Theory and not truly what’s happening over the length - * see below)
By transposing this, two other formulae tools help understand the issues. I= E/R and R= E/I. In your example 220 less 185 gives you a Voltage Drop of 35 Volts by time it reaches your point of measurement. In any case this is a considerable loss of about 16%. A 10% loss is significantly large but 16% well let's say it needs to be addressed.
In any case a good solution would be increasing the wire size so the resistance over the length is reduced, reducing the resistance and resulting voltage drop.
To look at some reasonable numbers to plug in the above formulae. (The following assumes you’re the only LOAD on the supply line from the transformer.)
Most older homes in the US have 200 Amp services (I really don’t know Europe’s standards). So let's find the wire resistance R=E/I. You’re getting a Voltage drop of 35V and the load is max’ed out at 200 Amps. R = E/I = 35/200 = 0.175 Ohm of wire resistance. If you increase the wire thickness that results in 0.09 ohms in turn your IR Drop would be reduced to 1/2 the 35 volt loss or (E=IR) E= 200 * 0.09 = 18 V or about 8% loss, which is acceptable. Let look at it this way, a 16% voltage loss is large. This is saying there is insignificant supply to the load. Over loaded wires heat up, and could fail. Under supplied devices are starved of power, and will be damaged over time, including lack of performance over their shorten life cycle.
As I said there could be other issues involved. One mentioned is being down stream from other user’s LOAD(s). The company controls the supply maybe reducing their voltages due to high demand. Others major factors that come quickly to mind is poor connections such as loose or corroded connections.
If you are familiar with Tesla vs Edison regarding AC vs DC in transmission of electricity over great distance. The wire length doesn’t affect the AC as it does with DC. Please keep in mind I used the DC formulae as a SIMPLE WAY to paint a picture without the very complicated math involved in the AC theory, bring reactance into play. Still there is loss. In any case a 35 V drop should be addressed by a professional to remedy it.
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