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Recently my brother has decided to change conductors in the home's electrical circuit.

We had aluminium conductors and now we are going to replace it with copper conductors, but my friend told me that we'll pay more for electricity with copper conductors.

I can't understand whether is it true and, if so, why.

According to my opinion, the less energy loss is (due to heat dissipation), the more effective (cheaper) the circuit is; energy loss is proportional to material's resistivity.

My friend told me that with the resistivity loss i get increased electrical current and higher costs.

Please explain, will I get increased electrical current and higher price and why.

Thank you!

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  • \$\begingroup\$ Where is your energy meter placed with respect to the new cabling? Did you change wires also upstream from the meter? \$\endgroup\$ Commented Nov 18, 2017 at 8:57
  • \$\begingroup\$ We live in apartments, so I guess energy meter is placed near the door; we are going to totally replace wires inside apartments. \$\endgroup\$
    – metacube
    Commented Nov 18, 2017 at 9:03

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An extreme simplification of your situation is represented in the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

where \$V_s\$ represents the voltage delivered by the power company, \$R_w\$ the wiring resistance and \$R_{load}\$ your home appliances.

If the energy meter is placed before both \$R_w\$ and \$R_{load}\$, it will measure the power consumed by both resistor:

$$ P = \frac{V_s^2}{R_w+R_{load}} $$

If you lower \$R_w\$ the power \$P\$ will be higher and so will the recorded energy consumption by your household (but see below).

What will change in your favor is the efficiency of your electricity bill: more of the consumed power will go into powering your appliances.

In other words, there will be less drop on \$R_w\$ hence \$R_{load}\$ will experience an higher voltage across it. This means that, for example, traditional light bulbs will shine a little brighter, giving out more lumens, i.e. more light. Other appliances will give a little better performance too: an electric heater will warm up to the set temperature in less time, so it will turn off earlier (thus compensating in part your higher power consumption from the grid).

Of course the load of the electrical system is not necessarily purely resistive. Electronic loads with no PFC (power factor correction) will behave differently than a simple resistor. Anyway, if they get a slight more voltage across them, chances are that the current they will draw will be lower (sort of constant-power load), thus again compensating for your "theoretical" higher power draw from the grid.

EDIT (to integrate a relevant comment by Neil_UK)

As Neil has said in its comment, it is unlikely that the difference in your bill will be much different due solely by the different resistivity of the wires.

I'll add that most probably, the best rationale behind converting aluminium wiring to copper wiring is reliability. The wiring system will heat-up less, hence its insulation will be less stressed. Moreover copper is more flexible than aluminium and will endure mechanical stress (e.g., from thermal expansion) much more nicely. Even from a corrosion standpoint copper is better, because its oxide is mildly conductive whereas aluminium oxide is quite a good insulator.

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  • \$\begingroup\$ Thank you very much for explanation! Am I properly undestand that wire heat loss will decrease, appliances will have better performance and overall recorded energy consumption will be increased? \$\endgroup\$
    – metacube
    Commented Nov 18, 2017 at 9:29
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    \$\begingroup\$ While your bill would increase, you would not notice. It would be possible by measurements of the wiring to calculate whether it would rise by 0.0001%, or 0.1%. In either case you really wouldn't notice the difference. Leaving your TV on for an hour once during a 3 month billing period would swamp any difference due to the wiring. \$\endgroup\$
    – Neil_UK
    Commented Nov 18, 2017 at 10:03

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