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Does thickness of wire affect the electricity bill, for same load? Thicker Wire has lower resistive losses ,does this reduce power bill? Does a longer length of the same diameter wire increase bill by the same reasoning?

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    \$\begingroup\$ No chance you will see the difference in the bill. Just be careful, thin wires tend to burn under load. \$\endgroup\$ – Gregory Kornblum May 25 '16 at 18:09
  • \$\begingroup\$ In some places, you get charged a fixed distribution fee based on the size (rating) of your main breaker. In that sense, you'd actually pay more for a "thicker wire" coming into your house. Other than that, the difference in your bill will be absolutely negligible. \$\endgroup\$ – Richard the Spacecat Jun 24 '19 at 8:40
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You are REQUIRED by code to use specific wire sizes for specific loads and in no case can you use a wire that is rated for less than 125% of a continuous load. So if you follow the law, your wire heating is minimal. If you increase the wire size to be larger than necessary, it will technically reduce the losses, but at a much lower rate of increase in benefit so there is almost never a return on that investment. The energy savings may be only a few dollars per year, whereas the cost of the larger wire for an entire house may result in hundreds of dollars in increased cost. You would not likely live long enough for the energy savings to pay for itself.

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"Power consumption drops when line resistance increases" is true when you have a resistive load, but not true when you have an active (regulating) load. And it is not interesting to do so.

Let's assume that you have an apparatus that needs 100W to work. For example your fridge that requires 100W to cool down to the set temperature level, or your computer that uses and adaptor and consumes 100W to do its work.

Suppose that your fridge or adaptor can run at 100V and also at 10V. So at 100V the current would be 1A, and at 10V it would be 10A. If the line voltage is 100V, then at 0Ohm of line resistance, the apparatus gets 100V and all delivered power is used by the apparatus. If the line resistance is changed to 9 Ohms then there would be a 90V loss in the lines at 10A and the apparatus would get 10V and consume the 100W it needs. However the loss in the lines would be 900W - 9 times the usefull power.

In practice you will not have such a big difference, but it will always work the same way.

For a constant "usefull" power, the lower the line resistance, the lower the consumption and the lower the bill.

If you accept the principle of getting less usefull power, the bill will be lower with higher line resistance, but you will also get a lower amount of light from light bulbs and the cost per useful Wh will be higher.

With (partially) inductive and capacitive loads, the computation is more complex but the line resistance will consume a lot with respect to the power actually consumed in the load. This is why there is a lower limit on what is called the "Power factor", but that is another (related) story.

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Power consumption drops when line resistance increases - it's a case of more resistance in series with the load takes less current and hence power is less. However, if you are boiling a kettle it will take longer to boil so, you maybe be taking less power but you are taking it for a longer time and so the power loss in the cable (due to \$I^2R\$) will add a bit on your bill.

Remember you are not billed on power but power x time i.e. energy.

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Thinner wire will have higher resistance and will reduce current and voltage to the load. This will reduce power consumed but a higher percentage of that will be lost in the wires.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1 (a) Low-resistance wire and (b) high resistance wire.

Figure 1 shows a 10 Ω load fed from 100 V supply. With zero wire resistance \$ I = \frac {V}{R} = \frac {100}{10} = 10~A \$. Power is given by \$ P = I^2R = 10^2 \cdot 10 = 1000~W \$.

(a) has 0.2Ω series resistance. \$ I = \frac {V}{R} = \frac {100}{10.2} = 9.8~A \$. Power in load is given by \$ P = I^2R = 9.8^2 \cdot 10 = 961~W \$. Power lost in wire is given by \$ P = I^2R = 9.8^2 \cdot 0.2 = 19.2~W \$. Overall, we've saved a little on our bill and lost 20 W to the wire.

(b) has 2Ω series resistance. \$ I = \frac {V}{R} = \frac {100}{12} = 8.33~A \$. Power in load is given by \$ P = I^2R = 8.3^2 \cdot 10 = 694~W \$. Power lost in wire is given by \$ P = I^2R = 8.3^2 \cdot 2 = 132~W \$. Overall, we've reduced our bill by 174 W (which means that the equipment won't run as intended) and lost 132 W to the wire.

Summary table
R        Load power    Loss in wire    Total (billed)
0 Ω      1000 W          0 W           1000 W
0.2 Ω     961 W         19 W            982 W
2.0 Ω     694 W        132 W            826 W

For AC (Alternative Current)?

DC or AC. Same rules apply.

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    \$\begingroup\$ That assumes resistive loads. Many loads today are constant-power loads. (PCs, TVs, anything running off of a switching supply.) Probably lighting and other resistive type loads will dominate, but not as much as they did years ago. \$\endgroup\$ – John D May 25 '16 at 18:47
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Theoretically the thinner the wire and longer it is, the higher it's resistance and the overall load seen from the outside (power plant?). Some of the energy entering your house will just dissipate as heat on the wires and will not produce any useful work, while you are still billed for it. But practically it should be unnoticeable on your bill.

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  • \$\begingroup\$ Higher resistance == higher power consumption?? \$\endgroup\$ – Wouter van Ooijen May 25 '16 at 19:01
  • \$\begingroup\$ Sorry, it's inaccurate indeed. Will try to rephrase. \$\endgroup\$ – Eugene Sh. May 25 '16 at 19:05

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