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Since Power = V*I increasing voltage reduces current and so it is better to have a higher voltage. My question is, since voltage is potential difference i.e. joules per metre, increasing the voltage also increases the heat lost so why is this beneficial?

In addition to this, since we are trying to make the transfer of electricity more efficient and this means moving more electricity with less heat loss. Power is the heat loss per second and since we're dealing with a fixed power wouldnt we want to increase current rather than voltage so that we move more electricity in less time?

I know that my reasoning is wrong somewhere so can someone clarify why this is wrong and provide background information about voltage/current etc. as well as any relevant maths to back it up. Thanks in advance

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  • $\begingroup$ Careful with your units: voltage is measured in $[Volt]=[J/C]$. A potential difference only performs work when it moves charge, hence $P=UI$, i.e. both a potential difference and moving charges (i.e. a current) are needed. It is the current which causes a voltage difference along a conductor (rather than between two conductors), which leads to resistive aka "$I^2R$" losses. That's why we are trying to decrease the current when we are transmitting power over long distances. $\endgroup$ – CuriousOne May 21 '16 at 20:44
  • $\begingroup$ Does this answer your question? physics.stackexchange.com/questions/248229/… $\endgroup$ – Farcher May 22 '16 at 16:46
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Power does indeed equal V*I, but the "more correct" form of the equation should be Power = delta-V * I. The power used by a circuit component is related to the voltage drop across the component, NOT the absolute voltage that the component is experiencing.

Second point: the transmission of electrical power is done in a way that minimizes losses in the transmission lines, because the customer cannot be billed for the energy that heated the electrical lines before the power got to his/her house. Since the transmission lines have a fixed resistance for the length of line that it takes to get electricity to an individual customer, and since the power involved in heating those lines is equal to I^2 * R, the power transmission strategy involves minimizing I, which means maximizing V by transforming the power to very high voltage, transmitting the power to the customer, then transforming the power to low voltage at the customer's location.

Another way to look at this problem: According to Ohm's law, V = IR, but again, this is more correctly expressed as delta-V = IR. The resistance of the transmission line is fixed, so the voltage drop and the power consumption are both related to the current flow. Minimizing current in the transmission lines minimizes power loss before the power gets to the customer.

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  • $\begingroup$ what exactly is the power station transporting to houses, current, voltage or something different? $\endgroup$ – KingJ May 22 '16 at 1:28
  • $\begingroup$ The power station is transmitting power, which is current times voltage drop across all of the circuit elements in the residence. $\endgroup$ – David White May 22 '16 at 2:07

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