Transmission Line

Walter Lewin talks about this in one of his MitX lectures, but also I saw this Khan Academy India on “Transformer currents”.

Transmission lines use very high voltages over long distances to avoid losing as much energy as possible.

We use step-up Transformer (Device)s from the generator to the transmission line (increasing voltage, ex: 100V → 1000V), and step-down Transformer (Device)s from the generator to the transmission line (increasing voltage, ex: 100V → 1000V), and step-down Transformer (Device) from transmission line to the final destination.

Because over long distances, the wire having a tiny resistance, power will be lost as $P=Vi$. Power is preserved between the transformers, we have $P_i=P_o$ $V_i{i}_i=V_0{i}_0$

Since we are increasing the voltage, the current decreases. Since $P=i^{2}R$, and the current is lower, we have much less energy lost as heat as power is being transmitted across the transmission line.

You might think, wait a second, but we also have the formula for power $P = \frac{V^2}{R}$. Doesn't the power increase because we are increasing voltage? Well, think about where you are measuring the voltage. Measure it across the resistor, $V=iR$.
 
I am still a little confused, but you should find that $V = i^2R^2$.

There are practical limitations to increasing the voltage, but yes, theoretically, the higher voltage, the more perfect the transmission (no loss of energy).