Does it take more energy to transmit electricity uphill? And if so, how much energy would have to be expended before you could possibly measure the loss (since gravity is very weak)? Let's say the load is 1000m above sea level, and the power source is at sea level?
 A: It does not, for two reasons.
First, for every electron that you push uphill through the wire, another electron flows back downhill in the return wire and any work expended in lifting the one electron is returned by the electron that is coasting downhill.
Second, even if this were not true, the gravitational force is fundamentally smaller than the electromagnetic force by a factor greater than 30 orders of magnitude. This would make any such effect far, far too small to detect.
A: I interpret the question as "Is a charged battery battery heavier than a discharge one?". A modern battery easily holds 0.4 MJ/kg, corresponding to a mass of 0.44 10$^{-11}$ kg/kg by the famous relation $E=mc^2$. So a fully charged battery weighs 4.4 trillionth more than an empty one.
A: One kilogram contains 9*10^16 Joules.
Lifting a kilogram, or 9*10^16 Joules, 1000 meters takes 10 kJ.
So it looks like 1/9000000000000 of energy is lost when transmitting it from sea level to 1000m above sea level.
Here "lost" means "converted to potential energy".
A: If you are using alternating current there is no loss since there is no net flow up.
If it is a direct current gravitation acts as if there was a voltage resisting the flow up (and enhancing it down) corresponding to $m_e g/q\approx 5.6\cdot 10^{-11}$ V/m. This is exceedingly small.
In a general relativity framework there would be a gravitational time dilation effect slowing the frequency, but AC power delivered seems to be independent of frequency.
