Does sending current upwards reduce the voltage? I was thinking about relativistic situation of having a vertical wire on a neutron star or high-gravity environment and then say if I send current upwards (away from the ground) will the voltage be reduced just like light is red-shifted as it moves away from gravitational environment? 
 A: You don't need to use relativity to see what will happen in to a current in a gravitational field. We assume a wire of constant cross section with length in the vertical direction , and a constant current flowing through it. This must be done by applying an electric field $\mathbf{E}$ along the length of the wire, producing the current density (according to the classical Drude model):
$$\mathbf{J} = k(q\mathbf{E} + m\mathbf{g})$$
where $\mathbf{g}$ is the local acceleration due to gravity, and $q$ and $m$ are the charge and mass of the charge carriers of the material. As $\mathbf{J}$ is constant everywhere, we get:
$$\mathbf{E} = \frac{1}{qk}\mathbf{J} - \frac{m}{q}\mathbf{g}$$
We see that the contribution due to gravity is an additional voltage of $-\frac{m}{q}\Phi$, where $\Phi$ represents the gravitational potential difference along the length of the wire.
Now, as $g = \vert \mathbf{g} \vert$ decreases with height, $E$ decreases with height for an upward current. This obviously means that the voltage drop per unit length also decreases with height.
As this result holds for Newtonian gravity, we also expect it to hold in the limiting case of weak gravity in relativity. I can't imagine any simple calculation of this kind using general relativity, but I would expect the qualitative result to be similar.
To see the parallel with gravitational redshift in this limit, we use a semi-classical model for a photon (following Feynman in his Lectures on Physics) of energy $U = h\nu$, where $\nu$ is its frequency. From the conservation of energy, after traversing a (Newtonian) gravitational potential of $\Phi$, its energy must change by $\Delta U \approx -\frac{U}{c^2}\Phi$ in the limit of a weak gravitational field, thereby giving a frequency shift of:
$$\nu' \approx \nu\left(1-\frac{\Phi}{c^2}\right)$$
Or in terms of wavelength $\lambda$
$$\lambda' \approx \lambda\left(1+\frac{\Phi}{c^2}\right)$$
$$\implies z = \frac{\Delta\lambda}{\lambda} \approx \frac{\Phi}{c^2}$$
This is in fact the weak field limit of the result obtained from the Schwarzschild metric (see, for example, here).
A: Current is a measure of charge flowing in some unit of time. The gravitational field will dilate time. The current will be reduced in the observer's point of view. $I = V/R$. Since resistance is constant, this means that voltage must have decreased in the observers point of view.
