Under the hood
The physics of laserproduced ionized wires.
There are a number of ways to use a laser to ionize the air  ultrashort laser pulses will self focus to produce filaments of light interacting with a narrowcore ion channel, highly intense near ultraviolet beams can cause twophoton ionization of air, and tuning an ultraviolet beam to just the right energy will allow single photon ionization of air that is not absorbed too quickly to be useful. The problem is choosing your beam parameters so that you can deliver current to your target without incinerating him.
Desired electrical properties
There are a few parameters that we know. Modern electromuscular disruption devices such as the Taserbrand electroshock devices deliver 0.0019 A for 100 μs at a pulse rate of 19 per second. This tells us that the duty cycle is 0.0019. The taser uses an initial voltage spike of 50 kV to ionize a conductive path from its prongs through intervening clothing and air gaps, and then drops its voltage to 1200 V to drive the immobilizing current.
Since this is known to work and is usually safe, we will attempt to reproduce it for the RES device.
Let us examine the initial electrical contact in more detail. A pair of ionized air channels contact the target, one charged to +25,000 V and the other to 25,000 V. Electrical conduction across dry skin usually encounters a resistance of approximately 100,000 Ω, but for wet skin this can decrease to 1000 Ω, and the resistance through the body below the skin is about 500 Ω.
The voltage applied is enough to drive a current across and into the body.
However, often the beams will encounter clothing, leaving an unionized air gap between the beams and the skin. When this happens, current flow in the beams momentarily stops since it has nowhere to go.
The high voltage of the positively charged beam will draw negatively charged ions in the blood and other body fluids to it while the negatively charged beam will attract the positive ions.
This flow of ions screens the interior of the body from the electric field, so that the entire 25,000 V drop is across the short air gap.
Air will undergo cascade breakdown into an ionized spark at fields of 3,000,000 V⁄m.
Consequently, this voltage can strike an arc across a gap of about 8 mm to produce a conductive channel through clothes, air, and other insulative materials.
At this point, current resumes flowing and the voltage drop is redistributed through the beams and body according to their innate resistance.
A higher voltage will be able to arc across a proportionally greater air gap.
The voltage, and the breakdownfield of air, put a limit on the minimum separation of the ionized air channels. If the channels are closer than the voltage divided by the breakdown field, the current will arc across the channels and short them out. With a handheld weapon, it will be difficult to get the channels more than a few cm apart, which in turn limits the voltage to a few times 30,000 V.
Local electrical behavior of an air ion channel
The resistance per unit length of an ionized air channel with crosssectional area A, charge carrier density n and electron mobility μ is
^{dRa }
 =
 ^{1}
 =
 ^{1}
 .

dx
 A n μ
 λ μ

where λ is the ionization per unit length.
The electron mobility depends on the air the current travels through, for Earth air at sea level, μ ≈ 3×10^{19} m² A ⁄ V for dry air and 1×10^{19} m² A ⁄ V for humid air.
The power per unit length lost to resistive heating is
where I is the conducted current and d is the duty cycle, or fraction of time the current is running.
The energy per unit length required to ionize a section of the current path is
where ε is the energy to ionize a single molecule. For nitrogen, it is approximately ε ≈ 2.5×10^{18} J.
But we don't need to just ionize the air, we need to keep the ionization channel open for the duration of the curent pulse. Unfortunately for us, ions in air tend to recombine quickly, especially at high ionization density.
We will need to replenish the ionization channel as it decays.
In air the recombination rate dn⁄dt is
^{dn}
 = α n² = α
 ^{λ²}
 .

dt
 A

For air, α ≈ 1×10^{11} m³/s if humid or α ≈ 1×10^{12} m³/s if dry.
To keep the ion channel open we will need to supply a time average power per unit length of
^{dPi}
 = α 
 ^{dn}
  ε = α A n² ε =
 ^{λ²}
 α ε =
 ^{αε}
 .

dx
 dt
 A
 A μ² (dR⁄dx)²

We see that for a given ionization per unit length needed for good conductivity, a wide area beam is desired.
This distributes the ions far enough away that they do not encounter each other as frequently so that recombination is slower and the ionization channel lasts longer, thus reducing the power needed to ionize the channel.
Selffocused laser filaments
Selffocused laser filaments have an area of their ionized core of about A ≈ 1×10^{8} m², an ion density of n ≈ 1×10^{20} m^{3}, and a radiation power of P ≈ 5×10^{9} W (for 100 femptosecond pulses, this corresponds to E ≈ 5×10^{4} J).
An individual filament thus has a resistance of R_{a} ≈ 125×10^{6} Ω per meter of length, and maintaining the filament for 100 μs would require 500×10^{3} J.
Since most of this energy will still be propagating in the filaments when it reaches the target, this will result in serious burns.
Multiple channels could decrease the resistance, but would increase the delivered power and cause worse burns.
The problem here is that all filaments are the same, with too small of a diameter and too much energy for typical engagement ranges.
Twophoton ionization
An alternate approach is twophoton ionization using a rapid series of short ultraviolet flashes during the pulse. This allows us to control the beam area and ionization density.
At any point along the beam, the rate of decrease of the beam intensity is
which has the solution for an initial intensity I_{0} at position x = 0 in a beam of constant area
I(x) =
 ^{ I0 }
 .

1 + k I_{0} x

such that
^{dI}
 (x) =
 ^{dI⁄dx(x=0)}

dx
 (1 + k I_{0} x)²

By altering the energy, duration, and area of the beam we can control the ionization density.
The instantaneous beam power is equal to P' = I A, so that
The ionizing power per unit length is equal to the instantaneous beam power derivative times the fraction of time the beam is on during the pulse.
It is, therefore, proportional to the intensity derivative, and the total power is proportional to the intensity
P(x) =
 ^{ P(x=0) }
 .

1 + k I_{0} x

For a beam of constant width, the beam power lost to ionization is proportional to the change in intensity
^{dPi}
 (x) =
 ^{dPi⁄dx(x=0)}
 .

dx
 (1 + k I_{0} x)²

We can use the relation between resistance per unit length and ionization power per unit length to find
^{dRa}
 (x) = [
 ^{dRa}
 (x=0)] (1 + k I_{0} x).

dx
 dx

This has the solution
R_{a}(x) = [
 ^{dRa}
 (x=0)] (x + ½ k I_{0} x²).

dx

We start by choosing the energy E_{0} radiated into the beam of each pulse, the pulse duration δt, the distance to the target x', the current J, the beam diameter d, and the fraction f of the original intensity incident on the target.
We then know
and can solve for the necessary product k I_{0}
k I_{0} =
 ^{1⁄f  1}
 .

x'

We then know that
^{dI}
 (x=0) = I_{0}
 ^{1⁄f  1}
 ,

dx
 x'

^{dPi}
 (x=0) = P(x=0)
 ^{1⁄f  1}
 ,

dx
 x'

^{dRa}
 (x=0) = √(
 ^{ α ε }
 ),

dx
 A μ² ^{dPi}⁄_{dx}(x=0)

for beam energy E, and
R_{a}(x') =
½ x' (1 + ^{1}⁄_{f})
[
 ^{dRa}
 (x=0)].

dx

If the target is at x', he will be subject to the power f E_{0} spread out over an area A.
Because you need two beams  one for the current to go out and another for the current to return, the resistance, radiated power, and radiated energy are double the single beam values given above. For the resistance of a human body denoted R_{h} ≈ 100,000 Ω, we have a total resistance of
a dissipated resistive power of
a total energy draw per pulse of
E_{t} = 2 E_{0} + P_{r} δt,

a discharge voltage of
and the target will be subject to heating of 2 f E_{0} from the beams and R_{h} J² δt from the current.
Example parameters
Take
E_{0} = 0.1 J

δt = 100 μs

J = 0.0019 A

f = ½

x' = 10 m

d = 3 cm.

Then, we proceed to find
^{dPi}
 (x=0) = 100 W⁄m,

dx

^{dRa}
 (x=0) = 19000 Ω⁄m,

dx

R_{a}(x') = 285,000 Ω,

R_{t}(x') = 670,000 Ω,

U = 1270 V.

At 20 pulses per second, and 2 beams per pulse, the target will absorb a total of 2 W, with the gun using 4 W on creating the beam and 0.0004 W on resistive heating.
4 W spread out over two 3 cm diameter spots will not cause burns, and at this wavelength in the ultraviolet it may cause localized sunburns but can't get through the lens and cornea to cause blindness.
Let us keep the parameters the same except for the distance, which we increase to 100 m to explore longer range behavior. We alter the discharge time or energy for each beam flash that keeps the ion channel open during the pulse to keep f = ½.
We now have
R_{a}(x') = 9,000,000 Ω,

R_{t}(x') = 18,000,000 Ω,

U = 34,000 V.

The target is exposed to the same amount of beam and resistive energy.
The peak resistive power dissipation increases to 65 W, but due to the short duration of the pulses is only 0.13 J.
This is approaching our maximum assumed discharge voltage of 50,000 V, so 100 m is likely approaching the maximum range of this device.
Range will decrease in humid or wet conditions, since electrons recombine with ions faster in humid air, leading to fewer charge carriers and a higher resistance.
At longer ranges, the current will decrease in accordance with Ohm's law, with the result that there may be insufficient current to cause the target's muscles to seize up.
Note that the resistance of the ion channels to the target do not matter for the purpose of establishing the arc across the last bit of air gap caused by the beams stopping at clothing  as long as the air gap prevents current from flowing all the voltage drop occurs across the air gaps.
Burn Safety
Using the above figures of 0.1 J per beam per pulse spread out over a 3 cm diameter circle, we find that each pulse delivers a fluence of 0.014 J ⁄cm².
This is ultraviolet light, which can cause sunburn, so let's look at how many pulses are needed to start a sunburn.
From http://www.pacificu.edu/optometry/ce/courses/15719/uvradiationpg2.cfm we find that one UV index unit is 25 mW⁄cm², and at UV index 7 the fairest skin person will take at least 10 minutes to develop a sunburn.
From this we conclude that a total fluence of a bit more than 100 J ⁄cm² is needed to initiate a sunburn; for dark skinned people it can be more than ten times this value.
Comparing this to the fluence of a single pulse, a victim would need 7500 pulses to the same spot to start a sunburn.
There is also the possibility of thermal burns.
From http://www.fas.org/nuke/guide/usa/doctrine/dod/fm89/1ch3.htm#s3 we see that it typically takes between 20 and 350 J ⁄cm² of radiant fluence to ignite fabrics.
From http://nuclearweaponarchive.org/Nwfaq/Nfaq5.html, we find that fluences of 10 to 20 J ⁄cm² will cause first degree burns, 20 to 35 J ⁄cm² will cause second degree burns, and 35 to 50 J ⁄cm² is the threshold for third degree burns.
Thus, somewhere over 700 pulses would need to be delivered to the same location to risk fires or thermal burns.