I understand. Don't worry. In a little briefing:
2S2P is the cell configuration of your mech, with 2 in series sections (2S) of 2 cells in parallel (2P). Resistance adds in series and lessens in parallel so, in this case:
R
batt = 2 × (R
cell / 2)
Coil, battery and overall device resistances (battery contacts, switch, etc) are all in game. Preferably keep contacts clean and batteries tight fastened.
P is power (watts), a combination of voltage times amperage. You can think of it like torque and rpm. In this case, a 2S2P pack can deliver both twice the voltage and the amperage, for 4x power output. P = V × I (volts per amps,
Electric power - Wikipedia).
By Ohms' law:
I = V / R, thus P = V × (V / R) = V² / R
V = I × R, thus P = (I × R) × I = I² × R
The coil is our circuit's “load”. All of the voltage drops in the circuit, each part/component having a share proportional to its resistance. Because of this, we should aim to minimize all other resistances besides the coil's one for efficient performance.
V = I × R = I × (R
batt + R
mod + R
coil)
The bigger is R
coil with respect to R
batt + R
mod, the more efficient the power transfer. That's why you need a good mod plus good batteries to drive a low resistance coil hard without risks.
If, for example, we estimate our R
batt to be ≈25mΩ and R
mod about ≈10mΩ, means our R
total is ≈235mΩ (R
batt + R
mod + R
coil). The coil's voltage share is equal to R
coil / R
total (200 / 235), ≈85.11% in this case. Summarizing:
R
total = R
batt + R
mod + R
coil
V
coil = I × R
coil = (V / R
total) × R
coil = V × (R
coil / R
total)
P
coil = V
coil² / R
coil
If, for example, our battery is at ≈7.7V (≈3.85V/cell with no load), estimated power at the coil:
P
coil = (7.7V × (200 / 235))² / 0.2Ω = ≈214.72W
Remember, this is just an example. It may come close to your actual figures.
Cheers