I can tell we are looking at this differently....In the interest of me learning something new, and to be on common ground, let's use the commonly portrayed electronic/hydraulic analogy........
Volts represents the water pressure available (potential) from a finite tank (the Battery) .....Amps represent the volume of water flowing through.....Ohms represent the pipe diameter resistance to flow rate....And Watts represents the amount of power (work) it takes to push a given volume through........
Are we agreed on the analogy so far?
IF so, lets add a valve at the far end of our pipe ...A Closed valve = open circuit as NO volume flows (NO Amps), pressure equalizes throughout the system (at the greatest Voltage potential the "tank"/Battery is currently capable of), NO resistance to flow as there is no flow (NO Ohms), and NO power being used (NO Watts).....
Still in agreement?
Now, let's open the valve (close the circuit).......Unless the water in the tank is being replaced as quickly as we are draining it out of the valve, the pressure WILL slowly drop as the water level decreases in gravity fall height, (likewise our voltage reading will reduce slightly).....Since the tank represents the battery, (the power used to pump the water, and the lesser volume of water pumped back to the tank are represented as P/I), and unless otherwise replaced, the remaining energy (stored as Voltage potential) will diminish ....We will still have whatever the remaining tank pressure entering the system is, as our new maximum pressure (Voltage potential)......
Still good?......
Here's where our points of thought differ.....You posit that upon entering/passing through the 3 Ohm resistor the VOLTAGE is reduced by 7.8 Volts.....This would leave only 4.3 Volts to enter the coil, (nothing in between to increase the voltage) and IF it drew @ 2.625 Amps, there would be NO remaining power (P=V x A and you've reduced one of those to zero) to carry through the last leg back to the battery......This is easily disproved as a simple voltage and/or amperage reading between coil output (ground side) and B- will show......
I posit that there is (nominally) 12 Volts everywhere in the circuit and the resistances change the current flow speed (amperage) and thus the Wattage used......Employing the wattage used and amperage of the components, you can calculate the net voltage potential loss at the Battery, which while small is cumulative and is why a "total-loss" system usually lasts less than an hour.........(remember, Amps are actually Ampere Hours)
I don't think I'm wrong, but IF I am, somebody please correct me.......and explain why......