Since I’m only used to AC track circuits, how does the DC track circuit work? I understand at one end (the feed, or, start end) +12VDC is placed on one rail. This can travel as much as 14,000 feet or more to energize the coil of relay on the far end of the circuit (what we call the relay end). Just how much voltage and current would be left after traveling so far down the rail to energize a relay coil?? And, if only one rail is used to send the + side of the 12VDC, what is the other side of the relay coil connected to (earth ground, or is it the negative side of the Dc power supply via cable)? I love learning how other signal systems work!. If anyone wants to reply in depth, feel free to e-mail me, too!
In the UK, track circuits are fed at 2VDC and the nominal realy pick up voltage is 0.5 Volts. Both rails are used so the negative side of the relay is connected to the other rail.