1995 Track circuit calculations

[Archie]

Hello there, I am trying to get my head around the 'track circuit calculations' and in particular I have been looking at 'TC calc 1995 Q1.' I just wanted to know how the ballast resistance of 3.75 Ohms was achieved depicted on the diagram of the simplified circuit, it does show an equation but I can't quite make it out if someone could shed some light on this it would be much appreciated.

Many thanks
Archie

[Peter]

Hello there, I am trying to get my head around the 'track circuit calculations' and in particular I have been looking at 'TC calc 1995 Q1.' I just wanted to know how the ballast resistance of 3.75 Ohms was achieved depicted on the diagram of the simplified circuit, it does show an equation but I can't quite make it out if someone could shed some light on this it would be much appreciated.

Many thanks
Archie

I think you are referring to the answer given in this attachment. If so, all they have done to get the 3.75 ohms is to have combined the two parallel resistances of the resistance due to ballast leakage (5 ohms calculated from the given ballast resistance and the length of the track) and the relay resistance (given as 15 ohms). The basic electrical formula for resistors in parallel is 1/R1 +1/R2 = 1/R. In the attachment, they have written something like 5 // 15 = 3.75 ohms which is short hand for 5 ohms in parallel with 15 ohms.

You should note that in the simplified circuit, this does not represent the "ballast resistance" - it is merely "some resistance" in the circuit model for the given state of the TC to allow the required value to be calculated.

[Archie]

Hello there, I am trying to get my head around the 'track circuit calculations' and in particular I have been looking at 'TC calc 1995 Q1.' I just wanted to know how the ballast resistance of 3.75 Ohms was achieved depicted on the diagram of the simplified circuit, it does show an equation but I can't quite make it out if someone could shed some light on this it would be much appreciated.

Many thanks
Archie

I think you are referring to the answer given in this attachment. If so, all they have done to get the 3.75 ohms is to have combined the two parallel resistances of the resistance due to ballast leakage (5 ohms calculated from the given ballast resistance and the length of the track) and the relay resistance (given as 15 ohms). The basic electrical formula for resistors in parallel is 1/R1 +1/R2 = 1/R. In the attachment, they have written something like 5 // 15 = 3.75 ohms which is short hand for 5 ohms in parallel with 15 ohms.

I have just looked over my old 'electrical principle' notes, and realised I was being slightly thick, I just never had seen shorthand like that before due to my lack of experience but achieved 3.75ohms with 15/4.

You should note that in the simplified circuit, this does not represent the "ballast resistance" - it is merely "some resistance" in the circuit model for the given state of the TC to allow the required value to be calculated.