Hi, readers hope you all are enjoying learning. Now let me take you to the basics of railways, dealing with what are the power schemes and reasons behind their modification work by exploring deep concepts which we all might have studied in our school or college.

Have you ever thought about being an electrical background? And

- How does the locomotive get power?
- Does it work on AC or DC?
- How are Indian Railways powered?
- Is there any difference between the Delhi Metro

Let me brush up on some basic concepts to keep you engaging in this. We all are aware of the basics of electricity like:

- Phase, Neutral, and earth for single phase.
- Current always follows the least resistance path.
- In Parallel connections, the voltage remains the same, and the current is affected and vice versa.

There are various types of schemes that are followed by Indian Railways:

Indian railway traction system-

- 25 kV AC conventional OHE with no return conductor
- 25 kV with return conductor
- 25 kV with booster transformer and return conductor
- 2X25 kV (Latest)

In this article, we would explore the first 2 power schemes, and rest we would explore them in our next article.

Now, in railways for getting phase, we use a pantograph which you must have observed on the top of the locomotive, a V-shaped structure, which basically acts as an instrument to obtain phase supply from the wire (contact) running above which generally has 600 A of current as per RDSO and calculations. Now, we know without neutral the circuit remains incomplete so from where we get the neutral? The neutral is obtained via rails on which the locomotive moves.

Now, you must think then you must receive some electrical sensation while crossing via rails or touching the rail.

Now to avoid these Engineers have come up with the solution of reducing the rail voltage.

So, as per the EN 50122-1(European Standard) for a short duration, the value of the voltage at rail must be maintained at 60 V for a long duration and 480 V depending upon the time like for 300 milliseconds it is 480 V you may refer to the standard for more detail. Which is approx. 1/416 times the 25 kV value now as we all know by ohm law?

**Fact: **Voltage is the shocking element, not the current.

Voltage (V) = ( I * Z )

Where,

I is the current

Z is the impedance.

Current is reduced via creating parallel paths to via the process known as ‘**Logitudenal Bonding**‘. You can refer to figure 2 for clarification and to decrease the impedance value the negative rails of adjacent tracks are bonded in certain intervals which are known as **‘Intertrack Bonding’**.

Now, let us explore the laws which are most important in the railway industry, and during our studies we might haven’t thought of it in such a manner:

They are

Ampere circuital law which states:

The line integral of the magnetic field surrounding closed-loop equals the number of times the algebraic sum of currents passing through the loop.

On simplifying the equations we can obtain the formula :

Magnetic field (B) = UI/2Πr

Where, U is the permeability

I is the current flowing

And r is the radius.

Now, using the same principle in understanding the difference between normal OHE and OHE with return conductor.

As previously explained the catenary system carries the current which is 2I and the rails carry the return current. Now, using the right-hand thumb rule to know the magnetic field direction inside or outside, you may observe that magnetic flux is dense in between the phase and neutral and canceled out on the outer ends, It implies there is the dense magnetic flux in between the contact and the return rail.

And we all remember by Faraday Law of Induction that the change in magnetic flux will result in EMF generation and which implies all those S &T equipment possess the induced EMF which might get a regular failure during bad weather or any fault.

So, we engineers decided to find the solution using the Ampere Circuital Law.

We know,

- B directly proportional to I
- B inversely proportional to r

So, to reduce the magnetic field we can opt for 2 options.

- Either we can reduce the current which means reducing the number of trains.
- We can place all the S &T equipment away from the track resulting in the least interference.

But, both of them are impossible to feasibility.

So, what if we can place the return conductor on top so that the magnetic flux would be dense in that area and there is no interference, this system is known as a 25 kV system with a return conductor. This system is more energy efficient than a conventional OHE system.

But, now S & t equipment has been advanced and is interference compliant as it might not be feasible to change all the whole power system.

And its advanced version is adopted in Delhi Metro and is known as a Return conductor with booster transformer (will explain in depth in the next article), Observe while traveling. As this article is quite long would discuss more in the next article, till then keep me querying as it helps me to explore and learn more.

I completed my BTech in electrical engineering in 2020, believe in the philosophy of ‘FIRST YOU LEARN THEN YOU EARN’. I am currently working as a GET in the ‘Railway Electrification’ project.