Battery V/C

what i mean is how is that the cutoff voltage of 2 v is 1.75 and the 12v/c is 10.5?

is their any formula or standard ?

Is there a diode in the circuit?

This is the safety "threshold cutoff" for all of the solid state components in the circuitry.

Basically the Transistor-Transitor-Logic (TTL) operation will become very unstable when the voltage supply gets below the threshold cutoff value and can result in "runaway" circuit operation.

Gates will "set-reset" randomly, OP amps will not function correctly and will oscillate or randomly ramp up or down, etc.

In communications the frequency will vary dramatically and in the case of a PLC or DCS system, the I/O will go "crazy".

I presume you are talking about single lead-acid battery cells (2V nominal) and 12V lead-acid batteries. Your figure of 1.75V for a cutoff is arbitrary, and some would put it at 1.8V. This value is not derived from a formula but from observation of how the battery voltage declines as charge is drawn from it, and also, more importantly, how many discharge-recharge cycles the cell will tolerate if allowed to discharge to low values each time. There are different constructions of lead-acid cells, some allowing deeper discharges than others, but the point remains. If you habitually discharge a lead-acid cell to 1.6V, you may have problems with the function of the associated circuits and you will certainly have problems with reduced battery life.

I believe you're asking why you can't discharge below the cut-off voltage.That depends upon the battery chemistry and its internal design. If you deep-discharge a battery beyond the manufacturers' limits you will shorten its life, do it often enough and you'll never even get back to 2.0 volts and reduce its ability to accept a full charge. Time to Google on "rechargeable batteries" for lots more info.

I think that you have it right.

Sadly, if its a lead acid battery (the OP has not defined the battery type), discharging to these levels will "start" to damage to the battery.

How much damage actually takes place depends upon battery construction/type, how long the battery remains discharged, temperature and and and.......

If NOT a lead acid type, and NOT LI-ON type (I am not up to speed on this battery type myself), generally speaking discharging most Battery types below 1.0 to 1.1v per cell promotes damage.....

Its difficult to be exact with so little useful infos from the OP.....

Not everything in the world has a formula.
Moreover, not every formula is derived from some other formula(s).
Some are just empirical; in this case V_cutoff = 0.875 V_nominal , which may have been estimated from actual battery performance experience.

The formula for today is Similac^®_.

I believe this is normally associated with a current draw, i.e., amp-hours available before the cutoff is reached.

Finding the battery which meets the required specifications for providing operating current before cutoff is reached (and cell damage) is part of the (engineering) trick. Taking the charge/discharge cycle into account is also important, i.e., is the battery constantly being driven to cutoff?