For reduce in Ohmic loss transmission of electric power at very far distance of electric power is done at very high voltage ?

Explanation

`rArr` To reduce Ohmic loss transmission of electric power is done at very high voltage.
`rArr` Generally power station which produce electricity are very far from homes and factories.
`rArr` Power is transmitted by cables.
`rArr` Length of cables is very large hence their resistance is also very large `[R alpha l ]` . Consider resistance of this cable be `R_(C)`.
`rArr` Consider resistarnce a device be R. and voltage applied be V and current flowing through it is I then power,
P = VI
`rArr` Power dissipated in cable be `P_(c)` then ,
`P_(c) = I^(2) R_(c) " " `....(1)
But P = VI , then I = `(P)/(V)`
`therefore P_(c) = (P^(2) R_(c))/(V^(2)) " " ` [ From equation (1) ]
`rArr` Thus , power dissipated is inversely proportional to `V^(2) ` hence to minimize this power `(P_(c))` voltage is transmitted at very high value then power in cable will be reduced.
`(V)/(I) = R_(eq) = R_(S) = ` equivalent resistance
`R_(S) = R_(1) + R_(2)`
This can be extended for .n . number of resistors with different value.
`R_(eq) = R_(S) = R_(1)) + R_(2) + .... + R_(n)`
`therefore R_(eq) = sum_(i=1)^(n) `Ri
`rArr` If .n. resistors of equal value (R) are connected in series, then equivalent resistance `R_(eq) = R_(S) = ` nR
`rArr` Thus, in series connection of resistors equivalent resistance is larger than largest value of resistance.