A series battery of six cells each of `emf 2 V` and internal resistance `0.5 Omega` is charged by a `100 V dc` supply. What resistance should be used in the charging circuit in order to limit the charging current to `8 A`. Using this relation , obtain (a) the powwer supplied by the dc source, (b) the power dissipated as heat , and (c) the chemical energy stored in the battery in `15 min`.

Explanation

Given : number of cells , `n= 6` , emf of each cell , `E = 2 V`, internal resistance of each cell , ` r = 05 omega , charging voltage , ` V = 100 V` . Let `R` be the resistance used in the series of the circuit while charging the cells . Then current in the circuit will be
` i = (V - n E)/( nr + R)`
or `R = (V - nE)/( i) - nr = ( 100 - 6 xx 2 )/(8) - 6 xx 0.5`
` = 11 - 3 = 8 Omega`
Power supplied by `dc` source is `V xx i = 100 xx 8 = 800 W`
(b) Power dissipated as heat is
`i^(2)(R + nr) = 8^(2) ( 8 + 6 xx 0.5)`
`= 704 W`
(c) Rate at which the chemical energy is stored is
`800 - 704 = 96 W`
Therefore, chemical energy stored in `15 min` is
`96 xx 15 xx 60 = 86400 J`