A battery of emf E and internal resistance when connected across an external resistance of 12 ohm produces a current of 0.5 A. When connected across a resistance of 25 ohm, it produces a current of 0.25 A. Determine the (i) emf and (ii) internal resistance of the cell.

Explanation

`I = (E )/(R_(1) + r)`
`0.5 = (E )/(12 +r) 6 + (r )/(2) = E`,
`2E-r = 12` …(A)
Also `I = (E )/(R_(2) + r)`
`0.25= (E )/(25+r)`
`4E-r = 25` …(B)
Subtracting (B) from (A), we get (A) and (B) get the value of E or r
`4E-r -(2E-r) = 25-12`
`rArr E= (13)/(2)V`
Substituting the value of E in (A), we get
`r=1 Omega`