When the terminals of a cells are connected by a wire of resistance `4 Omega` , the potential difference across the cell is 1.6 V . If a wire of the same resistance is connected in parallel with the first, the potential difference becomes 1 . 33 v . The emf and internal resistance of the cell are respectively

Options

A. `1 V, 1 Omega`
B. `2 V, 1 Omega`
C. `1 V, 2 Omega`
D. `2 V, 2 Omega`

Answer

B

Explanation

`E = V + Ir`
Given, V = 1.6 V
`R = 4 Omega`
`I_(1) = (16)/(4) = 0 . 4 A`
E = 1 . 6 + 0 . 4 r . . . (i)
When a resistance of `4 Omega` is connected in parallel, the resultant resistance
`(1)/(R.) = (1)/(4) = (1)/(4) = (2)/(4)`
`R. = 2 Omega `
`I_(2) = (1.33)/(2) = 0 .66A`
So, E = 1.33 + 0 66r . . . (ii) 1.6 +0 . 4r = 1. 33 + 0 . 66 r
1.6 - 1.33 = 0 .66 r - 0 . 4 r
0 . 27 = 0 . 26 r
`r = (0.27)/(0.26) = 1 `
Substituting of r in Eq. (i) , we get
`E = 1 . 6 + 0 . 4 xx 1`
= 2 V