In the circuit `E,F,G,H`are cell of `emf` `2,1,3`and `1 Omega` respectively .The potential difference across the terminal of each of the cell `G` and `H` are

Options

A. `1.46V, 1.61V`
B. `1.61V, 1.46V`
C. `2.52V, 1.61V`
D. `1.61V, 2.25V`

Answer

b

Explanation

The distribution of current in the network following Kirchhoff's first rule has According to Kirchhoff's second rule in the closed circuit `DCBD`
`3 - 1= 3 I+ 1 I+ 2 I_(1) or 2 = 4 I + 2 I_(2)`
or `1 = 2I+I_(1)`….(i)
In the closed circuit `DBAD`
`2 - 1 = 2 I_(1) + 2(I_(1)+I_(2)) + 1(I - I_(1))`
or `1 = 3 I- 5I_(1)`....(ii)

Multiplying (i) by `5` and adding with (ii) we get
`5+1=10 I or I = 6//3A`
Terminal pot. diff. across `G`
(cell is appliying current)
`= epsilon = Ir = 3 - (6)/(13) xx 3= 1.61V`
Terminal pot. diff. across `H`
(cell is charged)
`= epsilon' = Ir'= 1+ (6)/(13) xx 1= 1.46V`