Two cells of unequal emfs `epsilon_1` and `epsilon_2`, and internal resistance `r_1` and `r_2` are joined as shown. `V_A` and `V_B` are the potential at `A` and `B` respectively.

(i) One cell will continuously supply energy to the other
(ii) The potential difference across both the cells will be equal
(iii) The potential difference across one cell with be greater than its emf.
(iv) `V_A - V_B = (epsilon_1 r_1 + epsilon_2 r_2)/(r_1 + r_2)`.

Options

A. (i),(ii)
B. (ii),(iii)
C. (i),(iv)
D. all

Answer

D

Explanation

Assuming cell of emf `E_1` supplying current

`V_A - V_B = E_1 - ir_1 = E_2 + ir_2`
`i= (E_1 - E_2)/(r_1 + r_2)`
`V_A - V_B = E_1 - ir_1 = E_1 - ((E_1 - E_2)/(r_1 + r_2)) r_1`
=`(E_1(r_1 + r_2)-(E_1 r_2 - E_2 r_1))/(r_1 + r_2)`
=`(E_1 r_2 + E_2 r_1)/(r_1 + r_2)`.