Two ideal batteries of emf `V_1 and V_2` and three resistances `R_1, R_2 and R_3` are connected as shown in the figure. The current in resistance `R_2` would be zero if

Options

A. `V_1=V_2 and R_1= R_2=R_3`
B. `V_1=V_2 and R_1=2R_2=R_3`
C. `V_1=2V_2 and 2R_1=2R_2=R_3`
D. `2V_1=V_2 and 2R_1=R_2=R_3`

Explanation

From loop 1 `R_2+V_1=0`
`implies iR_1+i_1R_2=V_1`……(i)
From loop 2
`i_1R_2-(i-i_1)R_3=-V_2`
`i_1(R_2+R_3)-iR_3=V_2`………(ii)
Solving equation (i) and (ii)
`i_1(R_2+R_3)-iR_3=-V_2 times R_1`
`i_1R_2+iR_1=V_1 times R_3`
`i_1(R_1R_2+R_1R_3+R_2R_3`
We get `i_1=(V_1R_3-V_2R_1)/(R_1+R_2+R_1R_3+R_2R_3`,
If `i_1` is zero
`V_1R_3=V_2R_1`
It is possible in followin options
(a) `V_1=V_2 implies R_1=R_3` (b) `V_1=V_2 implies R_1=R_3`