A 6 V cell with `0.5Omega` internal resistance , a 10 v cell with `1Omega` internal resistance and a `12Omega` external resistance are connected in parallel . The current (in amper) through the 10 V cell is

Options

A. `(0.60)`
B. 2.27
C. 2.87
D. 5.14

Answer

C

Explanation

`E_(eq) = (12-2m)E`
The current through the circuit , when batteries aid each other
`3= (12-2m)E+2E)/(R )`
When they oppose each other
`2= ((12-2m)E-2E)/(R )`
Dividing (1) with(2) . We get ,
`[((12-2m)E+2E)/(R]/[((12-2m)E-2E)/(R )]= (3/2)`
`Rightarrow (14E-2mE)/(R)/(10E-2mE)/R = (3/2)`
`14-2m/10-2m = 3/2`
`2(14-2m)= (10-2m)3`
`28-4m = 3--6m`
`Rightarrow 2m = 2`
`therefore m=1`