In the circuit shown in figure a `12 V` battery with unknown internal resistance `r` is connected to another battery with unknown emf E and internal resistance `1Omega ` and to a resistance of `3Omega` carrying a current of `2A`. The current through the rechargeable battery is `1 A` in the direction shown. Figure the unknown current `i` internal resistance r and the emf E.

Answer

A, B, C

Explanation

Applying loop law equation in upper loop, we have
`E+12-ir-1=0`……….i
Applying loop law equation in lower, we have where
`i=1+2=3A`
`E+t6-1=0`……………ii
Solving these two equations we get
`E=-5V` and `r=2Omega`