Find the emf (`V`) and internal resistance `(R)` of a single battery which is equivalent toa parallel combination of two batteries of emf `V_1` and `V_2` and internal resistances `r_1` and `r_2` respectively, with polrities as shown in figure

Answer

A, B, C

Explanation

a. Equivalent emf (`V`) of the battery
`PD` across the terminals of the battery is equal to its emf when current drawn from the battery is zero. In the given circuit,

Current in the circuit
`i=("Net emf")/("Total resistance")=(V_1+V_2)/(r_1+r_2)`
Therefore, potential difference between `A` and `B` would be
`V_A-V_B=V_1-ir_1`
`:. V_A-V_B=V_1-((V_1+V_2)/(r_1+r_2))r_1=(V_1r_1=V_2r_1)/(r_1+r_2)`
so the equivalent emf of the battery is
`V=(V_1r_2-V_2r_1)/(r_1+r_2)`
Note that if `V_1r_2=V_2r_1:V=0`If `V_1r_2gtV_2r_1:V_A-V_B`= Positive i.e. `A` side of the equivalent battery willl become the positive terminal and vice versa.
b. Internal resistance (r) of the battery
`r_1` and `r_2` are in parallel. Therefore , the internal resistance `r` will be given by
`1//r=1//r_1+1//r_2`
or `r=(r_1r_2)/(r_1+r_2)`