Three ammeters `A,B`, and `C` of resistances `R_A,R_b` and `R_C` respectively are joined as shown. When some potential difference is appllied across the terminals `T_1` and `T_2` their readings are `I_A,I_B and I_C` respectively Then,

Options

A. `I_(A) = I_(B)`
B. `I_(A)R_(A)+ I_(B)R_(B)= I_(C)R_(C)`
C. `I_(A)/I_(C) = R_(C)/R_(A)`
D. `I_(B)/I_(C) = R_C/(R_A+R_B)`

Answer

A, B, D

Explanation

a., b., d.
As A and B are connected in series, hence IA = IB. The potential
difference acorss both the branches will be the same.

`I_(A)R_(B) + I_(B)R_(B) = I_(C)R_(C) or I_(B)/I_(C) = (R_C)/(R_A + R_B)` .