For ensuring dissipation of the same energy in all these resistors `(R_1, R_2, R_3)` connected as shown in the figure, their values are related as

Options

A. `R_(1)=R_(2)=R_(3)`
B. `R_(2)=R_(3) and R_(1) =4R_(2)`
C. `R_(2)=R_(3) and R_(1)=1//4R_(2)`
D. `R_(1)=R_(2)+R_(3)`

Answer

C

Explanation

As the voltage in `R_(2) and R_(3)` is same therefore, according to,
`H=(V^(2))/(R ) cdot t, " "R_(2)=R_(2)`
Also the energy in all resistance is same.
`therefore" "i^(2)R_(1)t=i_(1)^(2)R_(2)t`
Using `i_(1)=(R_(1))/(R_(2)+R_(3))i=(R_(3))/(R_(3)+R_(3))i=(1)/(2)i`
Thus `i^(2)R_(1)t=(i^(2))/(4)R_(2)t" or, "R_(1)=(R_(2))/(4).`