Two resistors of resistance `R_(1)` and `R_(2)` having `R_(1) gt R_(2)` are connected in parallel. For equivalent resistance R, the correct statement is

Explanation

Let `v_(0)` be the terminal velocity attained by the rod PQ (in the steady state). If `i_(1)` and `i_(2)` be the currents flowing through `R_(1)` and `R_(2)` in this state, then current flowing through the rod PQ is `i = i_(1)+ i_(2)` (see the circuit diagram) as shown in Fig.

`:.` Applying Kirchoff.s loop rule, yields.
`i_(1)R_(1) =Bv_(0)l` and `i_(2)R_(2) = Bv_(0)l`
`:. i_(1)+i_(2)=Bv_(0)l((1)/(R_(1))+(1)/(R_(2)))` .....(i)
Given that, `P_(1) = i_(1)^(2)R_(1)=(B^(2)v_(0)^(2)l^(2))/(R_(1))` .....(ii) and `P_(2) = i_(2)^(2)R_(2) = (B^(2)v_(0)^(2)l^(2))/(R_(2))` .....(iii)
Also in the steady state, the acceleration of PQ = 0.
`implies mg =B (i_(1)+i_(2))l` (or) `mg = B^(2)l^(2)v_(0)((1)/(R_(1))+(1)/(R_(2)))` .....(iv)
Multiplying both sides by `v_(0)`, we get
`mgv_(0) =B^(2)l^(2)v_(0)^(2)((1)/(R_(1))+(1)/(R_(2)))=P_(1)+P_(2)` (From Eq. (ii) and (iii)]
`:.` The terminal velocity is `v_(0) = (P_(1)+P_(2))/(mg)`
Substituting for `v_(0)` in Eq. (ii),
`P_(1)=(B^(2)l^(2))/(R_(1))((P_(1)+P_(2))/(mg)) implies R_(1)=[(Bl(P_(1)+P_(2)))/(mg)]^(2) xx (1)/(P_(1))`
Similarly from Eg. (iii) `R_(2) =[(Bl(P_(1)+P_(2)))/(mg)]^(2) xx (1)/(P_(2))`