Three equal resistace, each of `R` ohm, are connectedd as shown in the figure. A battery of `2V` and of internal resistance `0.1` ohm is connected across the circuit. The value of `R` for which the heat generated in the circuit maximum will be

Explanation

Let `E` be the end and `r` be the internal resistance of the cell and `R` be the resistance of network of resistance shown in the circuit
Current in the circuit `I = (E )/(r +R')`
Heat produced `(H)` in the circuit `t` time seconds will be `H = I^(2)R' t = E^(2)/((r+ R')^(2))R t`
Heat produced will be maximum if `(dH)/(dR') = 0 :. (dH)/(dR') =(E^(2)r)/((r +R')^(2) - 2E^(2R's)/((r + R)^(2))) = 0`
`(E^(2)r)/((r +R)^(2)) = (2E^(2)Rt)/((r + R)^(2)) or I = (2R)/(r + R') or r = R'`
Since the time resistance each of `R Omega` are connected in parallel therefore
`(1)/(R) = (1)/(R) + (1)/(R) + (1)/(R) = (1)/(R) or R = (R)/(3)= r or R = 3r = 3 xx 0.1 = 0.3 Omega`