Three equal resistances, each of `ROmega` are connected as shown in figure. A battery of emf 2 V and internal resistance `0.1Omega` is connected across the circuit. Calculate the value of R for which the heat generated in the circuit is maximum?

Options

A. `0.2Omega`
B. `0.3Omega`
C. `0.4Omega`
D. none of these

Answer

B

Explanation

The given network is a parallel combination of three resistances. Combined resistance `R// = R//3`
Current (I) =`(E)/(R//3+r)`
Power (P) = `((E )/(R//3+r))^(2)(R)/(3)`=`(E^(2)R//3)/([(R)/(3)-r]^(2)+(4Rr)/(3))`
For maximum power, `(R)/(3)`-r=0 or R=3r=`0.3Omega`