When two identical batteries of internal resistance `1Omega` each are connected in series across a resistor R, the rate of heat produced in R is `J_1`. When the same batteries are connected in parallel across R, the rate is `J_2= 2.25 J_1` then the value of R in `Omega` is

Explanation

In series `,i=(2E)/(2+R) implies J_1=i^2r=((2E)/(2+R))^2`. R in parallel `i=E/(0.5+R) implies J_2=i^2R=(E/(0.5+R))^2.R`
`J_1/J_2=2.25=(5(0.5+R)^2)/(2+R)^2 or 1.5=(2(0.5+R))/((2+R))` On solving we get `R=4 Omega`