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_2` then the value of R in `Omega` is

Answer

4

Explanation

Given `J_(1)//J_(2) = 2.25`
`((2E)/(2r+R))^(2) R = J_(1) and ((2E)/(r+2R))^(2) R = J_(2)`
`:. ((r +2R)^(2))/((2r+R)^(2)) = (J_(1))/(J_(2)) = 2.25`
or `(r + 2R)/(2r+R) = 1.5 = (3)/(2)`
On solving we get
`R = 4r = 4 xx 1 = 4 Omega`