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InterviewSolution
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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 |
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Answer» Correct Answer - 4 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` |
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