A cell of emf E and internal resistance r supplies currents for the same time t through external resistance R 1 = 100 Ω a n d R 2 = 40 Ω separately. If the heat developed in both cases is the same, then the internal resistance of the cell is given by

Question

A cell of emf E and internal resistance r supplies currents for the same time t through external resistance R 1 = 100 Ω a n d R 2 = 40 Ω separately. If the heat developed in both cases is the same, then the internal resistance of the cell is given by, 28 . 6 Ω, 70 Ω, 63.3 Ω, 140 Ω

MARKS App · Accepted Answer

Current drawn from the cell in resistance R 1 will be I = E / ( R 1 + r ) Therefore, the heat produced in R 1 i.e. H 1 = E 2 R 1 t R 1 + r 2 Heat produced in R 2 , H 2 = E 2 R 2 t R 2 + r 2 As per question H 1 = H 2 or E 2 R 1 t R 1 + r 2 = E 2 R 2 t R 2 + r 2 On solving we get; r = R 1 R 2 = 100 × 40 = 63.25 Ω

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