NTA Abhyas JEE Main2020PhysicsCurrent ElectricityPractice
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
Options
- A28 . 6   Ω
- B70   Ω
- C63.3 Ω
- D140 Ω
Correct answer
C. 63.3 Ω
Step-by-step solution
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