3.15 (a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. What are the current drawn from the supply and its terminal voltage?

(b) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Ω. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?

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    Answered by

    Payal Gupta | Contributor-Level 10

    5 months ago

    3.15 (a) Number of secondary cells, n = 6

    emf of each secondary cell = 2 V

    Internal resistance of each secondary cell, r = 0.015 Ω

    Resistance of the resistor, R = 8.5 Ω

    Current drawn from the supply, I is given as I = TotalvoltageR+totalinternalresistance

    6×28.5+6×0.015 = 1.396 A

    (b) Terminal voltage = I ×R = 1.396 ×8.5 = 11.87 V

    After long use emf = 1.9 V

    Internal resistance of the cell, r = 380 Ω

    Maximum current can be drawn = 1.9380 = 5 ×10-3 A

    No, the cell cannot drive the starter motor of the car as it requires high current.

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