The ends of a resistance are connected to 19 cells in series, each of internal resistance 0.1 ohm.The current is found to be 2A. The number of cells is reduced to 15 and an extra resistance of 9.5 ohm is connected in series to the given resistance. The current becomes half.Find the given resistance and the emf of each cell.

Asked by SUHAS MELMURI | 17th Feb, 2014, 01:10: PM

Expert Answer:

Let ? be the emf for each cell and r is given resistance.

For 19 cell connected in series,

Total emf V = 19 ?

I = 2 A

Internal resistance for each cell = 0.1 Ω

Total internal resistance for 19 cells = 0.1 x 19 = 1.9 Ω

V = IR

19 ? = 2 (1.9 + r)          ………………………  (1)

Now the number of cells is reduced to 15 and an extra resistance of 9.5 ohm is connected in series to the given resistance and the current I’ becomes half.

V’ = I’ R’

15 ? = 1 x [(0.1 x 15 = 1.5) + 9.5 + r]

15 ? = 11 + r        ………………………………. (2)

Solving eq(1) and (2) we get,

? = emf for each cell = 1.65 V

 r = 13.77 Ω

Answered by  | 17th Feb, 2014, 06:25: PM

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