A cell gives a current of 0.15A through a resistance of 8(Omega) and 0.3A. When…

A cell gives a current of 0.15A through a resistance of 8(Omega) and 0.3A. When the resistance is changed to 3(Omega) the internal resistance of the cell is

  • A.
    0.05(Omega)
  • B.
    1.00(Omega)
  • C.
    1.50(Omega)
  • D.
    2.00(Omega)
  • E.
    2.50(Omega)
Correct Answer: Option D
Explanation

The E.M.F of a circuit is given by 

(E=I(R+r))

Hence, E = 0.15(8+r)  = 0.3(3+r)

1.2 + 0.15r = 0.9 + 0.3r

1.2 – 0.9 = 0.3r – 0.15r

0.3 = 0.15r

r = (frac{0.3}{0.15})

r = 2(Omega)

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