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
0.05\(\Omega\)
1.00\(\Omega\)
1.50\(\Omega\)
2.00\(\Omega\)
2.50\(\Omega\)
Correct answer is D
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\(\Omega\)