When a resistance R is across a cell, the voltage across the terminals of the cell is reduced to two-third of its normal value. The internal resistance of the cell is
\(\frac{2}{1}\)R
\(\frac{1}{2}\)R
\(\frac{3}{2}\)R
5R
Correct answer is B
IR = \(\frac{2}{3}\) I(R + r)
r = \(\frac{1}{2}\)R