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
The correct answer is: B
Explanation
IR = \(\frac{2}{3}\) I(R + r)r = \(\frac{1}{2}\)R