

InterviewSolution
Saved Bookmarks
1. |
In a potentiometer arrangement for determining the emf of cell, the balance point of the cell in open circuit is 350 cm. When a resistance of `9 Omega` is used in the external circuit of cell , the balance point shifts to 300 cm. Determine the internal resistance of the cell. |
Answer» Given `l_(1) = 350` `l_(2) = 300` `R = 9Omega` ltbrlt The internal resistance can be be calculate as . ` r= (R(l_(1) - l_(2)))/(l_(2))` ` r= 9((350-300)/(300)) = (9xx50)/(300) = 1.5Omega` |
|