Saved Bookmarks
| 1. |
Fig. shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5Omegais used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell. |
|
Answer» Solution :Here balancing length of potentiometer wire in opencircuit of given cell `l_1`= 76.3 cm. The balancing length in closed circuit `l_2`= 64.8 cm and external RESISTANCE JOINED with cell R = 9.5 `OMEGA` ` THEREFORE ` Internal resistance of the given cell `r = (l_1 - l_2)/(l_2) R = (76.3 - 64.8)/(64.8) xx 9.5 = 1.7 Omega` |
|