Saved Bookmarks
| 1. |
Twelve cells each having the same emf are connected in series and are kept in a closed box. Some of the cells are wrongly connected. This battery is connected in series with an ammeter and two cells identical with each other .The current is 3A when the cells and battery aid each other and 2A when the cells and battery oppose each other . How many cells ae wrongly connected ? |
|
Answer» Solution :Let m cells wrongly connected in the battery and `xi_("net") = (12-2m)xi` When two cells aid the battery , then current. `3=((12- 2m) xi+ 2xi)/(R) ""…..(i)` where R is the total resistance of the CIRCUIT. When two cells OPPOSE the battery , then `2=((12-2m)xi - 2xi)/(R) "".....(ii)` Solving above equations , we get m = 1 HENCE ONE cell is wrongly connected in the battery. |
|