Saved Bookmarks
| 1. |
Whena drya cell is connectedacrossa bulband the circuitis switched on , then thebrightness of bulbgoes ondecreasing . Explain. |
Answer» Solution : If theemf ofcell is `epsi`and its internal resistanceis r, then currentthroughthe bulb of RESISTANCER is `l= (epsi)/(R + r)` . EMF of thecell `epsi= V_(+) + V_(-)` is thedifferenceof chemicalpotentialsof electrodes,whichis constantfor a givecell ANDIS independent of timeof use . Internalresistanceof the cellis inverselyproportionalto charge density). Withpassageof timeof use, iondensityin theelectrolyte (as resistivelyis inverselyproportionalto charge density ). With passageof timeof use ,ion densitygoes ondecreasingso internalresistancer of cellgoes onincreasing . THUS,the currentl through the bulbgoes ondecreasing . |
|