Saved Bookmarks
| 1. |
A battery of emf 10 V and internal resistance 3Omegais connected to a resistor. If the current in thecircuit is 0.5 A, find (i) the resistance of the resistor, (ii) the terminal voltage of the battery. |
|
Answer» SOLUTION :Here emf`epsi`= 10 V, internal resistance r = 3`Omega` and current drawn I = .5 A (i) As `I = (epsi)/(R + r) ` HENCE `0.5 = (10)/(R+3) rArr R = 17 Omega` (ii) The internal voltage of BATTERY `V = epsi - IR = 10 - 0.5 xx 3 = 8.5 V` |
|