A rechargeable battery has a constant emf of 13.2 V and an internalresistance of 0.92 . It ischarged by a 14.7-V power supply for a time interval of1.68 h. After charging, the batteryreturns to its original state as it delivers a constant current toa load resistor over 6.01 h. Find theefficiency of the battery as an energy storage device. (Theefficiency here is defined as the energy delivered to the loadduring discharge divided by the energy delivered by the 14.7-Vpower supply during the charging process.)
Explanation / Answer
Energy supplied to the battery during charging Ein = VinIintcharge Iin = V/Rin Iin = V/Rin Ein= Pin t = Vin(V/Rin ) tcharge Ein= 13.2 ((14.7 - 13.2)/0.92)1.68 Ein= 36.2 W-h Ein= 13.2 ((14.7 - 13.2)/0.92)1.68 Ein= 36.2 W-h Energy wasted during charging by the battery Ewbis Ewb= Iin2 Rintcharge ; since Iin=(V/Rin )= ((14.7 - 13.2)/0.92 )=1.63 A Ewb= (1.63)2 0.92 x1.68 Ewb= 4.11 W-h The the available energy Eav is Eav = Ein - Ewb =36.2 - 4.11 Eav = 32.1 W-h Energy delivered to the load EL is thedifference between available energy Eavand energy requred to overcome the internal resistanceRin during time t load is EL = Eav -Eint ButEav = [V2/(Rin +RL )] t load The RL is notgiven Then RL is RL= V2t load /Eav - Rin EL = [I2/RL ]tload and I= V/(Rin +RL ) EL = [V2/[(Rin +RL )2 RL ]tload where V= 13.2 V Efiiciency eff is eff = (EL/Ein )100% Just plug and play Please let me know if you have any questions.