Question

An alkaline cell is discharged at steady current of 4A for 12 hours , the average terminal voltage being 1.2V to restore it to its original state of voltage ,being 1.44V Calculate the A-H and W-H efficiency in this particular case​

Answer 1

Answer:

QUESTION:

An alkaline cell is discharged at steady current of 4A for 12 hours , the average terminal voltage being 1.2V to restore it to its original state of voltage ,being 1.44V Calculate the A-H and W-H efficiency in this particular case______________________________

ANSWER:

An alkaline cell is discharged at a steady current of 4 A for 12 hours, the average terminals voltage being 1.2V. To restore it to its original state of charge, a steady current of 3 A for 20 hours is required, the average terminal voltage being 1.44 V. Calculate the ampere-hour (Ah) efficiency and Wh efficiency in this particular case

Ah efficiency = Ah of discharge/Ah of charge = (12 x 4)/(20 x 3) = 0.8 or 80%

Wh efficiency = Ah effi. x (Av. volts on discharge)/(Av. volts on charge) = (0.8 x 1.2)/1.44 = 0.667 or 66.7%

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