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Question

An alkaline cell is discharged at a steady current of 4A for 12 hours. To restore it to its original state of charge, a steady current of 3A for 20 hours is required. Calculate Ah efficiency.

This question was previously asked in
RRB ALP 2018 CBT 2 Fitter Question Paper (21-Jan-2019) (Shift 3)
The correct answer is
80%

Calculating Alkaline Cell Ah Efficiency

This solution explains the calculation for Ampere-hour (Ah) efficiency of an alkaline cell based on given discharge and charge parameters.

Discharge Calculation

First, calculate the total Ampere-hours (Ah) delivered during the discharge phase.

  • Discharge Current (\(I_{discharge}\)): 4A
  • Discharge Time (\(t_{discharge}\)): 12 hours
  • Calculated Discharge Ah: \(Ah_{discharge} = I_{discharge} \times t_{discharge}\)
  • \(Ah_{discharge} = 4 \text{A} \times 12 \text{ h} = 48 \text{ Ah}\)

Charge Calculation

Next, calculate the total Ampere-hours (Ah) supplied during the charging phase to restore the cell.

  • Charging Current (\(I_{charge}\)): 3A
  • Charging Time (\(t_{charge}\)): 20 hours
  • Calculated Charge Ah: \(Ah_{charge} = I_{charge} \times t_{charge}\)
  • \(Ah_{charge} = 3 \text{A} \times 20 \text{ h} = 60 \text{ Ah}\)

Efficiency Calculation

Finally, calculate the Ah efficiency using the formula:

  • \(Efficiency_{Ah} = \frac{\text{Ah delivered during discharge}}{\text{Ah supplied during charge}} \times 100\%\)
  • \(Efficiency_{Ah} = \frac{48 \text{ Ah}}{60 \text{ Ah}} \times 100\%\)
  • \(Efficiency_{Ah} = 0.8 \times 100\%\)
  • \(Efficiency_{Ah} = 80\%\)

The Ah efficiency of the alkaline cell is 80%.

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