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Trickle charging and Ah/Wh efficiency (2.2 V, 2.44 V)

Short answer

Trickle charging supplies a small continuous current, about 1% of capacity, to offset self-discharge and keep a standby battery full. For the data, Ah output is 48 and input is 60, so Ah efficiency is 80%; Wh output is 105.6 and input is 146.4, so Wh efficiency is 72.13%.

Explain trickle charging method for a battery charging system. A lead acid cell is discharged at a steady current of 4A for 12 hours, the average terminal voltage being 2.2V. To restore it to its original state of charge, steady current at 3A for 20 hours is required, the average terminal voltage being 2.44V. Calculate the ampere-hour efficiency and watt-hour efficiency in this particular case.

Trickle charging

Trickle charging keeps an idle battery fully charged by supplying a very small continuous current equal to its self-discharge loss.

  • The current is about 1% of the Ah capacity or less (for a healthy battery, as little as about 1 mA per Ah).
  • The charger (step-down transformer + rectifier + series resistance/regulator) remains permanently connected.
  • The battery feeds the load only when the mains fails; after that it is recharged normally and trickle charging resumes.
  • Used for emergency lighting, fire alarm, DG starting and switchgear batteries.
AC ─►[Transformer]─►[Rectifier]─►[R]─► + Battery −

Numerical

Given: discharge 4 A × 12 h at 2.2 V average; charge 3 A × 20 h at 2.44 V average.

Ampere-hour efficiency

Ah output = 4 × 12 = 48 Ah
Ah input  = 3 × 20 = 60 Ah
η_Ah = 48 / 60 × 100 = 80 %

Watt-hour efficiency

Wh output = 48 × 2.2  = 105.6 Wh
Wh input  = 60 × 2.44 = 146.4 Wh
η_Wh = 105.6 / 146.4 × 100 = 72.13 %

Answer: Ampere-hour efficiency = 80 %, watt-hour efficiency = 72.13 %.

✓ Answer checked by Claude Opus 5.5 (high effort)

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