Skip to main content

Trickle charging and Ah/Wh efficiency (4 A, 12 h, 1.2 V)

Short answer

Trickle charging keeps an idle battery fully charged with a small current matching its self-discharge. For the lead-acid cell, ampere-hour efficiency is 80% (48 Ah out / 60 Ah in) and watt-hour efficiency is 66.67% (57.6 Wh out / 86.4 Wh in).

Explain trickle charging method for battery charging system. A lead acid cell is discharged at a steady current of 4 A for 12 hours, the average terminal voltage being 1.2 volt. To restore it to its original state of charge, a steady current at 3 A for 20 hours is required the average terminal voltage being 1.44 volt. Calculate the ampere-hour efficiency and watt-hour efficiency in this particular case.

Trickle charging

Trickle charging means connecting an idle, fully charged battery permanently to a charger that supplies a very small current, equal to its self-discharge (local action) loss, so that the battery remains fully charged and ready.

  • The trickle current is small, about 1% of the Ah capacity or less (for a healthy battery, as little as about 1 mA per Ah).
  • Normally the battery does not feed the load; it is called upon only when the mains fails.
  • Applications: emergency lights, fire alarm panels, DG set starting batteries, telephone exchanges, switchgear control batteries.
  • The charger is a small transformer–rectifier with a series resistance or regulator to set the current.
  • Too high a current causes gassing and water loss; too low a current lets the battery sulphate. The specific gravity should be checked periodically.
 230 V AC ─►[Step-down]─►[Rectifier]─►[R]─┐
                                          + │
                                      [Battery]
                                          − │
 ─────────────────────────────────────────┘

Numerical

Given: discharge 4 A for 12 h, average voltage 1.2 V; charge 3 A for 20 h, average voltage 1.44 V.

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 × 1.2  = 57.6 Wh
Wh input  = 60 × 1.44 = 86.4 Wh
η_Wh = 57.6 / 86.4 × 100 = 66.67 %

(Check: η_Wh = η_Ah × 1.2/1.44 = 0.8 × 0.8333 = 0.6667.)

Watt-hour efficiency is lower than Ah efficiency because the cell is charged at a higher voltage than it gives on discharge.

Answer: η_Ah = 80 %, η_Wh = 66.67 %.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Industrial Power Distribution and Illumination chapter 9 questions →