Skip to main content

Trickle charging and Ah/Wh efficiency (30 A, 6 h)

Short answer

Trickle charging maintains a standby battery at full charge by supplying a small continuous current equal to its self-discharge loss. For the given cell, ampere-hour efficiency is 90% (180 Ah discharged / 200 Ah charged) and watt-hour efficiency is 72% (360 Wh discharged / 500 Wh charged).

Explain the trickle charging method for a battery charging system. Calculate the (i) ampere-hour efficiency and (ii) watt-hour efficiency of a secondary cell which is discharged at a uniform rate of 30 A for 6 hr. at an average terminal voltage of 2 V. It is then charged at the uniform rate of 40 A for 5 hr. to restore it to its original condition. The terminal voltage during charging is 2.5 V.

Trickle charging

Trickle charging is the method of giving a battery that is not in use a small, continuous charging current just large enough to make up for its self-discharge (internal losses), so it always stays fully charged.

  • The current is very small, about 1% of the Ah rating or less (e.g. about 1 A or less for a 100 Ah battery); a healthy lead-acid battery may need as little as about 1 mA per Ah.
  • The charger is permanently connected; the battery does not supply the load in normal times.
  • Used for standby batteries: emergency lighting, fire alarms, engine-starting batteries of DG sets and switchgear tripping batteries.
  • Over-trickling causes gassing and water loss, so the current must be adjusted.
 AC ─►[Transformer+Rectifier]─►[R]─► + Battery −
                  (small current)

Numerical

Given: discharge 30 A for 6 h at 2 V average; charge 40 A for 5 h at 2.5 V average.

(i) Ampere-hour efficiency

Ah on discharge = 30 × 6 = 180 Ah
Ah on charge    = 40 × 5 = 200 Ah
η_Ah = 180 / 200 × 100 = 90 %

(ii) Watt-hour efficiency

Wh on discharge = 180 × 2   = 360 Wh
Wh on charge    = 200 × 2.5 = 500 Wh
η_Wh = 360 / 500 × 100 = 72 %

(Check: η_Wh = η_Ah × V_dis/V_ch = 0.9 × 2/2.5 = 0.72.)

Answer: Ampere-hour efficiency = 90 %, watt-hour efficiency = 72 %.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Industrial Power Distribution and Illumination chapter 9 questions →