Battery Runtime Calculation
How long will battery power the system without alternator?
t (hours) = C_battery (Ah) × V_battery / P_system
Adjusted for Peukert's law (high discharge rate reduces capacity):
C_actual = C_rated × (I_rated / I_actual)^k
Where k = Peukert exponent (1.1–1.3 for lead-acid, ~1.0 for lithium)
Worked example:
System: 2000W RMS, 70% Class D efficiency → 2857W draw Battery: 100Ah AGM at 12V Average music usage: 30% of peak → 857W average → 71.4A average
t = 100 / 71.4 = 1.4 hours (without Peukert)
Peukert correction (k = 1.2, rated at 5A):
C_actual = 100 × (5 / 71.4)^1.2 = 100 × (0.070)^1.2 = 100 × 0.045 = 4.5 Ah?
Wait — this gives an absurd answer because the Peukert formula scales C relative to the rated discharge rate. Let me restate:
At 5A (20-hour rate), battery holds 100 Ah. At 71.4A, actual capacity:
t = C_rated × (I_rated/I_actual)^(k-1) / I_actual
t = 100 × (5/71.4)^(1.2-1) / 71.4
t = 100 × (0.070)^(0.2) / 71.4
t = 100 × 0.617 / 71.4
t = 0.86 hours = 52 minutes
Bottom line: 52 minutes at 30% usage factor from a 100Ah battery. Add a second battery: ~104 minutes. This is why serious competition systems carry large battery banks.