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Inverter Size Calculator: How many watts does the inverter in my RV need?

A 1,200 W coffee maker needs an inverter rated at about 1,500 W – and draws over 110 A from a 12 V battery while running. From your 120 V appliances this calculator determines the recommended inverter size, the surge peak for motors and compressors, the battery current and the energy that losses and idle cost every day.

Units

Appliances that should run at the same time

Only enter appliances that really run at the same time. Start-up factor: 1 for heating appliances/power adapters, 2–3 for motors and compressors.

Advanced settings

Don't run the inverter at its limit continuously; 20–30% reserve is common.

Result

Recommended inverter continuous rating

1,500 W

continuous load 1,200 W + 25% reserve

Required surge rating
1,200 W
for start-up currents
Battery current at continuous load
111 A
Battery current at surge peak
111 A
Energy from the battery per day
232 Wh
of which losses + idle 52 Wh
  • Wire gauge, fusing and BMS discharge current must match the battery current. This calculator does not replace installation planning by a professional.
Show calculation
  1. 1Continuous load: 1,200 W (sum of appliances)
  2. 2Surge peak: 1,200 W (power × start-up factor)
  3. 3Recommended continuous rating: 1,200 W × (1 + 25%) = 1,500 W → 1500 W
  4. 4Battery current at continuous load: 1,200 W ÷ (90% × 12 V) = 111 A
  5. 5Daily losses: 180 Wh ÷ 90% − 180 Wh + 8 W × 4 h = 52 Wh

 

How it's calculated

The inverter must continuously supply the sum of all appliances running at the same time – plus a reserve so it doesn't work at its limit all the time. Motors and compressors (cooler, air conditioner, power tools) briefly draw a multiple of their rated power at start-up; the inverter's short-term surge rating must cover this peak.

Continuous rating ≥ continuous load × (1 + reserve) Surge rating ≥ Σ (power × start-up factor)
Example: 1,200 W × 1.25 = 1,500 W

Battery current

On the 12 V side a multiple of the 120 V current flows. Under load the battery voltage drops, and the inverter has losses – both increase the current:

Battery current [A] = power [W] ÷ (efficiency × battery voltage [V])
1,200 W ÷ (0.90 × 12 V) = 111 A

Losses per day

Losses [Wh] = energy ÷ efficiency − energy + idle consumption [W] × hours switched on
An inverter with 8 W idle that stays on 24 h costs 192 Wh per day – often more than the actual consumption.
Wire gauge, fuse and the battery's maximum discharge current (BMS) must match the battery current. Manufacturer specifications and codes apply to the installation; this calculator only provides guidance.

Worked example

In the morning a coffee maker (1,200 W) should run while a compressor cooler (60 W, start-up factor 3) is connected. 12 V system, 25% reserve, 90% efficiency, inverter switched on four hours a day.

Inputs

  • Appliances that should run at the same time:
  • Kaffeemaschine: 1,200 W × 1 × × 0.15 h/day
  • Kompressor-Kühlbox 230 V: 60 W × 3 × × 10 h/day
  • Battery voltage: 12 V
  • Power reserve: 25 %
  • Efficiency: 90 %
  • Idle consumption: 8 W
  • Switched on per day: 4 h

Result

2,000 W

Recommended inverter continuous rating

Required surge rating
1,380 W
Battery current at continuous load
117 A
Battery current at surge peak
128 A
Energy from the battery per day
899 Wh

Calculation

  1. Continuous load: 1,260 W (sum of appliances)
  2. Surge peak: 1,380 W (power × start-up factor)
  3. Recommended continuous rating: 1,260 W × (1 + 25%) = 1,575 W → 2000 W
  4. Battery current at continuous load: 1,260 W ÷ (90% × 12 V) = 117 A
  5. Daily losses: 780 Wh ÷ 90% − 780 Wh + 8 W × 4 h = 119 Wh

The variables explained

Power (W)
Power draw per the nameplate. For microwaves the electrical input (higher than the cooking power).
Start-up factor
Brief extra demand when switching on: heating appliances 1, power adapters 1–1.5, motors and compressors 2–3, some air conditioners up to 5.
Power reserve (%)
Safety margin to the continuous rating. Prevents overload shutdowns and keeps efficiency in the good range.
Efficiency (%)
Typically 85–93% at medium to high load.
Idle consumption (W)
Power draw with no load connected, per the data sheet 5–15 W (much less with eco/search mode).

Common mistakes

  • Using only the microwave's cooking power: 800 W cooking power means 1,200–1,400 W input.
  • Ignoring start-up currents: coolers and air conditioners trigger the overload shutdown if the inverter is too small.
  • Buying too big: a 3,000 W inverter has more idle consumption and worse partial-load efficiency than a properly sized 1,500 W inverter.
  • Forgetting the battery: many 100 Ah LiFePO4 limit the discharge current to 100 A – not enough for 1,500 W.
  • Choosing modified sine wave: coffee makers with electronics, power adapters and induction need pure sine wave.

Assumptions and limits

  • Battery voltage under load set conservatively at 12.0 V or 24.0 V.
  • Efficiency is assumed constant; at small loads it is lower.
  • Fuse sizes and wire gauges are deliberately not issued as approvals – manufacturer data, installation method and codes decide.

Frequently asked questions

What inverter do I need for a coffee maker?

For a 1,200 W drip coffee maker, a pure sine wave inverter with 1,500 W continuous rating. The battery current is around 110 A (12 V) – the battery must be able to deliver that, and the feed cable typically needs 2 AWG to 1/0 AWG (35–50 mm²) (wire gauge calculator).

Is a 1,000 W inverter enough for induction?

Only with a cooktop that can be limited to under 800 W. Common single-burner induction cooktops draw 1,800–2,000 W and need a 2,000–2,500 W inverter – and a battery that can deliver 170 A and more.