Watt Hours Interactive Calculator

If you want a battery bank to run a device for any useful length of time, you need to know exactly how many watt-hours the load will draw, and how much usable energy your battery can really deliver. All of this comes down to a handful of linked variables: power, time, voltage, actual efficiency, and how much you plan to drain from your battery (depth of discharge). This calculator will do the math for total energy use, available runtime, how much battery you need, and estimated cost — just plug in what you know. These numbers are relevant to off-grid solar, backup systems, electric vehicles, and anything where storage is your power source. Below, I lay out the core math, a real example with numbers, and notes on what actually matters in the real world.

What is a watt-hour?

A watt-hour (Wh) is a basic measure of electrical energy. It tells you how much energy is used or stored when something runs at a given power level, for a set amount of time. One watt-hour means a device could draw 1 watt steadily for 1 hour — that's it.

Simple Explanation

Think of watt-hours like the size of a fuel tank. If you burn fuel quickly (high wattage), your tank empties faster. For example, a 60 W light bulb running for 5 hours uses 300 Wh — that's the same energy as a 300 W device running for just 1 hour. Big battery, small load: long runtime. Small battery, big load: short runtime.

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System Diagram

Watt Hours Calculator

How to Use This Calculator

1. Pick your calculation mode — energy, power, time, battery runtime, cost, or required capacity.
2. Enter the numbers you know (power, hours, voltage, battery size, efficiencies) into the boxes provided.
3. Edit efficiency or depth of discharge only if you have firm numbers for your exact setup. The defaults are typical, but the real world varies.
4. Click Calculate and check the result.

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Equations & Variables

Simple Example

A 60 W light bulb runs for 5 hours.
Energy = 60 W × 5 h = 300 Wh (0.300 kWh)
At $0.12/kWh, cost = 0.300 × $0.12 = $0.036
On a 12V 100Ah battery (80% DoD, 90% efficiency): runtime = (12 × 100 × 0.80 × 0.90) / 60 = 14.4 hours

Use the formula below to calculate energy in watt-hours.

Use the formula below to calculate usable battery energy.

Use the formula below to calculate battery runtime.

Use the formula below to calculate required battery capacity.

Use the formula below to calculate energy cost.

Theory & Practical Applications

Fundamental Physics of Energy and Power

Watt-hours give you the total, not just the instantaneous, picture. If you only look at moment-to-moment power (watts), you're missing how much energy is really being burned over time. The E = P × t equation seems straightforward, but applying this in the real world means watching for changing loads, guessing at how efficient your hardware is, and knowing that most loads aren't constant.

Real-world loads bounce all over the place. Motors are a good example: maybe they pull 500 W once running, but will briefly draw 1800 W at startup for a few seconds. That short surge still adds extra watt-hours to your battery demand over a day. Ignore it and you’ll undersize your battery for anything that cycles on and off. Resistance heaters can look stable, but even these drop a bit in draw as they heat up—actual long-run totals often end up 8–12% below your naive calculation.

Battery Chemistry and Depth of Discharge Considerations

How much energy you can safely use from a battery depends on its chemistry. Lithium iron phosphate handles deeper regular discharges, up to 80–90%, without major lifespan loss. Lead-acid batteries hate being drained below 50% — go deeper and you slice cycle life fast. It's not a straight line, either: a lead-acid may claim 500 cycles at 50% discharge but will slump to 200 cycles at 80%. Designing for a deeper discharge gives you more runtime per cycle but costs you in faster battery replacement.

Temperature makes this all worse. Battery output shrinks about 1% for each degree below 20°C. So that 12V 100Ah lead-acid battery that gives you 1200 Wh at room temperature may only cough up 720 Wh at minus 15°C. People miss this; cold weather can cut useable capacity nearly in half. If you're planning for outdoors, sizing for summer is very different than sizing for winter — active heating or larger banks may be needed just to keep up.

System Efficiency Losses in Real Applications

System efficiency isn’t theoretical; it's how much actually makes it from battery to load. Inverters usually lose 5–15%, and if you run them at low load compared to their rating, losses can jump (the “valley” where efficiency is worst). For example, a 2000 W inverter running a 150 W load is often stuck at about 75% efficiency, wasting 50 W just running itself. Sizing inverters and converters for your likely load—not just the theoretical peak—makes a real difference.

Wire losses through simple resistance (I²R) catch a lot of new designers. Running high power at low voltage means big currents and big voltage drops. At 12V, a 1000 W load means over 80A — even with heavy copper, that’s tens of watts burned up in the wire. Jumping to 48V cuts current by 4x, so wire heating (and loss) is 1/16th. This is why almost all bigger systems go for higher voltages even if the hardware is a bit more complicated.

Peukert's Equation and Non-Ideal Battery Behavior

Don't assume batteries give you their full rated amp-hours at all discharge rates. The Peukert effect comes into play as soon as the current goes up. For lead-acid, pull current 5–10x faster than the “20-hour” rate and you may only get 2/3 of the nameplate capacity. Lithium chemistries do better but aren't immune. In practical terms, heavy loads need bigger batteries than you'd predict by just dividing watt-hours by volts.

Worked Example: Off-Grid Solar System Design

Take an off-grid cabin needing: 45 W LED lighting (6 hr/day), 120 W laptop (4 hr/day), 750 W refrigerator running 35% of the time, and a 1500 W microwave (15 min/day). The client wants 3 days of battery, at 24V, with lithium, 85% overall efficiency, and only 80% discharge max.

Step 1: Calculate daily energy consumption per load

LED lighting: E₁ = 45 W × 6 h = 270 Wh/day
Laptop: E₂ = 120 W × 4 h = 480 Wh/day
Refrigerator: E₃ = 750 W × 24 h × 0.35 = 6300 Wh/day
Microwave: E₄ = 1500 W × 0.25 h = 375 Wh/day

Step 2: Sum total daily consumption

E_daily = 270 + 480 + 6300 + 375 = 7425 Wh/day

Step 3: Calculate three-day energy requirement

E_3-day = 7425 Wh/day × 3 days = 22,275 Wh

Step 4: Account for system efficiency

E_required = 22,275 Wh / 0.85 = 26,206 Wh (from battery)

Step 5: Calculate total battery capacity considering DoD

E_total = 26,206 Wh / 0.80 = 32,758 Wh total battery capacity

Step 6: Convert to amp-hours at 24V

C = 32,758 Wh / 24 V = 1365 Ah at 24V

Result: The system requires approximately 1400 Ah of 24V lithium battery capacity, typically implemented as eight 175 Ah cells in series (actual configuration: 3.2V × 8 = 25.6V nominal).

Step 7: Verify peak power capacity

Worst-case simultaneous load = 45 + 120 + 750 + 1500 = 2415 W peak
Peak current = 2415 W / (24 V × 0.85) = 118.5 A
C-rate = 118.5 A / 1365 Ah = 0.087C (well within lithium capabilities)

The bulk of the daily energy use above comes from the refrigerator—even though it isn’t the highest “wattage,” the hours add up. Get this wrong and your battery will always feel too small.

Industrial Applications and Load Profiling

Server UPS systems and electric vehicles live and die by careful watt-hour calculations. A server rack pulling 8.2 kW average with a UPS sized for 20 minutes needs E = 10,000 W × (20/60) h = 3333 Wh. For VRLA batteries at 48V, 50% DoD, and 90% efficiency: C = 3333 / (48V × 0.5 × 0.9) = 154 Ah at 48V. These calculations work for backup but rarely account for gut-punch loads like HVAC or longtime self-discharge.

EVs aren’t simple, either. Drag rises with speed, so energy use per mile jumps quick above about 50 mph. Example: A car might see 250 Wh/mile at 45 mph, 420 Wh/mile at 75 mph — so a “75 kWh” battery (with 90% actual usable) gives you 270 miles if you cruise slow, but only 160 or so at highway speeds. Regen helps only for stop-and-go.

Cost Optimization and Utility Rate Structures

If you face time-of-use electric rates, shifting load to low-rate periods or preheating/cooling can save real money, and batteries can help—just make sure you’re not losing more in inefficiencies or degraded battery life. For a 4500 W water heater, heating 50 gallons from 15°C to 60°C, you’ll use about 9.92 kWh per rise. Running this during peak instead of off-peak makes a several-dollar difference each time (see formula above). For grid-tied solar, keep an eye on seasonal deficit — winter sun hours can drop output nearly in half compared to summer, so don’t size for the best case only. The gap adds up, and missing it means you’ll pull from the grid more than planned. For more calculations, check out the engineering calculator library.

Frequently Asked Questions

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