Energy Savings Calculator
Dollars saved by replacing a device with an efficient one — annual, monthly, and kWh.
Updated
From your bill — the US average is about $0.175.
You need
$27.15/year
155.1 kWh saved a year
- Saved per month
- $2.26
- Saved per year
- $27.15
- Energy saved a year
- 155.1 kWh
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In short
How much does replacing a 100 W device with a 15 W one save?
85 watts saved for 5 hours a day is 155.1 kWh a year, about $27.15 at the 17.5 cents per kWh this tool defaults to, or roughly $2.26 a month. The formula is watts saved ÷ 1000 × hours × 365 × rate, so the same 85 watts saved on a device running 24 hours a day would be worth $130.31 instead.
The figure counts running cost only, so divide any extra purchase price by the annual saving yourself to get the years to break even.
How to use the energy savings calculator
Enter the old device power in watts, the new device power in watts, how many hours a day it runs, and your electricity rate, and you get the savings from the swap: the dollars saved a year, the saving a month, and the kilowatt-hours saved over the year.
The defaults price the classic upgrade, a 100 W incandescent bulb replaced by a 15 W LED on five hours a day, which comes to 155.1 kWh and about $27.15 a year. The rate is the input that swings the answer most.
The tool defaults to 17.5 cents per kWh, while the Energy Information Administration reported a US residential average of 18.11 cents year to date for 2026 and 18.44 cents in May 2026, so use the figure from your own bill.
85 W
Watts saved
100 W incandescent to 15 W LED
155.1 kWh
Saved a year
at 5 hours a day
$27.15
Yearly saving
at $0.175 per kWh
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The saving comes entirely from the gap in wattage, so getting both numbers right is what makes the answer real. The old wattage is whatever the device draws today: the nameplate on an incandescent bulb, the sticker on an old fridge, the label on a window air conditioner.
The new wattage is what the replacement pulls, and LED packaging lists it plainly while Energy Star appliance tags give a yearly kWh figure you can work back from by dividing by the hours it runs. A plug-in power meter beats both, because a nameplate usually states a maximum rather than a typical draw.
If the two devices do the same job at the same brightness, the same cooling, or the same throughput, then the difference between them is pure saving.
Swapping a whole house of bulbs?
This page prices one device at a time. The LED savings calculator multiplies a single bulb swap across every fitting you are replacing in one go.
Open the LED savings calculator →Do
- Take both wattages from the devices themselves or from a plug-in power meter.
- Check the replacement does the same job at the same output first.
- Set hours per day to genuine runtime rather than hours plugged in.
- Divide the extra purchase price by the annual saving for the payback years.
- Rank the always-on, high-hours loads first, since those return the most.
Don't
- Enter a refrigerator nameplate at twenty-four hours a day, because compressors cycle.
- Reverse the old and new wattage fields, which hides a real saving as zero.
- Read the figure as a bill prediction, since it counts running cost alone.
- Pay a large premium for an efficient model on a rarely-used load.
What a given number of watts saved is worth in a year, at four different runtimes. The formula section gives you one answer for one device; this gives you the whole landscape, so you can see at a glance that a small saving on something that never switches off beats a large saving on something that rarely runs, and rank your upgrade candidates before spending anything.
| Watts saved | Saving a year at 1 hour a day | At 4 hours a day | At 8 hours a day | At 24 hours a day |
|---|---|---|---|---|
| 5 W | $0.33 | $1.32 | $2.64 | $7.93 |
| 10 W | $0.66 | $2.64 | $5.29 | $15.86 |
| 15 W | $0.99 | $3.97 | $7.93 | $23.80 |
| 20 W | $1.32 | $5.29 | $10.58 | $31.73 |
| 25 W | $1.65 | $6.61 | $13.22 | $39.66 |
| 35 W | $2.31 | $9.25 | $18.51 | $55.53 |
| 50 W | $3.31 | $13.22 | $26.44 | $79.32 |
| 75 W | $4.96 | $19.83 | $39.66 | $118.98 |
| 85 W, the default swap | $5.62 | $22.47 | $44.95 | $134.85 |
| 100 W | $6.61 | $26.44 | $52.88 | $158.64 |
| 150 W | $9.92 | $39.66 | $79.32 | $237.97 |
| 250 W | $16.53 | $66.10 | $132.20 | $396.61 |
| 500 W | $33.05 | $132.20 | $264.41 | $793.22 |
| 1,000 W | $66.10 | $264.41 | $528.81 | $1,586.44 |
How much does runtime change what a swap is worth?
Hours per day is the multiplier that decides whether a swap is worth doing at all. The same 85 watts saved is worth $22.47 a year on a fitting that runs 4 hours a day and $134.85 on one that runs continuously — six times as much for the identical purchase.
Be honest about real runtime rather than how long the device is plugged in: a refrigerator compressor cycles rather than running flat out, a living-room lamp works a few hours an evening, and a basement bulb almost never. Set the hours to match the device you are actually pricing, and if you are unsure, run the calculation twice at the low and high ends of your guess to see how much the uncertainty is worth.
Read it: Runtime, not the wattage gap, decides where an upgrade dollar earns the most.
Row taken from the reference table above; the calculator itself defaults to 17.5 cents, about 3 percent lower.
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The formula, worked line by line
The saving is the power you no longer draw, turned into energy and then into money. Subtract the new wattage from the old to get the watts saved, divide by 1,000 to reach kilowatts so the units match the kilowatt-hours your utility bills, multiply by daily hours for the energy saved each day, then by 365 days and your rate for the yearly figure. Nothing else enters the calculation, because both devices are assumed to be doing the same job.
The shape of that expression is worth noticing, because it explains every counter-intuitive result the tool produces. Saving is proportional to watts and to hours in equal measure, so 20 watts saved on something that runs all day beats 200 watts saved on something that runs half an hour.
It is also floored at zero: if the replacement draws as much as the original or more, there is nothing to save, and the tool will not invent a negative saving to flatter a bad upgrade.
watts saved = old watts − new watts (floored at zero)
daily kWh saved = (watts saved ÷ 1000) × hours per day
annual saving = daily kWh saved × 365 × rate
payback in years = extra purchase price ÷ annual saving- Watts saved
- 100 − 15 = 85 W
- Daily energy saved
- 85 ÷ 1000 × 5 = 0.425 kWh
- Yearly energy saved
- 0.425 × 365 = 155.1 kWh
- Rate
- × $0.175
- Saving a year
- $27.15
The monthly figure the tool shows is 0.425 × (365 ÷ 12) × $0.175 = $2.26, using a month of 30.42 days so that twelve of them come to exactly the year above rather than five days short of it. Push the same 85 watts to a device that never switches off and the arithmetic gives 85 ÷ 1000 × 24 × 365 × $0.175 = $130.31 a year, nearly five times as much for an identical wattage gap.
Payback in one division
Payback follows from that annual figure in a single division. An LED costing two dollars more than the incandescent it replaces but saving $27.15 a year pays for itself in about 27 days. A more efficient appliance costing $150 more than a basic model and saving $40 a year breaks even in 3.75 years, which is worth doing if the appliance will plausibly last ten.
The comparison that fails is the pricey efficient model on a low-hours load: the same $150 premium against a $5 a year saving needs 30 years, which no appliance survives.
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Questions people ask
How much does replacing an incandescent bulb with an LED save?
Swapping a 100 W incandescent for a 15 W LED saves 85 watts. At five hours a day that is 0.425 kWh saved daily, 155.1 kWh a year, and roughly $27.15 at $0.175 per kWh, which is about $2.26 a month from a single bulb. Run that bulb ten hours a day and the saving doubles to about $54.29, and a bulb that never switches off would save around $130.31 a year. The more hours it is on, the more the swap is worth, so start with the lights that burn longest.
How do I work out the payback period on a more efficient device?
Divide the extra you paid for the efficient device over the cheaper, thirstier one by the annual saving, and you get the years to break even. An LED costing two dollars more than an incandescent but saving $27.15 a year pays back in under a month. A pricier efficient appliance costing $150 more than a basic model and saving $40 a year breaks even in 3.75 years, which is worth doing if the appliance lasts ten. The same $150 premium against a $5 a year saving needs 30 years and is not.
Is it worth replacing an old refrigerator with a new one?
Often yes, because a fridge runs around the clock, so even a modest wattage cut multiplies over 8,760 hours a year. A 50 watt reduction held continuously is worth about $79.32 a year at 18.11 cents per kWh. The trap is the input: a compressor cycles rather than running flat out, so entering its nameplate watts at 24 hours a day overstates the saving badly. Use the Energy Star annual kWh figures for both units, or a plug-in meter, and weigh the result against the new fridge price.
Which devices should I replace first to save the most?
The always-on, high-hours loads, because the saving is watts saved multiplied by hours and those rack up the most hours. A refrigerator, a porch or hallway light that never goes off, a well or pool pump, or a water heater beats a rarely-used appliance every time, even when the rarely-used one has a bigger wattage gap. The table on this page makes the ranking obvious: 20 watts saved continuously is worth more per year than 100 watts saved for one hour a day. Rank your candidates by yearly saving and work down.
Why does the calculator sometimes show no saving?
Because the saving is floored at zero. If the new device draws as much power as the old one, or more, there is nothing to save and the figure stays at zero rather than going negative, since an upgrade only pays off when the replacement is genuinely lower-wattage for the same job. Double-check that you have entered the old device draw in the old field and the new one in the new field, because reversing them turns a real saving into none at all and hides a good upgrade.
Sources
Where the constants and formulas on this page come from. Each line names the figure it backs.
The 18.11 cents per kWh year-to-date 2026 US residential average used to compute the reference table, and the 18.44 cents May 2026 figure.
Electric Power Monthly, Table 5.3 — Average price of electricity to ultimate customers by end-use sector — US Energy Information Administration, Year to date through May 2026
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