Step 1: audit your home
Make a table with four columns: appliance, power (W), hours per day, energy (kWh per day). The power is printed on the label of each device. Ask a family member how many hours it runs. Include small things like phone chargers and the TV standby light.
Energy = power × time. Divide by 1000 to change watt-hours into kilowatt-hours: E (kWh) = P (W) × t (h) ÷ 1000. One kWh is one unit of electricity on the bill.
Step 2: find the big users
Draw a bar for each appliance. Heaters and geysers have a high power, so even a short time uses a lot. A fridge has a small power but runs for many hours. A bulb has a low power and its bar is short. The biggest bars are where a saving is worth the most. Do not guess: measure and compute.
The total for the home per day, times 30, is the units per month. Units × price per unit = the monthly bill.
Step 3: write the saving proposals
- Use efficient devices: an LED bulb gives the same light with much less power; a star-rated fridge or fan uses less energy.
- Use for less time: switch off lights and fans in empty rooms, shorter geyser use.
- Stop standby use: unplug chargers and TV when not in use.
- Look after appliances: clean fridge coils and fan blades so they work better.
For each proposal, work out: units saved per month, money saved per month, and the payback time (cost of the new device ÷ money saved per month).
Step 4: present your plan
Write a short plan for your family: a table of before and after, the total saving in units and rupees, and which two ideas to do first. A good plan uses numbers, is easy to do, and is honest about cost.
Try it: use the buttons in the 3D to apply one idea at a time. Predict the new total before you press each button, then check.
Key formulas and definitions
- E (kWh) = P (W) × t (h) ÷ 1000
- 1 kWh = 1 unit of electricity = 3.6 × 10⁶ J
- Monthly bill = units per month × price per unit
- Payback time = cost of new device ÷ monthly money saved
Worked examples
1. A 60 W bulb runs 5 hours a day. How many units does it use per day?
E = 60 × 5 ÷ 1000 = 0.3 kWh = 0.3 unit.
2. An LED bulb of 9 W gives the same light. Find the units saved per month (30 days) by swapping, for 5 h per day.
Old: 0.3 kWh/day. New: 9 × 5 ÷ 1000 = 0.045 kWh/day. Saved per day = 0.255. Per month: 0.255 × 30 = 7.65 units.
3. The price is 7 per unit. Find the money saved per month in the example above.
7.65 × 7 = 53.55 per month.
4. A 2000 W geyser runs 0.5 h a day. If the time is cut to 0.25 h, find the units saved per month.
Old: 2000 × 0.5 ÷ 1000 = 1.0 kWh. New: 0.5 kWh. Saved 0.5 per day, so 15 units per month.
5. A home uses 3.62 kWh per day. Find the monthly units and the bill at 7 per unit.
Monthly units = 3.62 × 30 = 108.6. Bill = 108.6 × 7 = 760.2.
6. An LED costs 100 and saves 53.55 per month. Find the payback time.
100 ÷ 53.55 ≈ 1.9 months. After that, the saving is free.
Common mistakes
- Forgetting to divide by 1000. Watt-hours and kilowatt-hours are different by a factor of 1000.
- Mixing minutes and hours. 30 minutes = 0.5 h before multiplying.
- Looking only at big watts. A fridge with a low power can use more energy than a geyser, because it runs all day.
- Counting only the device price. A good proposal also checks the units saved and the payback time.