How do I calculate watts for household appliances?

[njuice8]

Hi, for a physics project we have to find the power usage of household appliances.

My charger says "60W inputAC 100-240V 1.5A output 16.5-3.65V"
Would the watts just be 60W? Or would I have to calculate the amount by using P=AMPsx120volts (provided by my teacher)?

My other charger says "input 100-240V output 5.1V=850mA 0.2A"
How would I calculate the amount of watts for this charger?
Would I use P=AMPxvolts?
I'm not sure what number to use for the volts and AMPs because it lists a few numbers (as I stated above)

Thanks!

 

[CWatters]

Labels on appliances and chargers can be missleading.

The label on a charger or adaptor will tell you what it can deliver not what the appliance actually draws. For example I have a mains adaptor that says 1A @ 12V. That means it can deliver upto 1A at 12V. However the appliance I use it with has a label that reads 0.75A at 12V. That means the appliance can draw upto 750mA at 12V. However even then the actual consumption might be lower. The appliance might normally draw 300mA with the ocasional peak draw of 750mA. This would be especially true for something like a printer which spends most of it's time consuming very little power but might have quite a high power consumption when actually printing something.

In the case of a battery powered appliance the power consumption of the appliance can be very different to that of the charger because the charger is designed to pump energy into the battery quickly (eg fast charge = high power for short time) while the appliance might be designed to run for a long time off the battery (low power consumption = long battery life).

So the answer you give depends on what level of detail the teacher wants?

If it's just to give you an idea of how much typical devices consume then I would look at the label on the actual appliance itself where possible.

In your case the first charger can deliver a maximium of 1.5A at 16.5V = 24.75W however it may draw more than that from the mains because it is not 100% efficient (some power is wasted as heat in the charger). The 60W figure refers to the maximuim that the charger might draw from the mains taking into account the efficiency of the charger. You could quote the 60W figure as your answer but that's likely to be an over estimate.

The second charger sounds like it is designed to deliver 5.1V at upto 850mA which is 4.3W. I believe the 0.2A may actually refer to the peak input current. eg it's the maximium it might draw from the mains. The wide input voltage range of 100-240V means it is designed to work from either 120V or 230V mains making it suitable for use in different countries. The maximium current draw is most likely to occur when the adaptor is run from 120V/60Hz (USA) rather than 230V/50Hz (European) because P = IV. If you reduce V then I has to increase to deliver the same power. So the peak input power is likely to be around 120V * 0.2A = 24W. Again that's quite a lot more than the 4.3W actually needed by the appliance so is likely to be an over estimate.

Sorry that's complicated.

PS: Perhaps best look at appliances such as a kettle, coffee maker or heater which have more clearly defined power consumption figures.

 

[njuice8]

Hi, thanks for helping me!
It made sense :)
I was looking at my oven, and it said 13.5kW@120V and it also said 10kW@120V.
How would I calculate this? Thanks!

 

[CWatters]

13.5kW@120V and it also said 10kW@120V

Sounds like the oven might use more than one phase of the mains? Here in England that would be unusual but those figures suggest you might be in the USA where I think it's more common?

In which case the peak power would be the sum eg 13.5 + 10 = 23.5kW. However it would only be that high if everything was being used at once.

In addition ovens only draw peak power while heating up. Once upto temperature the thermostat switches on and off to maintain the set temperature. Then the power consumption depends on how well insulated the oven is and how frequently you open the door to check if your food is cooked. On average the power you need to maintain a set temperature in the oven is equal to the power lost from the oven. An ideal oven would have no losses and therefore it would need no power to maintain it's temperature.

This effect also applies to a fridge or freezer. The rating label on the back may tell you how much the pump consumes when it's running but the average consumption over a year won't be as high. In Europe (and perhaps the USA?) data sheets tell you the expected average power consumption. For example for this small fridge...

http://www.johnlewis.com/231508095/Product.aspx

See "Energy consumption 153kWh/year"

In this case the average power consumption would be just...

153,000/(24 * 365) = 17W

However the when the pump actually running it might be 10 or 20 times that much.

The peak figure tells you what size wire you need to supply the appliance and any fuse needed. The lower average figure tells you what it will cost to run.

 

[Nickie]

To calculate Watts, you can use the formula P = VI, where P is power in watts, V is voltage in volts, and I is current in amperes. In the first charger, since the voltage and current are both given, you can simply multiply them to get the power in watts. In this case, it would be 100-240V x 1.5A = 90-360W. However, it is important to note that the input voltage range is 100-240V, so the power usage may vary depending on the voltage it is receiving.

For the second charger, you can use the same formula, but you will need to convert the units to be consistent. The current is given in milliamperes (mA), so you will need to convert it to amperes (A) by dividing by 1000. Similarly, the voltage is given in volts, so you can use it as is. The calculation would be 5.1V x 0.85A = 4.335W.

It is important to note that the power usage may vary depending on the efficiency of the charger and the actual voltage and current it is receiving. It is always best to check the specifications provided by the manufacturer for more accurate power usage information. I hope this helps and good luck with your project!