# What is the difference between "Input" and "Output" values on power adapters? Which one represents the amount of power being drawn from a circuit?

I'm trying to get a rough idea for how much electricity I consume in my home office.

I understand that there are a lot of ways to measure this kind of thing:

• Turn everything else off in all other rooms. Turn off other circuit breakers, and monitor power usage meter.
• Measure with a device (e.g. Kill-a-watt)

In the absence of that, I'd like to get a rough idea by checking the power adapters for various devices. When I check power adapters, I see things like "input" and "output". Which one is the one actually drawing from my house?

For example: Does this mean the device is drawing 0.24 amps? Or 1.5 amps?

I suspect that "output" is in terms of DC and not actually drawing that much power so I should probably be looking at "input".

• Going to leave Pigrew hanging on such an answer!? Feb 18 '17 at 19:16

Power adapters convert the electricity from one voltage to another voltage, and sometimes between AC and DC.

The quick answer to your question is to multiply the INPUT voltage by the INPUT current. In this case, you likely have 120 V input (if you're in the United States). So, the MAXIMUM power would be 120*0.24 = 29 W, though the actual power usage is likely less than this. With efficient power supplies, it will be better to look at the electrical requirement of your device rather than the power supply's maximum power rating. For example, if the device uses 1.2 A at 5 V DC, it would use a maximum of about 6 W. I would multiply this number by about 1.2 to account for power supply inefficiencies, so your system may use about 7.2 W max.

And here's some related information about power supply efficiency that is somewhat relevant to the question (that I wrote before carefully reading the question):

The INPUT is what sort of electrical system you need to supply to the adapter (i.e. what your power company supplies).

The OUTPUT is what is supplied to your device.

Note that he amount of DC electrical power is calculated by multiplying the current by the voltage (P=I·V). For AC, this product is the maximum power that could be used, though the actual amount could be lower because of the power factor (P=I·V·PF), and the power factor is typically between 0.7 and 1.0 (except for some motors).

Because of inefficiencies in the power adapter not all of the input power is able to be output. This extra power is turned into heat. So, the input power is always larger than the output power.

In your example, the power adapter is rated to use a maximum of 58 W (0.24*240), but can output only 5*1.5=7.5 W. So, worse case based on the label, it will use 58 W, but only supply 7.5 W to your device, so about 13% efficient. Efficiency is defined as the output power divided by the input power.

Power supplies usually do not draw their maximum power during usage: They try to draw only the amount of power the device wants. So, this adapter usually won't be drawing 58 W. In fact, modern power supplies will use less than 1 W when plugged in with their device turned off and have inefficiencies of >90%. So, knowing the maximum does not tell you their typical usage.

One quick test of the efficiency is to check the temperature of the power supply. The hotter the power supply (while plugged in), the less efficient it is.

When choosing a power supply to reduce energy use, pay attention to their efficiency and not their energy use. New supplies should be rated as to their energy usage using roman numerals (I,II,III,IV,V) based on an international standard. Class V is currently the best efficiency rating, and will mean that the drawn power will closest to the supplied power, in which case you should look at the devices required voltage and current to get an idea of the power used.

• The maximum current is for the minimum voltage: 0.24A@100VAC which is only 24W even so taking 24W and providing 7.5W of output is a lousy 30% efficiency. At 120V, the input current would be 0.2A which also results in 24W. At 240V, it would be 0.1A, but still 24W. Feb 2 '17 at 8:28
• @DanD., the difference between our calculations is that you assumed constant efficiency with respect to input voltage, and I assumed constant current. Based on the label, we don't know what is the real-life behavior. I made a worst case assumption, and you made a more realistic assumption. The actual performance will be somewhere in the middle, as the efficiency will generally decrease as the input voltage increases. Feb 2 '17 at 17:36

The input of the device (AC) is what you would use to calculate the electricity usage from your power company. It is a maximum though and you can never know how much the device is actually drawing at any given time because it varies. The only way to accurately calculate the cost of each device would be to buy a device such as a Kill-A-Watt and take averages. For example, my computer's power supply is 450W DC. At 100% efficiency you would think it would draw 3.75 amps at 120V to reach that 450W, but it actually uses only 90 watts (0.75 amps) because it isn't under full load all the time. In short: Get a kill-a-watt. They are amazing. I do not work for them but I like their product. It saved me from buying a new fridge to save \$1 a year.