Electrical – Calculate the electric bill of a phone charger (AC adapter input or output?)

ac-dcadapterbattery-chargingchargerwatts

I need to calculate the amount of money I pay for charging my phone with this charger

Samsung phone charger

I know that I need to calculate the watts then kWh, so I multiplied 5 V by 1.55 A to get 7.75 watts and then use that to calculate kWh. I'm so confused though because whenever I asked anyone in my school they would say, "you don't pay for the output, you pay for the input, so what you did is wrong".

Are they right? Why or why not?
What is the correct method?

Note: I'm not looking for a very accurate number because I don't wanna use a Kill-a-Watt. In addition, I watched a video of someone using it on their charger and it gave a number around 4 or 5, that makes me think my teachers are wrong.

Phone battery
I'm adding info about the battery since some of you asked for it.
It's a Samsung secondary Li-on battery. Nominal voltage 3.85 V / 11.55 Wh. Charge voltage 4.4 V / 3000 mAh.

Best Answer

Those numbers tell you the current that the supply can provide, not how much your phone needs. Your phone will only draw as much as it's designed for, and so you don't know how much power is coming from the mains. So, they are correct. You can't get anything meaningful from the adaptor text itself.

There are two ways to get this figure. You can measure the current going into the adaptor itself, or you can measure the current going to your phone. The latter means you also need to know the efficiency of the adaptor as there is some loss in the adaptor when it converts mains voltage to 5 V. You could potentially estimate it from the mains current input (0.3 A), but that doesn't say what voltage or frequency it's at.

Adding calculations for the second method Ok, so your phone is probably charged over USB which used to be limited to 500 mA. Also, that 1.55 A is the maximum the supply can provide, you don't want to have that all the time. So, let's assume (crudely) that \$ I = 500\ \mathrm{mA} \$. Power to your phone is \$P_{out} = IV = 0.5 \times 5 = 2.5\ W\$. Then, let's assume the adaptor is 80% efficient, so \$P_{in} = \frac{P_{out}}{0.80} = 3\ W\$. The multiply by the time to completely charge, and you're there.