A coffee maker, which draws 13.5 A of current has been left on for 10 minutes. What is the net number of electrons that have passed through the coffee maker?

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Note that current refers to rate of flow of charge. Its relation to charge (Q) in a given time (t) is:

`I=Q/t`

In the formula, the unit of the current (I) is Ampere, the charge (Q) is Coloumb and the time (t) is second.

So, convert the given time to seconds.

`t= 10 min * (60 sec)/(1 min)= 600 sec`

Then, plug-in I = 13.5A and t=600s to the formula of current.

`13.5=Q/600`

To solve for Q, multiply both sides by 600.

`13.5*600=Q/600*600`

`8100=Q`

Hence, the amount of charge flowing in the coffee maker is 8100C.

Now that amount of charge is known, compute for the number of electrons.

Note that in 1Coloumb, there are 6x10^8 electrons.

So,

`n u mber of e^- = 8100C * (6xx10^8 e^-)/(1C)`

`n umber of e^-= 4.86xx10^12` **Hence,` 4.86xx10^12` electrons flow through the coffee maker.**

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