Question

A current of 11.37 A passes through a wire. How many electrons pass through the wire in 18.19 seconds ?

Answer 1

`\begin{equation*} =1.29*10^(21)\text{ electrons} \end{equation*}`

Step-by-step explanation

An Ampere is in the SI base unit of electric current equal to one coulomb per second.

`\text{ 1 Ampere=}\frac{1\text{ C}}{s}`

and remember that the charge of an electron is

`Q_e=-1.6*10^{-19\text{ }}C`

Step 1

a) let

`\begin{gathered} I=11.37\text{ Amperes} \\ I=11.37\text{ A}*(((1C)/(s))/(A))=11.37(C)/(s) \\ also \\ time\text{ =}18.9\text{ s} \end{gathered}`

so

`\begin{gathered} I=(Q)/(time) \\ replace \\ 11.37A=(Q)/(18.19) \\ multiply\text{ both sides by 18.19} \\ 11.37A*18.19=(Q)/(18.19)*18.19 \\ 206.8206\text{ C}=Q \end{gathered}`

b) now , to find the number o f electrons , divide the total a}charge by the charge of an electron

`Number\text{ of electrons=}\frac{206.8206C}{\lvert{-1.6*10^{-19\text{ }}C}\rvert}=1.29*10^(21)\text{ electrons}`

therefore, the answer is

`\begin{equation*} =1.29*10^(21)\text{ electrons} \end{equation*}`

I hope this helps you