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

Sagot :

[tex]\begin{equation*} =1.29*10^{21}\text{ electrons} \end{equation*}[/tex]

Explanation

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

[tex]\text{ 1 Ampere=}\frac{1\text{ C}}{s}[/tex]

and remember that the charge of an electron is

[tex]Q_e=-1.6*10^{-19\text{ }}C[/tex]

Step 1

a) let

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

so

[tex]\begin{gathered} I=\frac{Q}{time} \\ replace \\ 11.37A=\frac{Q}{18.19} \\ multiply\text{ both sides by 18.19} \\ 11.37A*18.19=\frac{Q}{18.19}*18.19 \\ 206.8206\text{ C}=Q \end{gathered}[/tex]

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

[tex]Number\text{ of electrons=}\frac{206.8206C}{\lvert{-1.6*10^{-19\text{ }}C}\rvert}=1.29*10^{21}\text{ electrons}[/tex]

therefore, the answer is

[tex]\begin{equation*} =1.29*10^{21}\text{ electrons} \end{equation*}[/tex]

I hope this helps you