Question

During the 3.7 min a 7.4 A current is set up in a wire, how many (a) coulombs and (b) electrons pass through any cross section of the wire's width?

Answer 1

1,642.8 coulombs and 1.39*10²¹ electrons pass through any cross section the width of the wire.

Step-by-step explanation:

Electric current is the circulation of electric charges in an electric circuit, while electric current intensity (I) is the amount of electricity or electric charge (Q) that circulates through a circuit in unit time (t).

Then, the intensity of electric current is expressed as:

`I=(Q)/(t)`

Where:

• I is the intensity expressed in Amps (A)
• Q is the electric charge expressed in Coulombs (C)
• t is the time expressed in seconds (s)

In this case:

• I= 7.4 A
• Q= ?
• t= 3.7 min= 222 s (being 1 min= 60 s)

Replacing:

`7.4 A=(Q)/(222 s)`

Solving:

Q= 7.4 A* 222 s

Q= 1,642.8 C

A Coulomb represents about 6.24*10¹⁸ electrons, so you can apply the following rule of three: if 1 C represents 6.24*10¹⁸ electrons, 222 C how many electrons does it represent?

`electrons=(222 C*6.24*10^(18) )/(1C)`

electrons= 1.39*10²¹

1,642.8 coulombs and 1.39*10²¹ electrons pass through any cross section the width of the wire.