Question

A 5.00 A current runs through a 12 gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has free electrons per cubic meter. (a) How many electrons pass through the light bulb each second? (b) What is the current density in the wire? (c) At what speed does a typical electron pass by any given point in the wire? (d) If you were to use wire of twice the diameter, which of the above answers would change? Would they increase or decrease?

Answer 1

The answer to your problem is,
`3.125 * 10^(19)` in electrons per second

Step-by-step explanation:

5A current is passing through the copper wire and the light bulb; it means that 5 Coulombs of charge per second is passing through the wire (as current = coulombs/second). To find the electrons per second, the following formula is used:

Electrons per second =

`N_(e) = (5)/(e) = (5)/(1.60-10^(-19) )`

`= 3.125 * 10^(19)`

Thus the answer to your problem is,
`3.125 * 10^(19)`