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A 5.00-a current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. copper has 8.5 * 1028 free electrons per cubic meter. (a) how many electrons pass through the light bulb each second

2 Answers

6 votes

Answer:


(dN)/(dt) = 3.125 * 10^(19) electrons per second

Step-by-step explanation:

As we know that electric current is defined as rate of flow of electric charge

so here we will have


i = (dq)/(dt)

here we know that


q = Ne

now from above equation


i = e(dN)/(dt)

here we know


(dN)/(dt) = number of electrons passing per second

e = charge of an electron

i = 5.00 Ampere

now from above equation we have


(dN)/(dt) = (i)/(e)


(dN)/(dt) = (5)/(1.6 * 10^(-19))


(dN)/(dt) = 3.125 * 10^(19)

so above is the total number of electrons passing through wire per second

User Webduvet
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8.2k points
2 votes

The correct answer is:
3.125*10^(19) electrons/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\cdot 10^(-19))=3.125*10^(19)

User Justin Harvey
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8.2k points