If 6.25 x 10^18 electrons pass a point in 2 seconds, what is the current?

If 6.25 x 10^18 electrons pass a point in 2 seconds, what is the current?

• A. 1 Coulomb per second
• B. 6.25 x 10^18 Coulombs per second
• C. 6.25 x 10^18 Coulombs
• D. 6.25 x 10^18 A

Answer: (A)

Current is the rate at which charge passes a point: I = ΔQ / Δt. Each electron carries a charge of 1.602 x 10^-19 C. If 6.25 x 10^18 electrons pass in 2 seconds, the total charge is Q = (6.25 x 10^18) × (1.602 x 10^-19 C) ≈ 1.00 C. Now divide by the time to get the current: I = Q / Δt ≈ 1.00 C / 2 s ≈ 0.50 A. So the current is about 0.5 Coulombs per second (0.5 A). The magnitude is 0.5 A, and the direction would be opposite to the actual electron flow for conventional current. The other options either ignore the time interval or confuse total charge with current.

Current is the rate at which charge passes a point: I = ΔQ / Δt.

Each electron carries a charge of 1.602 x 10^-19 C. If 6.25 x 10^18 electrons pass in 2 seconds, the total charge is

Q = (6.25 x 10^18) × (1.602 x 10^-19 C) ≈ 1.00 C.

Now divide by the time to get the current:

I = Q / Δt ≈ 1.00 C / 2 s ≈ 0.50 A.

So the current is about 0.5 Coulombs per second (0.5 A). The magnitude is 0.5 A, and the direction would be opposite to the actual electron flow for conventional current. The other options either ignore the time interval or confuse total charge with current.