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A flashlight bulb operating at a voltage of 14.4 V has a resistance of 11.0 Ω . How many electrons pass through the bulb filament per second (e = 1.6 ´ 10-19 C)? (Give your answer to two significant figures)

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For a DC circuit, the following equation relates the voltage, resistance, and current:

V = IR

V is the total voltage supplied, I is the total current, and R is the total resistance.

Given values:

V = 14.4V

R = 11.0Ω

Plug in the values and solve for I:

14.4 = I×11.0

I = 1.309A

Since one electron carries 1.6×10⁻¹⁹C of charge, divide the current by this number.

1.309/(1.6×10⁻¹⁹) = 8.182×10¹⁸

Round this value to 2 significant figures:

8.2×10¹⁸ electrons per second.

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