Question

A 580-mm long tungsten wire, with a 0.046-mm-diameter circular cross section, is wrapped around in the shape of a coil and used as a filament in an incandescent light bulb. When the light bulb is connected to a battery, a current of 0.526 A is measured through the filament. (Note: tungsten has a resistivity of 4.9 × 10-8 Ω • m.). How many electrons pass through this filament in 5 seconds?

Answer 1

1.64 x 10^19 electrons

Step-by-step explanation:

The current is defined as I=ΔQ/Δt where ∆Q is the amount of charge flowing past a point in the filament. This charge is comprised of electrons that each carry charge of e = 1.602 × 10^-19 C. So ΔQ=Ne=IΔt and the number of electrons flowing through the filament in 5 s is N=IΔte=(0.526 A)(5 s)1.602×10^−19 C=1.64×10^19 electrons.

Answer 2

Answer: 1.64 *10^19 electrons

Explanation: In order to the explain this problem we have to consider the following:

The current= charge/time; so

as the electrons move in the tungsten wire we have:

0.526 C/s= N electrons per second* charge of electron=

N electrons/s= 0.526/1.6*10^-19= 3.28 *10^18 electrons/s

Then, during 5 seconds will pass:

3.28 *10^18 electrons/s*5 5s= 1.64 *10^19 electrons