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 ? • m.) How many electrons pass through this filament in 5 seconds?
What is the resistance of this filament?
What is the voltage of the battery that would produce this current in the filament?

By definition [tex]I=Q/t[/tex], so we can calculate the total charge that went through the filament since we know the electric current and the time. If the charge of one electron is e, then the total charge will be given by Q=Ne. Combining these equations we get the number of electrons as:

[tex]N=\frac{Q}{e}=\frac{It}{e}=\frac{(0.526A)(5s)}{1.602\times10^{-19}C}=1.64\times10^{19}[/tex]

Since the resistivity of tungsten (W) is [tex]\rho_W=5.6 \times10^{-8}\Omega m[/tex], then resistance can be calculated using the formula:

[tex]R=\rho_W\frac{L}{A}=\rho_W\frac{L}{\pi r^2}=(5.6\times10^{-8}\Omega m)\frac{(0.58m)}{\pi (0.000023m)^2}=19.54\Omega[/tex]

Where we already obtained the radius from the diameter. Finally the Voltage will be given by the formula:

[tex]V=RI=(19.54\Omega)(0.526A)=10.28V[/tex]

ap8997154 ap8997154 · Answer 2 · 2022-02-26T14:41:14+01:00

The voltage of the battery that would produce this current in the filament is 10.28 V.

What is Voltage?

Voltage can be described as the pressure from an electrical circuit's power source that makes the electron work by pushing the charged electrons (current) through a conducting loop.

In order to calculate the number of electrons passing through this filament in 5 seconds, we need to know the total charge that went through the filament,

[tex]Q = Ne[/tex]

We know that the charge can also be written as Q=It, therefore,

[tex]It = Ne\\\\N = \dfrac{It}{e}\\\\N = \dfrac{0.526 \times 5}{1.602 \times10^{-19}}\\\\N = 1.64 \times 10^{-19}[/tex]

B.) We know that the resistivity of the tungsten is 5.6 \times 10^{-8} \Ohm m, therefore the resistance can be written as,

[tex]R = \rho_w \dfrac{L}{A}\\\\R = \rho_w \times \dfrac{L}{\pi r^2}\\\\R = 5.6 \times 10^{-8} \times \dfrac{0.58}{\pi (0.000023)^2}\\\\R = 19.54 \Ohm[/tex]

C.) We know that the relation between the current, voltage, and the resistance can be written as,

[tex]V = IR\\\\V = 19.54 \times 0.526\\\\V = 10.28\ V[/tex]

Hence, the voltage of the battery that would produce this current in the filament is 10.28 V.

Learn more about Voltage:

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