The filament of an incandescent light bulb is made of tungsten which has a thermal coefficient of resistivity of α = 4.4 x 10-3 K-1. The filament is initially at room temperature, To = 293 K. When it is connected across a V = 140 volt power supply the filament quickly heats up to T = 2900 K at which time the current flowing through the filament is I = 0.95 amps.

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CarliReifsteck CarliReifsteck · Answer 1 · 2019-10-15T12:54:55+02:00

Answer:

The current when the filament at room temperature is 11.84 A.

Explanation:

Given that,

Thermal coefficient of resistivity [tex]\alpha=4.4\times10^{-3}\ K^{-1}[/tex]

Initial temperature = 293 K

Voltage = 140 Volt

Final temperature = 2900 K

Current = 0.95 A

Suppose we find the numerical value of the current when the filament is at room temperature

So, We use the equation of resistance of a material changes with temperature

[tex]R=R_{0}(1+\alpha(T_{2}-T_{1}))[/tex]....(I)

Using ohm's law for value of resistance

[tex]V = I R[/tex]

[tex]R = \dfrac{V}{I}[/tex]

Put the value of resistance in the equation (I)

[tex]\dfrac{V}{I}=\dfrac{V}{I_{0}}(1+\alpha(T_{2}-T_{1}))[/tex]

[tex]I_{0}=I(1+\alpha(T_{2}-T_{1}))[/tex]

Put the value into the formula

[tex]I_{0}=0.95(1+4.4\times10^{-3}(2900-293))[/tex]

[tex]I_{0}=11.84\ A[/tex]

Hence, The current when the filament at room temperature is 11.84 A.