Question

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.

Answer 1

Answer:

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

Explanation:

Given that,

Thermal coefficient of resistivity $\alpha=4.4\times10^{-3}\ K^{-1}$

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

$R=R_{0}(1+\alpha(T_{2}-T_{1}))$....(I)

Using ohm's law for value of resistance

$V = I R$

$R = \dfrac{V}{I}$

Put the value of resistance in the equation (I)

$\dfrac{V}{I}=\dfrac{V}{I_{0}}(1+\alpha(T_{2}-T_{1}))$

$I_{0}=I(1+\alpha(T_{2}-T_{1}))$

Put the value into the formula

$I_{0}=0.95(1+4.4\times10^{-3}(2900-293))$

$I_{0}=11.84\ A$

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