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6 days ago

8

The filament of the light bulb is made of tungsten. The resistance of the light bulb at room temperature (20∘C), measured by an

ohm-meter, is about 3.0 Ω. Based on this information, estimate the temperature of the light bulb at 12.0 V. Neglect the thermal expansion of the filament.

1 answer:

Ostrovityanka [2.2K]6 days ago

7 0

Explanation:

The formula illustrating the relationship between resistance and temperature is as follows:

R =

$R_{o} + \alpha [T_{2} - T_{1}]$

where, R = final resistance

= initial resistance $R_{o}$

= temperature coefficient of resistivity $\alpha$

= final temperature $T_{2}$

= initial temperature $T_{1}$

Given data as follows.

$T_{1} = (20 + 273) K = 293 K$

R = 36 ohm,

= 3 ohm $T_{2}$

= 0.0045 $R_{o}$

Substituting the provided values into the above formula gives us the following.

R = $\alpha$

36 =

$R_{o} + \alpha [T_{2} - T_{1}]$

=

$3 + 0.0045 \times [T_{2} - 293]$

= 7626.33 K

$T_{2}$ Thus, it can be concluded that $\frac{34.3185}{0.0045}$ the temperature of the light bulb at 12.0 V is 7626.33 K.

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Provided:

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