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Sagot :
Answer:
Following are the solution to the given points:
Explanation:
In point a:
Given:
[tex]Temperature =20^{\circ } \ C, \\\\Resistance (R_o) = 22 \Omega[/tex]
However, the resistance of the bulb is 120 V or 100 W.
[tex]R= \frac{v^2}{P}=\frac{120^2}{100} =144 \Omega[/tex]
Let Tungsten's standard temperature coefficient be = 0.0045 [tex]20^{\circ} \ C[/tex]
Calculating temperature:
[tex]\to R=R_0(1+ \alpha (T-T_0))[/tex]
[tex]\to 144=22(1+0.0045(T-20))\\\\\to 144=22+0099(T-20)\\\\\to T -20 1232.32\\\\\to T =1252.32^{\circ } \ C \\[/tex]
In point b:
Potential consequences for indirect failures on iridescent bulb power device atmosphere does have a minor impact on defects, humidity, pressure, and surface that light bulb within the glass as well as the sensor attributable to the mercury.
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