Answer :
The minimum safety power that is safely dissipated using a transistor of height 0.4cm and diameter 0.06cm is 0.04665
Calculation for the exposed area
[tex]\pi DL +\frac{\pi }{4} D^2\\\\Area = 3.14 * 0.6*0.4+\frac{3.14}{4} *0.6^2[/tex]
D = diameter = 0.6cm
Length = height = 0.4cm
Area = 1.0367cm²
minimum safety heat that is dissipated
hA (Ts - Tair)
30 * 1.0367 x 10⁻⁴(70-55)
= 0.04665
The minimum safety power that is safely dissipated is 0.04665W
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