If a signal is transmitted at a power of 250 mWatts (mW) and the noise in the channel is 10 uWatts (uW), if the signal BW is 20M
1 answer:
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Answer:
maximum value of the power delivered to the circuit =3.75W
energy delivered to the element = 3750e^{ -IOOOt} - 7000e ^{-2OOOt} -3750
Explanation:
V =75 - 75e-1000t V
l = 50e -IOOOt mA
power = IV = 50 * 10^-3 e -IOOOt * (75 - 75e-1000t)
=50 * 10^-3 e -IOOOt *75 (1 - e-1000t)
=
maximum value of the power delivered to the circuit =3.75W
the total energy delivered to the element =
Answer:
maximum allowable electrical power=4.51W/m
critical radius of the insulation=13mm
Explanation:
Hello!
To solve this heat transfer problem we must initially draw the wire and interpret the whole problem (see attached image)
Subsequently, consider the heat transfer equation from the internal part of the tube to the external air, taking into account the resistance by convection, and conduction as shown in the attached image
to find the critical insulation radius we must divide the conductivity of the material by the external convective coefficient
