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Nadusha1986 [10]
3 years ago
9

A Gaussian random voltage X volts is input to a half-wave rectifier and the output voltage is Y = Xu (X) Volts were u (x) is the

unit step function. Assume X has mean 0 V and variance σ^2 V^2. The output voltage Y is then applied across a (nonrandom) resistance of R ohms. The answers below should be expressed in terms of the Φ or Q functions or in closed form (no integrals).
(a) Find the probability that the current which flows through the resistor exceeds 1 Amp.
(b) Find the probability that the power which is dissipated in the resistor exceeds 1 watt.
(c) Find the mean and variance of the current which flows through the resistor.
(d) Find the mean and variance of the power which is dissipated in the resistor.

Engineering
1 answer:
adelina 88 [10]3 years ago
7 0

Answer:

Please look at attachment carefully.

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3 years ago
A school is playing $0.XY per kWh for electric power. To reduce its power bill, the school installs a wind turbine with a rated
Semmy [17]

Answer: Your question has some missing figures so kindly plug in the values into the solution provided to get the exact amount of money saved

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Explanation:

<u>Calculating  the amount of electric power generated by wind turbine</u>

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5 0
3 years ago
If the old radiator is replaced with a new one that has longer tubes made of the same material and same thickness as those in th
Nookie1986 [14]

Answer: hello some parts of your question is missing attached below is the missing information

The radiator of a car is a type of heat exchanger. Hot fluid coming from the car engine, called the coolant, flows through aluminum radiator tubes of thickness d that release heat to the outside air by conduction. The average temperature gradient between the coolant and the outside air is about 130 K/mm . The term ΔT/d  is called the temperature gradient which is the temperature difference ΔT between coolant inside and the air outside per unit thickness of tube

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3 0
3 years ago
Shows a closed tank holding air and oil to which is connected a U-tube mercury manometer and a pressure gage. Determine the read
damaskus [11]

Answer:

P_2-P_1=27209h

Explanation:

For pressure gage we can determine this by saying:

The closed tank with oil and air has a pressure of P₁ and the pressure of oil at a certain height in the U-tube on mercury is p₁gh₁. The pressure of mercury on the air in pressure gauge is p₂gh₂. The pressure of the gage is P₂.

P_1+p_1gh_1=p_2_gh_2+P_2

We want to work out P₁-P₂: Heights aren't given so we can solve it in terms of height: assuming h₁=h₂=h

P_1-P_2=p_1gh_1-p_2gh_2=(55)\cdot{32.2}h-845\cdot{32.2}h

P_2-P_1=27209h

3 0
3 years ago
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Answer:

A.

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