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.
(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.
1 answer:
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Answer:
Part 1: It would be a straight line, current will be directly proportional to the voltage.
Part 2: The current would taper off and will have negligible increase after the voltage reaches a certain value. Graph attached.
Explanation:
For the first part, voltage and current have a linear relationship as dictated by the Ohm's law.
V=I*R
where V is the voltage, I is the current, and R is the resistance. As the Voltage increase, current is bound to increase too, given that the resistance remains constant.
In the second part, resistance is not constant. As an element heats up, it consumes more current because the free sea of electrons inside are moving more rapidly, disrupting the flow of charge. So, as the voltage increase, the current does increase, but so does the resistance. Leaving less room for the current to increase. This rise in temperature is shown in the graph attached, as current tapers.
