The power dissipated in a resistor W is proportional to the square of the voltage V , r to be a constant; W = rV2 If r = 3, then V (with a very good approximation) is considered a normal random variable with a mean of 6 and a standard deviation of 1. Calculate the probability P (W> 12).
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Given that $r = 3$, we can rewrite the equation as $W = 3V^2$. Show more…
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Problem 1: The power W dissipated in a resistor is proportional to the square of the voltage V. That is, W = rV^2, where r is a constant. If r = 3, and V can be assumed (to a very good approximation) to be a normal random variable with a mean of 6 and a standard deviation of 1, find E(W).
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A voltage V is a Gaussian random variable with a mean (μ) equals 1 and a variance (σ^2) equals 2. (a) Find the probability that the voltage V is less than -1.5 (b) Find the pdf of the power dissipated by an R-ohm resistor (P=V^2/R)
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A current of $I$ amperes flowing through a resistance of $R$ ohms varies according to the probability distribution $$f(i)=\left\{\begin{array}{ll}6 i(1-i), & 0<i<1 \\0, & \text { elsewhere }\end{array}\right.$$ If the resistance varies independently of the current according to the probability distribution$$ g(r)=\left\{\begin{array}{ll}2 r, & 0<r<1 \\0, & \text { elsewhere }\end{array}\right.$$ find the probability distribution for the power $W=$ $I^{2} R$ watts.
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