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Greg A.

Math aficionado & electrical engineer at Bose

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Calculus

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Question:

Find the general solution for the following differential equation: y' + y = 2

Greg A.

Answer:

This is a first order differential equation, so we can use the integrating factor method to determine the general solution. Our p(x) term will be just be 1 since we are in standard differential equation form, so the integral of 1 with respect to x is just x. Therefore, our integrating factor is e^(x) We then multiply our entire equation by the integrating factor: (y'*e^{x} + y*e^{x} ) = 2* e^{x} The two terms on the left side combine to (y*e^x)' which can be shown using differentiation and the chain rule. Therefore (y*e^{x})' = 2*e^{x} Integrate both sides: y*e^{x} = 2*e^{x} + C Solve for y: y= 2 + C*e^{-x} Therefore, the general solution to this differential equation is y = 2+ C*e^(-x) where C is a constant that can be determined with an initial condition (if provided).

Electrical Engineering

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Question:

You design an RL circuit and apply an AC voltage source across the two components. If you use a DMM to measure 14 Vrms across the 100 ohm resistor and 12Vrms across the 50mH inductor, what is the source voltage supply?

Greg A.

Answer:

Contrary to what our gut tells us, Kirchoff's Voltage Law doesn't work as we expect it to. In other words, you can't just add the amplitudes together because we have to take into consideration the phase of these voltages. First let's determine the cutoff frequency (or -3dB frequency) which is where the magnitude complex impedance of the resistor is equal to the complex impedance of the inductor. In other words, Xr = Xl -> R = jωL since w is angular frequency, ω=2*π*f, so R=j*(2*π*f)*L. Find the magnitude |R|=|j*(2*π*f)*L| -> R = 2*π*f*L f=R/(2*π*L) f=(100ohm)/(2*π*50mH) = 318.47 Hz For any frequency much less than 318.47 Hz, or as we take the limit as f approaches DC/0 Hz, the impedance of the inductor becomes negligible which means its voltage leads the resistor's voltage by 90 degrees. Let Vr be the voltage across the resistor and Vl be the voltage across the inductor, then Vr = 14*sqrt(2)*cos(wt) and Vl = 12*sqrt(2)*cos(wt+90). As we know cos(wt+90) = -sin(wt), our source voltage can be simplified as 14*sqrt(2)*cos(wt)-12*sqrt(2)*sin(wt). The resultant peak amplitude can be found by either graphing and finding the peak voltage or by taking the square root of the sum of squares. In other words, sqrt[(14sqrt(2))^2 + (12sqrt(2))^2] = 26.077 Vp

Statistics

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Question:

Assuming test scores in Mr. Andrus's class are normally distributed, 16% of students scored above an 84 and 2.5% of students scored below a 42. What is the mean and standard deviation of this distribution?

Greg A.

Answer:

This question is where the 68-95-99.7 rule comes into play. What this says is that if a distribution is normal, approximately 68% of the observations lie within 1 standard deviation of the mean (+/- 1 σ), approximately 95% of the observations lie within 2 standard deviations of the mean (+/- 2σ), and approximately 99.7% of the observations lie within 3 standard deviations of the mean (+/- 3σ). Since we know 16% of students scored above an 84, that means that 84% did not. Normal distributions state that the median is equal to the mean, therefore we know the average is at 50%. Taking half of the 68% gives us the additional 34% of observations greater than the mean (50%+34%) which implies that μ + σ = 84. Similarly, we know that 13% of students scored below a 42. If we compute 50-2.5%, we find that the difference is 47.5% which is exactly half of the data contained within two standard deviations of the mean. Therefore, we know that μ - 2σ = 42. Using a simple system of equations can help us determine both the average and standard deviation: μ + σ = 84 -(μ - 2σ = 42) 3σ=42 σ = 14 Then we can solve for μ: μ = 84 - σ = 84-14 = 70

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