Recent content by SumDood_

Well, yes. That is what I am trying to find out. How does one determine whether any value of central tendency is appropriate or not for a given sample?

Then, does that mean that the mode becomes a relevant central tendency measure? I would say it isn't because it does not provide much information about the sample. If we already know the probability of each sample being drawn is equal, then what benefit does the mode value provide?

When taking a sample from a uniform distribution, every real number within 0 and 1 has equal probability of being drawn. I would think that there would be no mode, as no 2 values would be the same.

A study on strength properties of high-performance concrete obtained by using super-plasticizers and certain binders recorded the following data on flexural strength (in mega-pascals, MPa) from 28 tests: 6.1, 5.6, 7.1, 7.3, 6.6, 8.0, 6.8, 6.6, 7.6, 6.8, 6.7, 6.6, 6.8, 7.6, 9.3, 8.2, 8.7, 7.7...

$$\left| \frac{V_R}{V_S} \right| ^2 = \frac{R^2}{R^2 - 2R^2 \omega^2 L C + R^2 \omega^4 L^2 C^2 + \omega^2 L^2 }$$ Let's try one more time: \begin{align*} \frac{d}{dC}\left| \frac{V_R}{V_S} \right| ^2 &= \frac{d}{dC}\left(\frac{R^2}{R^2 - 2R^2 \omega^2 L C + R^2 \omega^4 L^2 C^2 + \omega^2 L^2...

@The Electrician I tried this: \begin{align*} V_R &= V_S \cdot \frac{R}{R + \left( j \omega L \right) \left(j \omega R C + 1 \right)} \\ \\ V_R &= R\cdot V_S \cdot \left( R-\omega^2RLC+j\omega L\right)^{-1} \\ \\ \frac {dV_R} {dC} &= -R\cdot V_S \cdot \left( R-\omega^2RLC+j\omega L\right)^{-2}...

You guy's help is extremely appreciated! \begin{align*} 1 &= \frac{R^2}{\left( R - \omega^2 LRC\right)^2 + \left(\omega L\right)^2} \\ \\ R^2 &= \left(R - \omega^2 LRC\right)^2 + \left(\omega L\right)^2 \\ \\ R^2 &= R^{2}-2\omega^2R^2LC+\omega^4R^2L^2C^2 + \left(\omega L\right)^2 \\ \\ 0 &=...

@The Electrician Tried again: \begin{align*} \frac{R}{R-(j\omega ^{2}RLC + j\omega L)} \cdot \frac{R-\omega ^{2}RLC-j\omega L}{R-\omega ^{2}RLC-j\omega L} &= \frac{R^{2}-\omega ^{2}R^{2}LC - j\omega RL}{(R-\omega ^{2}RLC)^{2}+(\omega L)^{2}} \\ \\ &=\frac{R^{2}-\omega ^{2}R^{2}LC}{(R-\omega...

@The Electrician So I tried rationalization, but ended up with an expression that has j in the denominator: \begin{align*} \frac{R}{R+(j\omega L)(j\omega RC+1)} \cdot \frac{R-(j\omega L)(j\omega RC+1)}{R-(j\omega L)(j\omega RC+1)} &= \frac{R^{2}-R(-\omega ^{2}RLC + j\omega L)}{R^{2}-[(-\omega...

$$1=\left| \frac{R}{R+(j\omega L)(j\omega RC+1)}\right|$$ I think I get it: $$(j\omega L)(j\omega RC+1) = -2R$$ OR $$(j\omega L)(j\omega RC+1) = 0$$ \begin{align*} (j\omega L)(j\omega RC+1) &= -2R \\ -\omega ^{2}RLC + j\omega L &= -2R \\ -\omega ^{2}RLC &= -2R - j\omega L \\ C &= \frac{-2R -...

##V_{S} = 120V ## ##f = 60Hz## ##L = 20mH## ##R = 10\Omega ## (a) Assume the capacitance is omitted. What is the resistor voltage? ##X_{L} = j\omega L = j2.4\pi## ##V_{R} = Vs\cdot \frac{R}{R+X_{L}} = 76.5067-j57.6847## ##V_{L} = Vs\cdot \frac{X_{L}}{R+X_{L}} = 43.493+j57.6847## (b) What...

It seems that everyone is finding this question to be weird. So, I'll post the solution that is at the end of the book: ##P = \frac{1}{2}VIcos(\psi)## ##Q = \frac{1}{2}VIsin(\psi)## Instantaneous power ##p=2Pcos^{2}(\omega t)+Qsin(2\omega t)##

You are right that this kind of circuit won't exist in the real world, but the question I've described is as stated in the book.

What I have provided is what is mentioned in the question. Let's assume that the current source is delivering power.

Help me understand what I am missing. So this is the circuit: Given: ##V_{RMS} = 120V## ##I_{RMS}= 10A## ##v=Re \{Ve^{jwt}\}## ##i=Re \{Ie^{j(wt-\psi)}\}## (a) Calculate and sketch real and reactive power P and Q as a function of the angle ψ (b) Calculate and sketch the instantaneous power...