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11

Corresponding wavelength is 22.11 meters long, but we want also to emit our EM waves into the environment. This means if we get a nice half-wave dipole antenna we would need it about 11 meters in length, $\lambda/2$. Which is quite large for mobile device. Ok, lets reduce size by using quarter-wave antenna as in WiFi, based on ideas of quarter-wave ...


8

I found a general, qualitative answer in David Blackstock's book Physical Acoustics, on page 46: Impedance is often described as the ratio of a "push" variable $q_p$ (such as voltage or pressure) to a corresponding "flow" variable $q_f$ (such as current or particle velocity). I also received a nice answer to this question on another Q&A site ...


7

I'll give the answer to this question using an unusual method that showed up in the American Mathematical Monthly's problem section perhaps in the late 1970s. This is not necessarily the easy way to solve the problem, but it works out nicely from an algebraic point of view. The way most people solve most resistance problems is to use series and parallel ...


7

I'll take it step by step here. First I'll write the answer for the first few cases with circuit analysis. Then I'll apply a reduction to show the pattern that the problem arrives at. N=1 $$Z = R+R=2R$$ N=2 $$Z = R+\frac{1}{\frac{1}{R}+\frac{1}{R + R}} = R \left( 1+\frac{1}{1+\frac{1}{1 + 1}} \right)=\frac{5}{3} R$$ N=3 $$Z = ...


6

For any given $n$, you can work it out via the rules for series and parallel resistors, but to get a general formula, valid for all $n$, doesn't look easy to me. The best way I know of is to get a recursive relationship giving the resistance of an $n$-step ladder in terms of an $(n-1)$-step ladder. If I'm not mistaken, the $n$-step ladder can be thought of ...


6

You might find the Yahoo "home_transistor" group a useful resource. There's also a series of videos on YouTube by Jeri Ellsworth including some where she makes transistors. In one, in particular, she takes the crystal out of a germanium point-contact diode and turns the crystal into a point-contact transistor (much like the Bell Labs transistor.) There ...


5

What makes it a good idea to use RMS rather than peak values The rms value, not the peak value, is the equivalent DC value that gives the same average power. Recall that power is the product of voltage and current: $p(t) = v(t) \cdot i(t)$ For a resistor, we have: $p(t) = R[i(t)]^2$ To find the average power, we must take the time average of both ...


5

While I agree with Alfred Centauri's answer, I am not sure it gives a direct answer to your specific question for the following reason. If there is a receiving antenna somewhere, there is always some reflection, however minute. If the antenna is connected to a circuit, the conditions of reflection will change, and the transmitter can "notice" that. In ...


5

Well, believe it or not, eels have been used to power Christmas trees (youtube link), so powering an electric motor isn't quite out of the question. However, Eels emit that 400 V at 1 A = 400 Watts (though the youtube video says that the eel was emitting 800 Watts). An electric car requires over ten thousand Watts of power to operate. So in theory, you ...


4

As suggested by Manishearth, one can perform a $Y$-$\Delta$ transform from $Y$-resistances $R_1$, $R_2$ and $R_3$, to $\Delta$-conductances $G_1$, $G_2$ and $G_3$ (using a $123$ symmetric labeling convention), cf. Fig.1 below. A x----x------x-----[3]-----x------x----x B | | | | [4] [2] [1] [5] | ...


4

The $A/W$ units refer to the current (in Ampère) produced per Watt of light incident on the photodiode. This current-production happens when the diode operates in the so-called photoconductive mode. Since your question wasn't on the inner workings of a photodiode, I won't expand on this, but Wikipedia contains some more information if desired.


4

As an intermediate step, consider a sinusoidal source driving an infinite transmission line with some characteristic impedance $Z_0 = 50\Omega$. The source "sees" a real impedance of $50\Omega$ and so, power is delivered to the line and, since the TL is infinitely long, the power is transported down the line, via an electromagnetic wave, without reflection. ...


4

The way every electric car works is by converting electrical energy into kinetic energy. The externally released energy of a discharge of Electrophorus electricus has been studied and has been found to be around $17 \mu\mathrm J$ per discharge. Since kinetic energy is $E_k = \frac{1}{2}mv^2$, if we assume a combined driver+vehicle weight of 1000 kg, then ...


3

Here is how I would do it, following the method outlined by kleingordon in a comment. This method is less cool but more general than Carl Brannen's answer, because it will work even in the case where there are crossing wires and you can't rearrange it into a single sheet of resistive material. Let the electric potential at $A$ be $V_A$ and that at $B$ be ...


3

To (hopefully) answer both your questions simultaneously, think of the concept this way: let's say I have an electrical circuit which consists of a battery (your EMF) connected by wires to some unknown electrical setup within a black box. Nothing appears to be melting or catching on fire within sight (which would imply the likely existence of a short ...


3

Resistors are generally used to dimension electrical devices to the ranges in voltage, current, time constants, what have you, that are needed. In this specific example the resistor is used to dimension the voltage drop in case one of the inputs has low voltage (lower than $V$), so that a current flows from $V$ to the input (it can only flow in this ...


3

I'm not sure why the resistor to ground from B is there, but you are incorrect at point D, the capacitor doesn't pass the DC level as you've indicated. It's a high-pass filter with C and R, so basically you need to move the DC-level on the Vd plot to ground - but keep the two transients like you've plotted them. That is, the curve should start at ground and ...


3

Uh... I liked this question. And a quick look at the Google or Google scholar did not give me much. So is written here more what I grasp people understand about impedance. Like Noldorin was saying I take impedance as how much is impeding power to be transferred from one place to another. In a homogeneous medium power is transferred as wave unimpeded but when ...


3

In many applications we are interested in the power. For example your electricity bill is based on the power you consume. For a DC source the power is: $$ W = VI = \frac{V^2}{R} $$ and for an AC source (assuming a resistive load so the voltage and current stay in phase): $$ W = V_{rms}I_{rms} = \frac{V_{rms}^2}{R} $$ So using the RMS values makes the ...


3

It's surprisingly difficult to find a nice simple description of how a transistor works. This description is from my old physics book - I suspect this may be oversimplified and I'm sure a complete description would run to lots of equations! Anyhow, this is what an NPN transistor looks like: so as you say, the collector-base junction is reverse biased and ...


3

Before I start, I should say, I am yet to meet a physicist who has a clue about the practical aspects of loudspeaker design. Likewise about audio engineers ;-) so the fact you are asking this, means you are well set up for making some good speakers. John Bird has written an electronics textbook which could be very useful for you which covers all the ...


3

Consider an ideal transformer for simplicity (from Wikipedia). (the voltages and currents shown in the picture are phasors) If you consider the above circuit (transformer) as a block in a larger circuit, because it is an ideal transformer you will not have any energy loss in the block. So you can write: $$I_P \times V_P + I_S \times V_S=0$$ and also, ...


2

The quotes you give do not really explain much. With a conductor one can short the static field of the atmosphere, one does not need oscillations for that. I think it was Benjamin Franklin with reaching the clouds who discovered that first hand? Lightening strikes when naturally induced upward leaders reach and meet the charged cloud downward leaders ...


2

Real LC circuits have some resistance, which wastes some of the energy as thermal radiation, and the cycling eventually dies. I think they also have some other non-idealities that allow energy to escape as far field electromagnetic radiation, correct? What are these non-idealities? Are they independent of the resistive component? ...


2

The small matter of the magnetic saturation of iron really ought to be mentioned in this discussion. In practise, the best iron cannot sustain a magnetic field of more than about 1 Tesla, or 1 volt-sec/m^2. We must then ask: how much current does it take to drive a field of that magnitude? That depends of course on the magnetic permeability of the iron and ...


2

Let's imagine that we live in beautiful world of brush-less motors without friction. Limitation must come from coil resistivity - the more power you pump into coils, the more losses you have due to their non-0 resistivity. At high RPM resistance also increases due to skin effect (reducing effective cross-section of the wire). So if one want to have more ...


2

There are some missing data in your question. What voltage does the batteries have, I'm going to assume 12V since it is common. Battery capacity, you typed it as 150A but I guess it was 150Ah. Please note that a normal car battery on a car like a VW Golf has approx 60Ah so 150Ah is a quite big battery. Output power, you state that you have 1000V (Volts) ...


2

Mechanical impedance matching does have an application in electrical transmission lines (or any elastic cable/structure vibration) because it helps describe how much of the wave gets through a discontinuity, and how much is being reflected. Mechanical impedance is force over velocity and along the cable it is equal to tension over wave speed. The ...


2

Your approach is not correct. Once you connect the two capacitors what happens is that the plates that are connected with a wire will have the same potential because they form a conductor. Suppose the initial charges are $Q_{1i}=Q$ and $Q_{2i}=0$. Because the potential difference across both capacitors is equal you get $$ ...



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