# Tag Info

3

No, this sadly will not work. You can of course get water up this way, but it won't disconnect from the glass tube. As capillary forces result from surface tension, to make the water fall back down, you will need to overcome this surface tension. This turns out to cancel the "won" energy. With a colleague I already discussed a more sophisticated way to try ...

1

How do voltage and voltage drops over a circuit relate to work done? The Volt unit is energy normalized to unit charge; Joule per Coulomb. Since the Amp unit is Coulomb per second, the product of the voltage across and current through a circuit element is the power associated with the circuit element. For a DC circuit, voltage and current are constant ...

1

But then that means that the electron in the 5 ohm circuit would have done 5x the amount of work (or work done on it) of the 1 ohm circuit over 5x the duration. You're confusing work with power here. Work has nothing to do with duration. If an electron crosses a potential difference of $V$ with any resistance in between, the work done is the same, ...

1

A few preliminary ideas which might help: It doesn't really matter what the speed of the electrons is - a current of 1 C/s (=1 A) just means that a coulomb worth of charge (equal to $6.2 \times 10^{18}$ electrons) passes each point in the circuit each second. Perhaps there is one electron travelling so fast that it does $6.2 \times 10^{18}$ laps of the ...

5

Voltage is similar to height. It plays the same role for electric charge as height*gravity does for a ball on a hill. So high voltage means high potential energy the same way a ball being high up on a hill means high potential energy. Voltage is not potential energy, the same way height is not energy. However, if you have a certain amount of charge $q$, you ...

1

First a minor quibble. Reactance is the imaginary part of impedance and is thus real. $$Z = R + jX$$ Thus, inductive reactance is: $$X_L = \omega L$$ Second, reactance is the imaginary part of impedance. So, to fully understand why this is, you must understand the notion of impedance. For the ideal inductor, we have that the voltage across is ...

0

Because the voltage across an inductor leads the current by $\pi/2$ in phase, i.e. it is a quarter of a cycle ahead. $\arg(Z)$, where $Z$ is the complex impedance, is the voltage's phase relative to that of the current. In using the phasor notation, one represents a quantity $x(t) =A \cos(\omega\,t+\delta)$ varying sinusoidally varying with time by the ...

1

To build on Jan Dvorak's comment and also user1512321's answer: If you mean one wire, then no. The device would be accumulating (or losing) electrons. Even one wire / AC is very inefficient (the environment has to serve as the second wire, but it's not connected in a conductive way) In true DC conditions, the electron flow is wholly analogous to ...

1

You could deliver a short burst, but not continuous power. Not without having a sink of some kind. In fact the example you gave about the heat also needs a sink of some kind. If you transmitted heat down a rod, the rod would heat up until the thermal gradient would disappear. Then the heat flow would cease. The exact same thing happens with electricity - ...

0

I've been thinking about this lately but I think that u missed something you all are saying that much energy will be needed to make the coil spin I've an objection on this it depends on the shape of the coil and its position also material used. I think if the coil was made to be wide and not long and just flought horizontal away enough from the earth ...

0

Why does a capacitor charge only upto the voltage of the source? Step by step: (1) When the capacitor voltage equals the source voltage, the voltage across the resistor in the series RC circuit is zero (2) By Ohm's Law, the current through the resistor must be zero too. (3) Because it is a series circuit, if there is zero current through the ...

0

Here's how I would convince myself of the correct answer. Draw a circuit diagram showing the voltage source, resistor, and capacitor. (I assume it's in a simple series circuit?) Next, write out Kirchhoff's loop rule. You should find something like $V_\text{source}-V_\text{resistor}-V_\text{cap}=0$. Note that this equation is true at any time, not only for ...

0

What kills you is the current not the voltage, as you read on your books. Of course that you to have a voltage difference so the current can flow, but it does not determine how strong the current will be. I do not know if I would die if I touch something with 1 kV. That's because the current will depend on the sum of the resistance between me and the ground ...

0

I always thought this rule of thumb was a bit silly - current kills because it was driven by a voltage, otherwise there would have been no current. The rule arises because of the variability of skin resistance. Little voltage applied internally across your heart will kill you, but the skin's variability means that it is impossible to say what external ...

-1

It is possible to arrive at the expression for Ohm's law by using a simple classical model. The simplest treatment I've seen of this happened to be on an optics book: Pedrotti and Pedrotti's "Introduction to Optics", Chapter "Optical Properties of Matter", paragraph "Conduction current in a Metal". Basically free electrons in metals can be thought to obey ...

1

The current you are going to get through your body depends on the voltage and on the resistance. You can touch a 110 V exposed cable using a piece of metal or a piece of plastic - in both cases the voltage is the same, but the resulting current - and hence the danger - is greater in the first (metal) case.

0

Ohms law is not a "law" as much as it is an excellent approximation of an unavoidable material property. Imagine a charge in vacuum with an contain electric field along z. This charge will continue accelerating, therefore the current is unbound. In this example the system has an inductance (current changes with time at constant applied potential), but no ...

-3

The resistor only limits the flow of current not with $\partial I/\partial t$ example inductor it will opposes the flow of current when it will change with time means time i.e $T=\frac1f$ so frequency changes take in inductor that's why resistor is frequency independent inductor frequency dependent. $V=IR$, but Inductor $\mathrm{e.m.f}=-L ... 1 There is no way to derive Ohm's "law" from simple definitions of electric field and current. You have to understand the dissipative mechanisms at work in a system to validate the notion of Ohm's law. Sam29's answer introduces these dissipations through the concept of the resistivity$\rho$or its reciprocal$\sigma$of a material. The effectiveness of ... 0 Let's start with$E=Fq$, like you have up top. Rearranging to solve for q, we have$q=E/F$. We know that a change in charge creates a current,$dq/dt=I$, so substituting$\frac{d}{dt}\frac{E}{F}$for$\frac{dq}{dt}$, we now have,$\frac{d}{dt}\frac{E}{F} = I = V/R$You could also use the relationship$E=-\bigtriangledown{V}$to obtain a relation. 0 There are a number of ways you can examine the law in a microscopic view. One of them is this: An applied voltage creates an electric field, which superimposes a small drift velocity on the free electrons in a metal conductor. This drift velocity is way smaller than the speed of transmission in a conductor. Now, the basic relations are:$$I=\frac VR\\ ... 1 Actually, cell phones do work in Faraday cages these days. What happens is that the conductor in the cage is not ideal, and there is some amount of leakage of electromagnetic radiation to and from the inside of the cage, specially at high frequencies. In order for the cage to be perfectly blocking it would need to have no holes at all (hence it is no longer ... -1 Wires are in fact resistors, but with VERY VERY tiny amounts of energy being thermally dissipated by the current due to EXTREMELY low drops in voltage over large lenths of the wire. Thus, current DOES flow through the neutral wire, but the drop in potential along a length is literally far too small for your voltmeter to detect. Review Kirchoff's Voltage Law ... 1 You're almost right. But... 1138 kilowatts power output will give you 1138 kilowatt-hours in, well... one hour, not 1 second. Just leave out the$\times 3600$It's better to avoid weird units(like kilowatt-hours as much as possible, so another longer way is this:$1138 \text{ kilowatts}$mean$1,138,000\text{ joules/sec}$So in one year you'll get ... 0 First of all, your measurement unit is wrong. Energy is measured in, say, kWh, not kW/h. Second, you should not multiply by 3600, as the final result is in kilowatt-hours. 0 Though it is not common to use Fresnel lenses for electricity generation but 100 MW power plant is nearing completion in Rajasthan state of India using linear Fresnel lens technology. So to say that this technology is not feasible for large scale use is not correct and time may come if that above mentioned power generation goes smoothly, the scene may change ... 1 A plane charge would be an infinite 2-dimensional sheet with constant charge density. Already in a line charge you have neglected edge effects, because the$1/r$dependence holds true only near the line provided you are far away from the end-points. Similarly, for a plane, the constant electric field holds true provided that you are much closer to the plane ... 3 Sure, many tall buildings are hit by lightning several times per year. It would be quite expensive to replace the cables each time. It has to withstand currents in the range between 10kA and 100kA for a few milliseconds. A copper wire with a diameter of several centimeters will survive lightning easily. A thin wire on the other hand is usually vaporized as ... 2 The analogy is wrong. A voltage source can only shock us if it is able to pass a considerable amount of current through our body ( ~ 250 mA or so, I dont know the exact value but you can Google it ). The circuit that you are trying to discuss, does indeed have 36 Amps of current flowing through it, but once you connect yourself to the circuit, you are in ... 0 The way I have heard it explained is not by the container but the water drops themselves. Statistically there is no way you can get a perfectly neutral water drop every single time. Eventually you will get a drop with a charge of 0.000000001 Farads. This tiny imbalance is enough to set the experiment in motion into a positive feedback system. You can think ... 0 With respect to power loss concept, when we say that the power is dissipated (or lost as you call it) it means that power was dissipated (or spent) as something else which might be useful (as an example power dissipated in a perfect lamp where all power is converted into radiation) or not useful (example is portion of power lost in heating the motor of fan, ... 0 well basically your missing some additional equations. The current coming out of the$10 \Omega$resistor is the same as the one going into it. Hence$I_3$in your equation would be just$0.5A$. Similar thing for$I_1$: The current going through the$20 \Omega$resistor is$0.2A$. Note that the power supply also doesn't alter the current only the voltage. ... 1 The reason why you got electric shock is because the way you kept it in the bucket is improper or device is faulty. You might have noticed that there is an aluminum hook along with the plate extending which marks min and max water level marking. It should be in contact with water. This plate and hook is connected to earth. It ensure water in the bucket is ... 3 This device looks like a inherently bad idea safety-wise, for the reason you found. I don't know what exactly is inside the handle, but you have to assume all it is doing is connecting wall power to a resistive heating element. I imagine the outside of the heating element is intended to be insulated from everything else. (By the way, this has nothing at ... 4 Numbers 2 and 3 are correct. That 150 kV rating is for open circuit. Once you put a load on it the voltage is going to drop fast because it can not deliver the kind of power your are suggesting by assuming both voltage (150 kV) and current (1.5 A) will be constant. The body will also start to conduct and the impedance will drop as well as different layers ... 5 Your assumption that Ohm's law is fully accurate for the stun gun + human circuit isn't correct. A stun gun uses a capacitor to store charge and the capacitor is constantly being recharged to deliver a series of high-voltage pulses. A capacitor has a finite amount of charge. Once it's charged, that's it, it can never deliver more charge than that until it ... 1 There are a lot of details missing in your question but I will try to get you in the right direction. First, whenever you write vector components, you need to be clear in what coordinate system you're using. I will assume that$\mathrm{OX}$points to the right in your figure,$\mathrm{OY}$points up, and$\mathrm{OZ}\$ points out of the screen (so that it is ...

1

The ground which locates in the non-inverting terminal of an Op-Amp is called virtual ground.

1

No, glass and indeed all amorphous materials do not exhibit piezoelectricity because piezoelectricity is intimately connected to the crystal structure of the material. Roughly speaking, if the charges within the unit cell are asymetrically distributed then when the crystal is mechanically deformed the positive and negative charges may be displaced by ...

0

The issue is that for a piezoelectric material you need a common orientation of the dipole moments. In normal glass there is no preferred direction and therefore an electric field will not create a bulk change in length. On the other hand it is possible to manufacture a piezoelectric glass, e.g. from a compound related to Strontiumtitanate (link to journal ...

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