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23

Suppose you are using a waterwheel to do some form of work (e.g. grind corn). You need a head of water to make the wheel move, and you could use either 1kg of water at a height of a million metres or you could use a million kg of water at a height of one metre. In both cases the water would do the same amount of work as it flowed through your wheel. The ...


10

firstly, you hooked up the red wire to negative and the black wire to positive--you got it backwards. secondly, computer fans operate at 12V, so you are short of several batteries. with luck, some fans might start spinning at 7V .


10

Perhaps I can clarify what I'm trying to get at with the famous waterwheel analogy 99 years ago, Nehemiah Hawkins published what I think is a marginally better analogy: Fig. 38. — Hydrostatic analogy of fall of potential in an electrical circuit. Explanation of above diagram In this diagram, a pump at bottom centre is pumping water from right to ...


9

does this mean that Ohm's law just fails in this case Ohm's law is not universal. The ideal resistor circuit element is defined by Ohm's law but not all circuit elements obey Ohm's law; Ohm's law only applies to ohmic devices. Physical resistors and conductors approximately obey Ohm's law but, for example, semiconductor diodes, transistors, ...


8

For static charges, the relationship is V (voltage) = Q (charge) / C (capacitance). Capacitance is a function of the shape, size and distance between objects, which are all continuous values. (Well, I suppose you could argue that shape and size are quantized to the atomic spacing of the object's material, but you can't say the same thing for distance.) So ...


8

Let me first comment that the statement electric fields cancel while the electric potentials just add up algebraically is not actually correct. Electric fields add due to the principle of superposition (see the section on superposition in the wikipedia article). However, when two electric field vectors are of the same magnitude but point in ...


8

1) If you are thinking of harmonics as sinusoidal waves, well yes, ALL waveforms are (can be described as) sum of harmonics. This is essentially the idea of the Fourier analysis. The problem is that to exactly reproduce a desired waveform you need in general an infinite number of harmonics. This is for instance the case of square waves. So in reality you ...


7

I've found this to be the case, too. Generally, my shop lights will flicker when turning on, especially when it is colder outside . There are two basic phases to this kind of light bulb: a start-up phase, and an operating phase. The start-up requires more voltage, because you are initiating the plasma stream between the terminals of the bulb. So, these ...


6

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 ...


6

Your analogy with water flow through a pipe is correct. In that analogy a voltage source corresponds to a pump that generates a specific pressure, and a current source corresponds to a pump that generates a specific water displacement (volume per second).


6

It is not the most straightforward viewpoint to say that a device "needs 1500 watts". This is more a consequence than a condition. What happens is that you create an electric circuit by plugging in a device into the outlet. That circuit follows Ohm's law: $$V = I R$$ So for a given voltage and resistance a certain current $I$ will flow. The power is simply ...


6

Voltage is a continuous function. If you are a certain distance from a (point) charge $q$, the potential is $$V=\frac{q}{4\pi\epsilon_0 r}$$ By adjusting the value of $r$ to anything you want (not quantized), you can get any potential you want. And so yes, when you do any analog-to-digital conversion, you will "destroy" a certain amount of information. ...


5

The term Resistance does not come into play while dealing with Photoelectric Effect. The latter is related to the emission of electrons when the surface of a metal(or any substance) is hit by photon particles(photon is the unit particle making up the light that we talk of). Here the more important concept is that of Work Function, i.e. the minimum amount of ...


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 ...


5

By Ohm's law, which states V = IR, where V is the voltage accross a resistor, I the current thru it, and R the resistance. The units work out so that no additional proportionality constant is required when V is in Volts, I in Amps, and R in Ohms. For example, if the 1.5 V battery is connected to a 47 Ω resistor, then 32 mA will flow. Of course you ...


5

EDIT: Put simply, potential difference is the work done by electrostatic force on a unit charge, while EMF is the work done by anything other than electrostatic force on a unit charge. I don't like the term "voltage". It seems to mean anything measured in volts. I'd rather say electric potential and electromotive force. And the two are fundamentally ...


5

When there is no resistance, as is the case with an ideal wire, any value of current satisfies Ohm's Law: $V = I R$ since both $V=0$ and $R=0$. UPDATE: But isn't V is like what causes the current? Perhaps a mechanical analogy of the resistor will help. Consider the dashpot where the velocity of the arm is analogous to current while the force acting ...


5

From a more general point of view, electric potential is not a scalar, but a component of a 4-vector (http://en.wikipedia.org/wiki/Electromagnetic_four-potential ) - it is not invariant with respect to boosts. Edit: in answer to question: Well the potential is just one component of the four-vector. Not the four-vector itself. So isn't it a scalar after ...


4

Voltage (Joule per coulomb) is a measure of electric potential energy gained by a positive test charge, or the work done in moving a positive test charge from infinity to a point in a positive electric field. This energy gained is due to the conservative electrostatic force between charge. When a charge gains potential it naturally does work equal to its ...


4

Voltage is voltage. Electrostatic voltage is still just voltage. Materials that can store electrostatic charges are insulators (or are insulated by something), so the charge cannot leak away. Because of this they can build up a high voltage, with relatively few electrons (charge). A conducting material on the other hand needs a power source to keep the ...


4

Suppose we have a source of electrical energy, say a battery, that puts out 100 Volts. It is connected through wires with a total resistance of 1 ohm to a heater with a resistance of 99 ohms. The battery sees a total resistance of 100 ohms, and thus pushes 1 Ampere of current through the circuit. The battery is delivering energy at 100 Watts The Power ...


4

Assuming you mean a macroscopic potential difference, the largest I know about was in the Nuclear Structure Facility accelerator at the Daresbury laboratory in the UK, and this was 30MV. The Wikipedia article on electrostatic particle accelerators claims this is about the highest possible in such devices.


4

Well, your lecturer certainly shouldn't have put it like this, however it's true that you have got a lot wrong here. It's stuff you definitely will need to understand better if you're studying power engineering. First, you seem to think that electrons are attracted by magnetic north poles. They aren't; in fact stationary charges and magnetic fields aren't ...


4

When you say "we see that if the current doubles then the potential difference is halved," you're assuming that $P$ is fixed, whereas when you say "doubling the current should increase the potential difference" you're assuming $R$ is fixed. But in fact, it isn't possible to change the current while keeping both of these things constant. Let us assume that ...


4

If you are wondering about causality, then I think that voltage difference $\Delta V$ is fundamental as it is the cause, and the current $I$ is the consequence. If you want to have current, you need movement of the charges. The most obvious way to move charges is to act upon them with electric field, and each electric field is accopmained with voltage ...


4

Electrons are accelerated by the constant applied electric field that comes from the external potential difference between two points, but are decelerated by the intense internal electric fields from the material atoms that makes up the circuit. This effect is modeled as resistance.


4

The battery has in both cases the same energy content, so it just depends on which method uses more energy per time. This power depends on the resistance $R$ you use to connect both terminals, with a given voltage $U$ derived from Ohm's law: $$P = U^2 /R$$ So, the smaller the resistance, the faster your battery will lose it's stored energy. The copper wire ...


4

Take whatever your wall supplies (110V, 220V, etc.), and multiply it by 1.8A. Note that this is maximum power usage, not constant.


4

That schematic is quite confusing; draw it with higher voltages at the top, lower voltages at the bottom, and signals flowing from left to right per standard procedure. It's also helpful to label components and nodes so that you can solve the equation symbolically. I haven't labeled some of the nodes here because they can be represented by the names of ...



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