Tag Info

37

In addition to the other answers, here is something for the intuition: $$V=RI$$ More "pressure" $V$ is required to keep the flow $I$ of charges constant when the flow is resisted by $R$. A thin wire has higher resistance than a thick wire, $R=\rho L/A$) analogous to a "bottleneck" in a traffic jam.

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 ...

20

The identity $$V = K \frac{dV}{dt}$$ is only guaranteed with a constant $K$ if your assumptions actually hold. The first identity $V=RI$ only holds for a resistor, while the other holds for a capacitor. So in this sense, the letters $V,I$ in these equations mean something else. In one of them, it's the current through (or voltage on) a particular resistor, ...

16

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 ...

14

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 ...

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 .

9

There are various ways to decide which of the assumptions are primary and which of them are their consequences but $E=VQ$ may be most naturally interpreted as the definition of the potential. The potential energy is a form of energy and the potential (and therefore voltage, when differences are taken) is defined as the potential energy (or potential energy ...

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, ...

9

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 ...

9

Think of plumbing for a close analogy. Voltage is how hard you are pushing, and current is how much flows. The relationship writes itself: why would you get more or less flow from the same pump? The measure of how much effort is used to get flow (it makes more sense as the reciprical: how much flows for a unit of effort) is the interesting property, and ...

8

This isn't really an answer, because your question doesn't provide enough information for an answer. However it explains what you need to do. In fact this is exactly what I did (back in 1983!) to measure the resistivity of evaporated silver films. If you have the film formed on some substrate you need to score two lines to leave a long narrow track like ...

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 ...

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 ...

7

What flows is not the voltage but the charge, and that flow is called current. Voltage can be without a current, if you have a single charge, that charge induces a voltage in all space, even if it's empty. Voltage, in the most physical way, is a scalar field that determines the potential energy per unit charge at every point in space. Now, you can't have ...

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 ...

7

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 ...

7

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. ...

7

This is true but strictly limited to RC circuits without external sources: that is, a resistor hooked up to a capacitor with nothing else in between. In that case, $V$ is indeed proportional to $\dot V$, with a crucial minus sign in between: $$\dot V=-\frac1\tau V,$$ where $\tau>0$ is some constant. This equation implies that $V(t)=V(0)e^{-t/\tau}$, ...

7

The physical meaning of the capacitance is precisely given by $\mathrm{d}Q=C\cdot \mathrm{d}V$: $C$ tells you how much charge there will be in the capacitor per voltage applied. For all capacitors, the linearity holds fairly well. Generally speaking, capacitance is given by a Q-V curve, which may consist of a linear region, a saturation region and a ...

7

Another possibility is that you are having to break through the oxide film that all Aluminium has in air. The voltage needed to do that can vary considerably. You might need to use sharp steel electrodes that will punch through the oxide and contact the metal directly.

7

Have you looked at Drude's model? I was taught something like that back at school and have kept it in mind as a intuitive way of understanding it. We want to understand why the current (rate of flow of charge) should be linear with the potential difference. The Drude idea is, as you noted, related to friction. Firstly, the EM field is linear in the ...

6

A negative voltage means that you have hooked your power supply across your device backwards. Purely resistive devices, like resistors and lamp filaments, don't care which way current flows through them, only how much current flows. Such devices will always have current-voltage curves which are "odd" functions, with $I(-V) = -I(V)$, as in your graph. It is ...

6

This comes from classical electrodynamics, there is no need to go to Standard model theory or quantum electrodynamics for this. The simple answer is that electric potentials, like electric fields, are just a way of characterizing the way charged particles interact with each other. So, charged objects create voltage analogous to the way that they create ...

6

Generally, no. You've written two equations. The first relates voltage and current for an isolated resistor. The second relates voltage and current for an isolated capacitor. Given only those two expressions, there's no reason at all that they can or should be combined. That is, you have no circuit, only isolated components. There are principles for ...

6

The instantaneous power expended in a resistor is proportional to $V^2$ (i.e. independent of its direction), so the average power expended is given by the mean square voltage! The square of the average absolute voltage will not yield the power expended in the resistor. Edit: And actually this close-to-duplicate question does have more extensive answers ...

6

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 ...

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

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 ...

Only top voted, non community-wiki answers of a minimum length are eligible