# Tag Info

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As you stated, we can think of a real battery as an ideal one with an internal resistance $R_i$. This battery is then connected to an external circuit with resistance $R$. Those 2 resistors form a voltage divider. If the EMF has a value of $V$ then the voltage measured across the external resistance is $V*R/(R+R_i)$. This voltage is equal to the EMF of the ...

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Diode laser with changeable wavelength, fan side with a prism and an array of sensors: wavelegth -> ampltude Laser and mirror to position the beam, fan side with array of sensors: position -> amplitude Laser with variable amplitude, fan side with electric thermometer: amplitude -> heat -> current -> amplitude Broken laser, fan side with regulator (work ...

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Sure, have multiple linear sensors, indicating amplitude.

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When we derive Ohm's Law using the Drude Model, we assume at one point of time that E=V/L, when is fact, E=dV/dL, unless E is constant, in which case the assumption E=V/L is true. But I don't understand why the electric field in a conductor must be constant as current flows. Generally, the electric field in a conductor does not have to be constant (in ...

1

An easy way to prove Ohm's law for electric fields that aren't constant is to first assume that the electric field is approximately constant over short lengths, just like $E=dV/dL$ suggests. Using that, you can derive Ohm's law for short lengths of material, $dV=IdR$. We'll assume that "current in = current out", which is true at steady-state. This allows us ...

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The current $I$ has a value in one point of the circuit, in contrast to the voltage $U$ which is always measured between 2 points. The definition of $I$ is the amount of charge $\Delta q$ that passes through a particular point in the circuit in the time $\Delta t$ (it's a quantity mathematically similar to the simple velocity in kinematics). So when you ...

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What is voltage? The way I see it, and this is only my concept, voltage represents electron pressure. because electrons repel each other,the more electrons you pack into a unit mass of a conductor the higher the voltage, such as in a metal foil capacitor. If you force electrons onto a non conductor, the room for electrons is very limited and the static ...

2

Two capacitors in parallel have the same voltage. Two capacitors in series have the same charge. Simplify the problem to two capacitors in series (each started life as two capacitors in parallel) - what is the ratio of their voltages. Then use $Q=CV$ to figure the charge on each pair; finally distribute the charge on the elements of each pair according to ...

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Assuming for now that this is homework, I'll provide this hint: the voltage on the 8.73 $\mu$C capacitor is not 21.9 V. Don't forget that that voltage has to be distributed among all of the components.

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In principle, it is possible, using, e.g., high-current relativistic electron beams - please see, e.g., the review http://arxiv.org/abs/physics/0409157 . @John Rennie offers reasonable arguments, but the very real problems he mentions can be overcome - I don't have time to describe the specific mechanisms (see the review). In experiments, propagation length ...

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The cathode ray tube has had the air pumped out. Electrons scatter off oxygen and nitrogen molecules so if you fired an electron beam in air it would be scattered in a short distance. The distance would depend on the beam energy, but it's a lot shorter than 100m. The range of electrons from beta radiation in air is around a metre. You could argue that ...

1

To start with one could have an ac current never grounded anywhere , for a household generator for example. The reason one grounds at the generator is for safety so the ground can pick up any miss chance, as it is a practically infinite sink for electrons. Only one of the two lines can be grounded of course :). It was found though that due to capacitences ...

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There is a LOT of capacitive coupling between the neutral wire and ground even if a DC current cannot flow. And we are talking about AC here.

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In my opinion the contribution of hysteresis losses in an iron wire can be important when comparing with conductive losses, but this is an issue strongly dependent of the iron alloy used according to hysteresis loop, conductivity and permeability. The analysis of the induced current distribution in conducting wires subjected to a harmonic axial voltage is ...

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There is a commonly used analogy for electric circuits called the hydraulic analogy. This imagines the electrons as water and the wires as pipes. The voltage is equivalent to the water pressure and the current is equivalent to the water flow rate. Start with a DC current and imagine the water is doing work by flowing through a water wheel: This is all ...

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At a beach the waves carry energy and momentum from the sea to the shore, even though the water in the waves moves back and forth. It is the same way with alternating current: what matters is the energy flow carried by the electric and magnetic fields, not the movements of the charges.

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If the voltage reverses doesn't the flow of electrons reverse? It depends. If the alternating voltage is across a diode then, no, the current through the diode doesn't (effectively) reverse but is instead unidirectional. However, a genuine alternating current periodically reverses direction - the electric charge 'sloshes' back and forth within the ...

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A metallic wire is electostaticly neutral the mobile negative charges equals the strongly Bounded pisitive charges , so resultant electric field is zero everywhere.

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If there's a way to solve its by assuming both ammeters to be identical and thus from your equations:$$0.2R+6=1.7R\\\implies R=4$$ because there's no other relation between them, or if you find one it'll be the same information disguised in other form. or it can be solved if we have any bit extra information; as to solve a linear equation in two variables ...

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From a circuit theory perspective, an inductor is effectively a short-circuit (wire) at DC while a capacitor is an open-circuit. Thus, any parasitic capacitance must be in parallel since, if it were in series, an inductor would be an open-circuit at DC. From an AC perspective, if the parasitic capacitance were in series, the inductor would appear ...

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The missing piece here is that the temperature of the resistor is a function of the current. Your equation should perhaps read $V = I\,R(T(I))$. Does that help?

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You have a function: $V(T, i) = i \cdot R(T)$ and you should get $\dfrac{dV}{di}$. $T$ doesn't change when you vary $i$ and $R(T)$ doesn't too, so it can be considered as a constant comparing to variable $i$. Fix $T$ at some generic value, for example $a$, doing this you get $R(T = a) = R_a$ So your function is reduced to $V(i) = i \cdot R_a$. Now you ...

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The question has already been answered but I would like to provide another from a slightly different perspective. Imagine that the unconnected battery terminals were connected together via another resistor: Now, there is an electric current circulating counter-clockwise and given by $$I = \frac{V_{BAT A} + V_{BAT B}}{R_1 + R_2}$$ Clearly, if we ...

3

No, there won't be any persistent current going through the resistor. There will only be some current for a tiny period of time. This will bring the poles of the left battery to potentials $0,V_1$, and those of the right battery to $V_1,V_1+V_2$. Note that the electrostatic potential of both poles of the two batteries that are connected to the resistor will ...

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No, there will be no current through the resistor. There is no potential difference between it's ends. There is a diference between the poles of each battery, but I see no reason why threre should be a voltage difference and current on the resistor. Both batteries create a potential difference relative to the resistor. It could be changed to have a ...

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The total voltage difference across the resistors ($V_3$) is, by design, a constant 5.4 volts1 (because it's being supplied by a power supply that pretty well approximates a constant voltage source). This total voltage difference must be dropped across the two resistors $R_1$ and $R_2$ in series: that is, $$V_1 + V_2 = V_3.$$ As the resistors are in ...

2

It depends what you're trying to do. The integral will give you the net current. So if you did the integral for the AC current going to the computer I'm typing this on you'd get the value zero. This is quite correct because no net current flows for kit powered by AC. On the other hand, if you're trying to work out how much power my computer consumes you ...

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I think is that on the outer diameter for a distance tending to zero, the electric field will be same as inside but when you move further outside of the cable towards larger distance, the field will be reducing.

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