# Tag Info

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I think you understand that the series has more resistance than the parallel, so more current should flow in the parallel case. Since there is more current in the parallel case, the battery has to supply more current so it is more stressed, and it gives out a lower voltage. So its $V$ (notice that $V$ is really just measuring the voltage drop across the ...

0

Suppose a conductor with cross sectional area $A$ has $n$ mobile charge carriers per unit volume, each carrying a charge $q$, which moves through the conductor at an average drift velocity of $v_d$. Now the total charge in a segment $\Delta x$ is: $\Delta Q = (nA\Delta x)q$ Now, the charge carriers moving at an average drift velocity of $v_d$ will move a ...

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I think the range of answers you have got from friend, teacher and web reflect that there is not a straightforward response. Without high voltage it would not be possible to drive the dangerous current through the body, but high voltage itself is not lethal - it depends how much current can be delivered at high voltage. Another question is how high does the ...

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Edit after rereading the quesiton. Agree that it is (all else being equal) easier for electrons to travel down a wire with a greater cross sectional area. (Not sure why this is not intuitive - for me it is easier to walk down a wide pavement than a narrow passage between two buildings - particularly if other people are about to provide some 'resistance' - ...

3

Actually, I am not sure you should expect a much lower resistance value. For example, with aluminum resistivity of 2.65x10^(-8) Ohm-m, a sample with length of 5 micron and cross-section area of 100 nm x 1 micron has a resistance of about 1.3 Ohm. Of course, you can question the rather arbitrary effective width of the sample - 1 micron, but I guess this ...

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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.

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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 ...

2

You are asking for a proof of Kirchhoff's voltage law or KVL, that the algebraic sum of voltage differences around any loop in a DC circuit is zero. KVL stems from Faraday's law, which essentially says that (unless there is time varying magnetic field, which, for an ordinary circuit is not true) the E field is curl-less, therefore the integral of E around ...

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The formula for power is $$P=IV$$ So one might be tempted to say that the average value of an AC current should produce the same power as a DC current. However, the formula for voltage across a load is $$V=IR$$ Therefore, the power across a load is $$P=I^2R$$ This means that if you want to find the equivalent DC power produced by an AC current, you find ...

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I found the solution to my problem : The equations can be written : $$I = V \times C_{ij} \times i\omega$$ Where $I$ is the vector of known currents going threw the electrodes, $V$ the potential of the electrodes and $\omega$ the rotational frequency of the current. In this case the potential of the transmitting electrodes is known and the current ...

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The drift velocity does not depend on the length or the cross sectional area of the wire, when dealing with a macroscopic (ordinary, everyday life) wire. However, if the wire is, say, too short, e.g. comparable to the average distance a charge carrier travels before undergoing a collision, then it might begin to depend on the wire length, but for all ...

1

You've begun with this: $P_l = I^2R$ $P_l = IV$ This is correct, but the $V$ here is not the line voltage, but instead the voltage drop across the resistor under consideration. Increasing the line voltage does not increase the voltage drop. Your diagram with a single resistance and a power station implies that the current in the line depends on that ...

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It depends on what is meant by "added to the lamp". If the resistor is in series with the lamp, then there will be a voltage drop across that resistor (since a current is flowing) and less of the potential is available for the lamp. But if the lamp is modified so it has higher resistance (a perfectly acceptable interpretation of the wording) then the lamp is ...

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If you add a resistors in series to another resistor (lamp), the voltage across the lamp will decrease. The new voltage will be: $V_{LAMP}=R_{lamp}V_{total}/(R+R_{lamp})$ The resistance in the lamp is a constant property. Regardless of the change the resistance is constant, it is the voltage across it that's different.

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In the equation $\mathbf{j} = \mathbf{v}\rho$, $\mathbf{v}$ is the average velocity of charges and the equation describes the net flow of charge. A black body spectrum describes a system at thermodynamic equilibrium and at thermodynamic equilibrium there is, by definition, not net transfer of heat. This means that the net energy flux through a surface is ...

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First of all, as the comments on the question point out, the current is infinite through a wire only if the entire wire is an ideal one. Meaning, the resistance of the entire wire is zero; then, not rigorously speaking, Ohm's law becomes I = V/0 resulting in an infinite current. If there is a simple ideal wire (a single loop) with a resistance R at some ...

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The voltage that drives the current is different in different parts of the circuit. The sum of the voltages driving current through each part of the circuit, wire and resistor must be equal to the voltage applied with a power source or battery. In the case of the circuit with the resistor and wire then nearly all the voltage will be pushing electrons ...

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Crudely, electrons repel each other and even out the charge. While the influence of the electrons travels at a good fraction of the speed of light the electrons themselves do not move much. From this link " The electricity that is conducted through copper wires in your home consists of moving electrons. The protons and neutrons of the copper atoms do not ...

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You're allowed to be confused on this issue - for many students it's a new phenomenon for which no analogy is perfect. Current is very easy to understand. As you mentioned it's basically just electrons flowing past per second. However, as it would be rather time-consuming always saying things like "the current is 1000000000000000000 electrons per second", ...

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my main question is, if the energy is not the kinetic energy of the electrons, what does in fact bring energy to the resistor and how does it heat up? We assume steady state operation. The drift velocity of the electrons entering the resistor must equal the drift velocity of the electrons leaving the resistor. This follows from the fact that the ...

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Yes, it is the electron kinetic energy gained between collisions with the lattice atoms which is transferred to the lattice atoms who heats the lattice. After collision the electron changes it direction essentially, so the drift velocity is small whereas the instant velocity is high. The drift velocity has nothing to do with the temperature. In absence of ...

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It is very simple, to pass current, it needs a path to follow unlike voltage (potential) which you can have with an open circuit. Think of it this way, electrons need to jump from atom to atom, this will happen until it reaches an end to the wire or conductor. At the end of the wire that ends, the electron stop and build-up to the highest voltage it can ...

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He "ignores" the magnetic field due to $I_2$ because he is interested only in the effect of the magnetic field caused by $I_1$ on the loop. So while there are indeed internal forces on the loop from the magnetic field that it creates itself, these are by his own admission not the forces he is interested in.

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Do not open the cone. Think of it in the profile view : You have an isosceles triangle. Now move along the axis of the cone, say a distance $x$ and take an element $\mathrm{d}x$. Somewhat like this : This small element is similar to the rectangle you described. With length as $2\pi r(x)$ and width $\mathrm{d}x$. You also know the velocity with which the ...

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The r in the Biot-Savart integral has to be taken from every current element to P, and for all the current elements that don't lie on that top line in your diagram (since you mentioned in a comment that the current is supposed to be present throughout the 3-dimensional wire), r should not be parallel to dl, the direction of current for that element, so in ...

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When you think electricity think water. Let's use a waterfall as an example for this analogy: The Water traveling from the high point to the low point of the waterfall is like electrons flow through a conductor. That is the current: current == flow. Voltage by definition would is the "difference of the potenticals" in the waterfall analogy the voltage ...

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AC can be made from DC, and vice versa. DC is a steady voltage, with AC the voltage fluctuates from negative to positive and back, many times per second. Most electricity generators generate AC. The reason for preferring AC is that, historically, it is easier change voltage with AC: all you need is a transformer. The generators generate a high voltage, ...

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