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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 ...


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.


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 ...


2

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 ...


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 ...


1

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 ...


1

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 ...


1

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 ...


1

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 ...


1

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 ...


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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