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In the formula for the resistance A is the area perpendicular to the length. It can also be argued that it is the area perpendicular to the direction of the current going through the resistor.

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But what happens with the current when the resistance equals 0? I mean... if Current = Voltage / Resistence ( 0 ) Couldn't be divided by Zero ._.

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In general, when formulae call for "area" they are talking about the surface area on which the force or whatever is occurring. Note that this is not always the entire surface area, it might just be the curved part of a cylinder for example. This is one such example, because we actually aren't interested in the end effects at the bases of the cylinder.

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Use delta star transformation to G , R1 and R2 That makes circuit simple and solve it

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Superconductivity is a phenomenon of exactly zero electrical resistance and expulsion of magnetic fields occurring in certain materials when cooled below a characteristic critical temperature. When you cool metal to its critical temperature it becomes a superconductor. In Metal there are Cooper pairs and it doesn't have enough energy to break these pairs. ...

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Use Thévenin's theorem and find the Rth (thevenin equivalent resistance) and Vth (Thevenin voltage) across the two terminals of Galvanometer. Just apply ohms law to get your answer.

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That's one way, assuming all else is constant. Another way is to measure the open circuit voltage and the short circuit current, keeping all else constant. The internal resistance is then $$R_S = \frac{V_{OC}}{I_{SC}}$$

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Yes, there will be increase in velocity of free electrons, when temperature increases. But the electrons will not be accelerated in a particular direction. Consider a conductor, when potential difference is applied across the two ends of the conductor, an electric field is set up. Under the effect of electric field, the free electrons accelerate and ...

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The increase in resistance in metals is mainly due to the increasing velocity of the thermic electron motion (I think $v_{\rm therm}\sim \sqrt{kT}$). This shortens the time $\tau$ of free motion of the electron. If $l$ is the mean free path length we have $\tau=\frac{l}{v_{\rm therm}}\sim \frac{1}{\sqrt{kT}}$. If the electron bumps into the next atom/ion it ...

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Initially when you attach the capacitor to the battery, said battery will act to create an electric field within the wire. On the side of the negative terminal this field will point perpendicular to the cross section of the wire toward the terminal of the battery (electric field points toward negative charge). On the side of the positive terminal the field ...

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The PF of a series RL circuit is found from $\tan(PF) = R/\omega L$ while the PF of a parallel RL circuit is found from $\tan (PF) = \omega L/R$. Thus the tangents of the power factors of the 2 circuits are reciprocals of each other. For example, if the series RL circuit has a power factor of 0.4 (or 21.8 degrees), the parallel RL circuit has a a power ...

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I would describe it as (example) 120 joules per coulomb (120 volts) divided by 60 coulombs per second (60 amps) equals 2 (ohms) of resistance "which means you have 1/2 or 2 times less the amperes then voltage". so maybe an ohm can be n of VpA (# of volts[SI] per amp[SI] or in this case, # of N Kg per charge for every charge per second). But that's still ...

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Not sure whether this is correct, but if you have to do it, I think you can say that it is: the work done by the conductor per unit charge per unit current through the conductor, or in terms of SI units, $\mathrm{\frac JC\cdot \frac1A}$ which is the same as: the work done by the conductor per unit current per unit time per unit current, $\mathrm{\frac ... 5 I think the short answer is, you don't. The reason we call the unit of force a Newton and not a kg m/s$^2$is because it is convenient and it expresses the relation you want to convey when used elsewhere (e.g.,$F=-kx$for a spring). Similarly, it is convenient to "hide" the MKS base units into a single term, the potential$V$in this case, so that the ... 3 I'm not sure there's much of a point to what you're asking. The intuitive way to understand an Ohm is to use$\Omega = V/A$. If you want to use SI units, you can, and the math indeed tells you that your other definition is correct, but you're not gonna get much out of it. Indeed, the most you could do is to separate it like this:$\$\begin{align}\Omega ...

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