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Point particles as the electrons (which are the charge carriers) move according to Newton's law $\textbf{F}=q\textbf{E}=m\textbf{a}$. Whenever an electric field is present it generates a difference of potential between two points $A$ and $B$ given by its differential form calculated between the two points $$V_A - V_B = \int_A^B \textbf{E}\cdot d\textbf{s}. ... 0 Voltage is a difference in electrical potential in a circuit. You can compare this to a ball sitting at the top of a hill - as it rolls downwards, it moves through a difference in gravitational potential. Because the ball has lower potential at the bottom of the hill, it will roll there spontaneously. The ball will never roll up the hill by itself, ... 0 If resistor is an ohmic one, on applying a higher volt battery, power will increase as power for a resistor is V^2/r. If it is non ohmic one, the answer may depend upon other factors as well. 1 Reducing eddy current does not change property of conductor or circuit Eddy currents (also called Foucault currents) are circular electric currents induced within conductors by a changing magnetic field in the conductor, due to Faraday's law of induction. Eddy currents flow in closed loops within conductors, in planes perpendicular to the magnetic field. so ... 3 Assume you have four waterfalls, and they connect one river to another. The waterfalls drop by five feet. If you put the waterfalls end to end, then the second river must be 20 feet below the first river. If, however, the waterfalls get put side by side, and still connect the two rivers, the second river is just five feet below the first river. Now, this ... 0 since voltage increases with resistance if we connect resistors in series the resistance too will increase with increase in voltage which is usually undesired so the votmeter is arranged in such a way that it gives least resistance and maximum current and that's why voltage gets divided In series. For parallel resistors since change of voltage of one ... 0 Best brightness probably is achieved at 2V - the original setting. The reasons are more complex than may be apparent at 1st glance. Neither 1V or 4V is very good and it depends on the characteristics of your light and your power supply. In most case the higher voltage would be better if the only choices were 1V or 4V. This is because the lamp will "load ... 2 In electric circuits there is concept called duality, which says that for any phenomenom that occurs related to voltage, there is a dual phenomenom related to current. Thus, voltage and current are equally important to understanding electric circuits. As for your light bulb problem, your light bulb requires 16 W of power to operate. Your supply is only ... 3 The charger accepts one voltage as its input, and produces another voltage as its output. This could be done with a transformer or any number of other conversion techniques - for example, in a switched mode regulator the voltage is connected very briefly to charge a capacitor, then disconnected again. It keeps repeating this to keep the output at the right ... 1 The main requirements are that Current flow into or out of "static sensitive" components be limited to below a level where damage is caused by I^2 x R heating. Devices which are damaged by voltage without current flow (such as Field Effect Transistors gate to drain voltages) do not have excessive differential voltage applied. If all "bodies" concerned ... 0 You are right for an ideal voltage source and an ideal wire Ohm's law doesn't work anymore. You can easily see that mathematically for fixed V and R=0 the equation V=IR has no solution for I. So Ohm's law doesn't work here, but since there is no ideal voltage source in the real world why bother looking for some generalized law that includes this ... 1 With a perfect conductor, there is no resistance in the wire. However, that does not mean there is no resistance in the circuit, as the battery itself has some internal resistance. The current through the wire will then be limited by that resistance, i.e.$$I=E/R_i The internal resistance is quite low. It has to be as otherwise its voltage would drop too ...

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For a truly ideal conductor, the voltage is identical at all points on the conductor in a steady state. So if you attached this to a real battery, the battery would shove charge along "trying" to maintain the voltage between the terminals. It wouldn't be able to do this. A very powerful battery might heat up trying to do this so much it damages itself or ...

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Do my statements make any sense whatsoever or am I just embarrassing myself? First of all, never be embarrased asking a question :) Though a quite long question, I will quickly go through your points: An electric field is a force that radiates away from protons and towards electrons. No, electric field and force is not the same. But you ...

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