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The motion of electrons in the wires and the voltages can't be "seen" by naked eyes so the whole science of electric circuits is automatically "harder to visualize" than mechanics. But all such laws and phenomena have mathematically similar analogies in mechanics. The voltage is analogous – not only mathematically but physically – to the slope of an ...


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I can see why you think physicists might have some insight into such a problem, because of the way it has been presented as an electrical circuit. Unfortunately we have little to add to what the Mathematicians and Computer Scientists can tell you. We certainly have no tricks which they do not already know about. I think this is a mathematical or ...


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Here is an answer I propose for the 2nd way. I'm for sorry for the bad paintings.


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Your mistake in the above is in what you call the force required to move the charge from across the potential difference $V$. You identify the force as $\vec{F}=Q_{test}\vec{E}$, when it should be $\vec{F}=-Q_{test}\vec{E}$. This makes physical sense -- the reason it takes work to move the charge is because you need to oppose the electric field present in ...


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It is like water in a hose. If the hose is full of water, water flows out the end immediately when you turn on the faucet. A drop of water at the faucet pushes a drop next to it, which pushes the next drop. Water doesn't flow that fast. If the hose is empty, it takes a while to reach the end.


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Because an LED acts like a diode, the negative current from the AC source will be clipped and the capacitor will always be charged and it will act like a second voltage source. Look at my picture below. You can see that the green line represents the voltage going through the capacitor and the blue line (it's a little hard to see since it's covered up by the ...


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Resistance can be interpreted in various ways depending on the circuit. It can be used to cause a potential drop, or it can be used as a heating device, etc. You are asking how resistance can change the current flowing through the circuit when connected in series. In that context, the resistance can be used to alter the total resistance of the circuit which ...


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Your error is to assume that only your red charges generate the heat, ie the red charges go through area $A$ and they are not replaced by any other charges. If that were the case then the factor of $\frac 12$ would be correct. However as the red charges move through the resistor black charges to the left of the red charges would move into the resistor and ...


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When dealing with inductors Sears & Zemansky state that "we need to develop a general principle analogous to Kirchhoff's loop rule". With an inductor present in the circuit they state that there is a non-conservative electric field within the coils $\vec E_n$ as well a conservative electric field $\vec E_c$. Assuming that the inductor has negligible ...


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Since you equate $W$ with $NqU$, that means $W$ represents the amount of energy dissipated as heat during the time interval $N$ charges passed through the cross section. That time interval is $t / 2$, so the resulting power is $P = {{UIt / 2} \over {t / 2}} = UI$.


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The regular spacing and sheer amount of holes in that image is a bit unusual, but I nevertheless suspect that most or all of them are vias: holes drilled through multiple layers of the PCB and plated with copper on the inside to provide an electrical connection between layers. A large number of regularly spaced vias in contiguous planes may be used to spread ...


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All of the AC motors will create a "back pressure" once power has been removed. I do however not believe you when you say you have a properly grounded piece of equipment. Please note...the grounding system must be complete ALL The way back to a earth ground (grounding rod ect)...grounding cables can have a tendency to create a thin film of oxidation on ...


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The inductor works the same as it does in AC circuits : the PD across it is related to the current through it by $V = L\frac{dI}{dt}$. You are correct : the current I through L increases in proportion to $1-e^{-kt}$ towards a maximum value. Meanwhile $\frac{dI}{dt}$ decreases from a maximum value to zero. When the maximum current flows it is constant, ...


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It depends on what you have. Sources can be modeled as a current source or a voltage source. If you treat it as a voltage source, it will always output that peak voltage and the circuit current will change according to the resistance changes. If you treat it as a current source, it will always output that peak current and the voltage drop across it will ...


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So shouldn't there already be a drop in voltage ( potential difference) before the charges even reach a resistor? Normally we view a battery or a cell as accumulator of charges in a manner that a potential difference is built up when we charge a battery with plates and electrolyte. Therefore if charges flow out or current is being drawn at certain voltage ...


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The traditional electric bell uses an electromagnet to move the arm that strikes the bell. Electromagnets require the core to be low coercivity as a high coercivity causes losses due to hysterisis, and iron happens to meet this requirement. There is nice explanation of the effect of coercivity in the answers to Properties to select suitable materials for ...


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As already said, you could just write Kirchhoff's equations assuming you know the voltage between A and B. Imagine you are given a two resistors connected in parallel. Then for a given voltage you can express currents through it and resistances using Ohm's law and charge conservation (this's Kirchhoff's equations). So the current would be proportional to a ...


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Long ago somebody decided that the direction of "conventional" current flow was the same direction as the direction of flow of positive charges. In that convention the flow of negative charge in one direction is equivalent to the flow of positive charge (and hence the conventional current) in the opposite direction. When introduced electricity usually ...


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The power consumed by your circuit determines how fast the battery drains. P = I * E: power (Watts) is found by multiplying the current (Amps) by the voltage (Volts). Since your battery has a (reasonably) constant voltage under normal operation, current is the variable here. I = E / R, amps = volts / ohms. If we combine these two equations, we get P = E ^ ...



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