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2

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


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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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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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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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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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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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