- #36
sophiecentaur
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Forget the meter. A battery is a Voltage Source - period.
I don't think your use of formulas was completely correct, but honestly I got pretty lost in what you were doing so I don't know what exactly you did wrong.Mr.Bomzh said:with my calculations I concluded that this current has to be 0.12mA, is that correct?
Assuming negligible resisance of the wires attached to the resistor , we can calulate the voltage drop over the resistor which is V=IxR which is 0.12x100=12volts.
I did made a mistake previously I should have just counted the wire resistanes in with the resistor because they altogether make one large resistor just with different resistivity along the way , smaller along the wire and higher along the resistor.
But my use of formulas was correct ?
As far as I can tell, this paragraph is correct. You should have done something more like this with the 100 Ω resistor and the 2 Ω wires. Find the voltage across each resistor exactly like you did here. Then, once you have the voltages, divide those voltages by the lengths of the elements to get the electric field in volts per metre.Mr.Bomzh said:So in a a series circuit with two resistors, one 4 ohm other 8ohm and a 12v battery, the current through each would be the same but voltage drops different ,For the current in the circuit I= V/R so 12/12 = 1 amp. of current , for the 4 ohm resistor V=1x4 and for the other resisor V=1x8, so the question is the current is the same through both resistor but the resistance of each differs so the electric field in the 8 ohm resisor is higher than the field in the 4 ohm resistor , assuming both have equal lengths? Ad the reason why it is higher in the 8 ohm resistor is because to force the same current through a higher resistance needs more emf which relates to higher voltage?
That is a very good idea.Mr.Bomzh said:Ok I have to lay down my cards , so that you folks don't get confused for my questions.
That's situation 1.the idea is more about that I need a high electric field from low voltage in a small area , which would be surrounded by a dielectric material ... it has to have high resistance so that I wouldn't need much current, yet I need the E field.
This is situation 2.I was just wondering what would happen if a dielectric placed between the plates of a capacitor had a resistive layer in between which would be attached to a circuit and current would pass in it , we know that the charges in the dielectric align themselves according to the charges on the plates oppositely of course.
This is an different situation again ... so you are, in fact, pondering three distinct situations.I was just wondering what would happen to the alignment of the dielectric charges if the dielectric had a e field inside of it.
No. That is not the problem!So we concluded after all these two thread talk that a capacitor in a circuit being in series with the pirmary could pump charges back and forth through the pirmary if it's capacity could be altered , the problem here ofcourse is how does one do that , not speaking about the rotors or any mechanical stuff but in a solid satte way.
OP has been repeatedly stating that too... I suspect the issue is a bit more than that - any mechanism that changes the capacitance, in the right way, using electricity, is already the kind of device he seeks to build with the capacitor - but without the capacitor.sophiecentaur said:Last sentence of that post is what I have been telling you in pretty well every post I (and others) have written here.That is the whole crux of the problem. Energy In is Energy Out.