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A typical voltmeter contains an internal Ohmic resistor with known and very high resistance $R$ (called the "input resistance" or "input impedance"), and an extremely sensitive ammeter that measures the current through that resistor. When the voltmeter is connected in parallel across some circuit elements, then ideally the internal resistor has resistance ...


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If voltage is all that you know, then the answer is No. If you know how much charge $Q$ in Coulombs is added, you only have to divide by the charge $e$ on each electron (in Coulombs). Otherwise, if you have a parallel plate capacitor and you know the capacitance $C$, you can work it out from $Q = CV$. Your suggestion that the extra electrons 'push more' ...


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I would say it is both! Because of the abundance of electrons, the electric field at the battery pole/boundary, at the instant of turning on the switch (t=t0), is quickly (within a few Debye lengths) screened and cannot possibly reach the electrons further down the wire. However, the electrons at the vicinity of the pole that do feel the effect of electric ...


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UPDATE : John : Thanks for data. Graph is ok. I note your intercept is E=3.94V but your calculations use E=4.5V. This explains the discrepancy in your results. If you use 3.94V you get r ranging from 1.59 to 1.76, close to slope value of 1.68 Ohms. ORIGINAL ANSWER : Your line of best fit gives an average internal resistance r based on all measurements. ...


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$Q=CV \Rightarrow \frac {dQ}{dt} = I = C \frac {dV}{dt}$ The voltage across the capacitor is equal to the voltage of the supply. So whatever the voltage of the supply does the voltage across the capacitor exactly follows. At time $t_A$ the capacitor is uncharged and the voltage across the capacitor is zero. However at this time the voltage of the supply ...


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in series the current is the same through each resistor. Not just the same, the current through each is identical. So that the lightbulbs will all have the same brightness but dimmer than if there was just one bulb on there. Well of course, series connected resistances add and so, the total resistance of two series connected bulbs is greater ...


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Light bulbs, or any loads, in series will all have the same current. This is unrelated to Ohm's Law - it's Kirchhoff's Current Law and it applies if the loads are ohmic or not. Assuming your source voltage stays the same, adding bulbs in series will increase the total resistance which will decrease the total current and make all the bulbs dimmer. The ...


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I can't understand at all everything after "As the resistance increases ...". Is that really how it was explained? Nonetheless there's an interesting point here. The analysis is not exactly the same as for an ohmic resistor, but not for the reason you suggest. The thing to consider is that the resistance of the light bulb depends on the temperature ...


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You are correct that light bulbs are non-ohmic (they don't obey Ohm's Law). But that makes no difference. The same current flows through each, even if they have completely different resistances. Electrical current (charge per second) is like the flow of a river. If there are no leaks, and no tributaries joining the river, then the volume of water per ...



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