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The usual hydraulic analogy for a capacitor is an elastic membrane: A capacitor doesn't allow current to flow across it, but you can push charge onto it by applying a potential. In the hydraulic analogy an elastic membrane across the pipe doesn't allow water to flow through it, but you can push some water through the pipe by elastically deforming the ...


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The resistor/capacitor in series arrangement is known as an RC circuit. The capacitor does charge to the potential of the battery however the resistor inflicts a time delay on the charging process. The equation for this is T=RC. t in seconds, r in ohms, and c in farads. The time result however is not the time that it takes for the capacitor to fully charge. ...


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Here, initially some current flows through R1 and C1 but only until C1 gets fully charged once C1 is fully charged no current flows through the resistor. If you charge up a capacitor through a resistor current will flow until the voltage across the capacitor is the same as the source. This is an exponential process and never really gets there but it comes ...


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What is the flaw in my thinking? The voltage across the capacitor in the series RC circuit given, assuming zero initial capacitor voltage, is given by $$v(t) = E\left(1 - e^{-\frac{t}{RC}} \right), t \ge 0$$ Note that $v(t) \rightarrow E$ as $t \rightarrow \infty$. The energy stored in the capacitor, as a function of time, is $$U(t) = ...


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Once the capacitor has fully charged the current in the circuit will be zero, so the voltage drop across the resistor is zero and hence the voltage across the capacitor is equal to the cell voltage. Having said this, the current falls exponentially with time so in principle the current takes an infinite time to fall to zero, and the voltage across the ...


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When the capacitor is fully charged, there is no current flowing through the resistor. From Ohms law, the voltage drop across the resistor is zero. That means the voltage drop across the capacitor is equal to the EMF of the cell. You can then find the energy in the capacitor.


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Yes, you can use the standard formula. the equations \begin{equation} I = C \frac{\mathrm{d}V}{\mathrm{d}t} \end{equation} or equivalently \begin{equation} Q = C V \end{equation} are really just the definition of capacitance. (similarly the equations for a resistor and inductor are just definitions of resistance and inductance) Any real system will have some ...


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If I wanted to calculate this current, should I use the same formulae No, since, for one, the capacitance $C$ of the apparatus changes as the capacitor is assembled. In the equation $$i = C \frac{dv}{dt}$$ $C$ is a constant. There are other considerations too but the bottom line is, no, you cannot use the above formula to determine the current in ...


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Disregarding conventional current. Whenever a capacitor is connected to a voltage source (battery) the electrons flow from high potential to low, now being that the capacitor is in series with the battery the electrons will start to pile on one side of the plate. But by doing this since the electrons have a natural electric field to them the electric field ...



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