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0

You want $\frac {dV}{dt}=\frac IC$, where $I$ is the current into the capacitor and $C$ is the capacitance. This is independent of the state of charge of the capacitor. The $I$ will vary depending on the state of charge. If you are charging from $5$ volts through a resistance $R$, the current will be $\frac {5-V}R$. Then when the input voltage goes to ...


1

One equation is for resistive circuit and the other is for capacitive circuit. Two can not be merged together.


6

This is true but strictly limited to RC circuits without external sources: that is, a resistor hooked up to a capacitor with nothing else in between. In that case, $V$ is indeed proportional to $\dot V$, with a crucial minus sign in between: $$ \dot V=-\frac1\tau V, $$ where $\tau>0$ is some constant. This equation implies that $V(t)=V(0)e^{-t/\tau}$, ...


1

Like any set of equations in physics, this is true when the conditions stated apply. Here, you are using two equations: $$ V(t) = RI(t) $$ when the current passes through an element that is a perfect resistor, and $$ C \frac{dV}{dt} = I(t) $$ when the current passes through an element that is a perfect capacitor. So, in a simple RC circuit consisting of a ...


5

Generally, no. You've written two equations. The first relates voltage and current for an isolated resistor. The second relates voltage and current for an isolated capacitor. Given only those two expressions, there's no reason at all that they can or should be combined. That is, you have no circuit, only isolated components. There are principles for ...


18

The identity $$ V = K \frac{dV}{dt} $$ is only guaranteed with a constant $K$ if your assumptions actually hold. The first identity $V=RI$ only holds for a resistor, while the other holds for a capacitor. So in this sense, the letters $V,I$ in these equations mean something else. In one of them, it's the current through (or voltage on) a particular resistor, ...


2

Of course you can charge a capacitor with AC. The problem is that you keep changing how it is charged. While you apply a positive voltage to one plate, it will get a positive charge; half a cycle later, it will attempt to get a negative charge; and so it continues. The capacitor is always a little bit behind - as your AC voltage is changing, the capacitor ...


0

Too many explanations, few of them are: $\\Z_c=\frac{1}{j\omega C}=\frac{1}{j0C}=\infty \\V_c=Z_cI=\infty $ In DC, the voltage across the capacitance does not change after full charging and remains equal and opposite to the DC Voltage across it and hence no current can flow through it.


2

AC has each wire positive half the cycle and negative the other half. It will charge the capacitor on one half cycle and discharge it on the other half. The net charge will be zero.


0

I think as we know E = V/d, and the field is same, so for field remains constant between the plates of the capacitor, while increasing the distance the potential also increases. In the same manner as that of distance so that the ratio of V and D is same always.


0

Depends on what you want to use the capacitance for. If plan to use it to store energy: the high losses due to the conductivity may not prove adequate for this use. The conductivity of water increases with the addition of ions. For example, Sodium or Chloride. Others are presented in this article. Condcutivity changes a a function of temperature too. ...


1

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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