Capacitor and resistor in parallel

[bp_psy]

Homework Statement

This is not an actual homework problem but this appears in a couple of more complicated problems that deal with RC circuits and Kirchhoff's laws.The connection consists of a capacitor in series with a resistor. My question is: will the capacitor be instantly charged when the circuit is closed or will it take a certain amount of time for it to charge? Will it remain uncharged?

Homework Equations

https://www.physicsforums.com/attachment.php?attachmentid=18127&d=1237862861
C=Q/V_ab
q=Q(1-e^-t/RC)
i=Ie^-t/RC

The Attempt at a Solution

From the equations and the fact once the circuit is closed the potential difference V_ab will be IR=E I am lead to believe that the capacitor will be instantly charged but the instantly part does not seem right to me.I do not think it is possible for the capacitor to remain uncharged since the potential difference between a and b will surely induce a charge in the capacitor.

 

[tiny-tim]

Hi bp_psy !

The connection consists of a capacitor in series with a resistor. My question is: will the capacitor be instantly charged when the circuit is closed or will it take a certain amount of time for it to charge? Will it remain uncharged?
…
i=Ie^-t/RC
…
From the equations and the fact once the circuit is closed the potential difference Vab will be IR=E …

No, the potential difference V_ab will be iR, where i=Ie^-t/RC

Kirchhoff's rules only apply to capacitors if you "invent" a "displacement current" flowing through the capacitor (equal and opposite to i) … see from the PF Library …

Displacement current:

No current ever flows through a functioning capacitor.

But while a capacitor is charging or discharging (that is, neither at zero nor maximum charge), current is flowing round the circuit joining the plates externally, and so there would be a breach of Kirchhoff's first rule (current in = current out at any point) at each plate, if only ordinary current were used, since there is ordinary current in the circuit on one side of the plate, but not in the dielectric on the other side.

Accordingly, a displacement current is deemed to flow through the capacitor, restoring the validity of Kirchhoff's first rule:

$$I\ =\ C\frac{dV}{dt}$$

and this linear displacement current $I$ (which might better be called the flux current or free flux current) is the rate of change of the flux (field strength times area) of the electric displacement field $D$:

$$I\ =\ A\,\widehat{\bold{n}}\cdot\frac{\partial\bold{D}}{\partial t}\ =\ A\,\frac{\partial D}{\partial t}\ =\ C\frac{dV}{dt}$$

which appears in the Ampére-Maxwell law (one of Maxwell's equations in the free version):

$$\nabla\,\times\,\bold{H}\ =\ \bold{J}_f\ +\ \frac{\partial\bold{D}}{\partial t}$$

Note that the displacement alluded to in the displacement current across a capacitor is of free charge, and is non-local, since it alludes to charge being displaced from one plate to the other, which is a substantial distance compared with the local displacement of bound charge in, for example, the presence of a polarisation field.

…

Inverse exponential rate of charging:

A capacitor does not charge or discharge instantly.

When a steady voltage $V_1$ is first applied, through a circuit of resistance $R$, to a capacitor across which there is already a voltage $V_0$, both the charging current $I$ in the circuit and the voltage difference $V_1\,-\,V$ change exponentially, with a parameter $-1/CR$:

$$I(t) = \frac{V_1\,-\,V_0}{R}\,e^{-\frac{1}{CR}\,t}$$

$$V_1\ -\ V(t) = (V_1\,-\,V_0)\,e^{-\frac{1}{CR}\,t}$$

So the current becomes effectively zero, and the voltage across the capacitor becomes effectively $V_1$, after a time proportional to $CR$.

 

[nlightner]

I would like to provide a more thorough explanation for the behavior of a capacitor and resistor in parallel. When a circuit is closed, the capacitor will not be instantly charged. It will take a certain amount of time for the capacitor to charge to its maximum potential, known as the steady state. This is due to the fact that capacitors have the ability to store charge, but they do not allow current to flow through them. Therefore, when the circuit is closed, the capacitor will initially act as an open circuit, allowing the current to flow through the resistor until the capacitor is fully charged.

The time it takes for the capacitor to charge can be calculated using the equations provided. The time constant, RC, determines how quickly the capacitor will charge. A smaller time constant means the capacitor will charge more quickly, while a larger time constant means it will take longer to charge.

It is also important to note that even though the capacitor is not initially charged, it does have the potential to hold a charge. This is because the potential difference between points a and b is present, and this potential difference will eventually cause a charge to build up in the capacitor.

In summary, a capacitor and resistor in parallel will not result in instant charging of the capacitor. It will take a certain amount of time for the capacitor to charge to its maximum potential, and it will never remain completely uncharged due to the potential difference present in the circuit.