# Capacitor and resistor in parallel

1. Mar 23, 2009

### bp_psy

1. The problem statement, all variables and given/known data
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?

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

3. The attempt at a solution
From the equations and the fact once the circuit is closed the potential difference Vab 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.

2. Mar 24, 2009

### tiny-tim

Hi bp_psy !
No, the potential difference Vab 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 $CR$.