Themes > Science > Physics > Electromagnetism > Electrostatics > Capacitors & Dielectrics > Capacitor : Storing Electrical Energy > RC Circuit

The charging and discharging of a capacitor takes place in a finite time, and we briefly discuss the salient aspects of these processes. Consider a $RC$ circuit as shown in Figure 7.14. When the switch $S$ is closed, the circuit is completed and current flows until the capacitor is fully charged, after which the current ceases to flow.
Figure 7.14: Circuit with a Battery, Capacitor and Resistor
\begin{figure} \begin{center} \epsfig{file=core/circuit2.eps, width=4cm} \end{center} \end{figure}
Since a capacitor gets charged in a finite amount of time, the first thing one does is to look for a constant which has the dimensions of time, and, within an order of magnitude, this would be the time taken to charge the capacitor. Consider the following combination.
$\displaystyle [R][C]$ $\textstyle =$ $\displaystyle \frac{V}{I}\times \frac{Q}{V}$ (7.47)
  $\textstyle =$ $\displaystyle \frac{Q}{I}$ (7.48)
  $\textstyle =$ $\displaystyle \frac{\mathrm{coulomb}}{\mathrm{coulomb/sec}}$ (7.49)
  $\textstyle =$ $\displaystyle \mathrm{sec}$ (7.50)

Again, the power of dimensional analysis! There is an inherent time scale in the problem, namely, $RC$. Suppose we have a battery of voltage $V$, the charging of the capacitor commences at $t=0$. Let the current at time $t>0$ be $I(t)$, and the charge on the capacitor be $q(t)$. The potential drop across the resistor, from Ohm's law, is $\displaystyle IR$, and the potential drop across the capacitance is $\displaystyle \frac{q}{C}$. At time $t$ the total drop across the $R,C$ circuit elements must equal the voltage of the battery, and is given by
\begin{displaymath} V=IR+\frac{q}{C} \end{displaymath} (7.51)

Also
$\displaystyle I=\frac{dq}{dt}$     (7.52)
$\displaystyle \Rightarrow \frac{dq}{dt}=\frac{V}{R}-\frac{q}{RC}$     (7.53)

The charge on the capacitor plates is zero when the charging starts, and at final time of $t=\infty$ the charge reaches the equilibrium value of $Q$. In other words

\begin{displaymath} q(0)=0 \mbox{\rm { ; }} q(\infty)=Q \end{displaymath} (7.54)

By direct substitution, one can verify that
$\displaystyle q(t)$ $\textstyle =$ $\displaystyle Q[1-e^{-t/RC}]$ (7.55)
$\displaystyle \Rightarrow I(t)$ $\textstyle =$ $\displaystyle \frac{dq}{dt}$ (7.56)
  $\textstyle =$ $\displaystyle \frac{Q}{RC}e^{-t/RC}$ (7.57)
  $\textstyle =$ $\displaystyle \frac{V}{R}e^{-t/RC}$ (7.58)

where in obtaining last equation above we have used $Q=CV$. Eqn.(7.71) is a typical case of an exponential dependence on time. For $t»RC$, that is, for time much larger than the characteristic time $RC$ for a capacitor, to a very good approximation the current is zero. One can conclude that for all practical purposes, the charging is over $RC$ seconds after the the charging starts. To discharge the capacitor, the same circuit as the one for charging is used, except that one removes the battery. The capacitor is first charged and then, at $t=0$ the switch $S$ is closed, connecting the positive and negative plates of the capacitor through a resistor $R$. It can be shown by an analysis similar to the one for charging the capacitor that the charge and current on the capacitor, at time $t>0$, is given by

$\displaystyle q(t)=Qe^{-t/RC}$     (7.59)
$\displaystyle I(t)=\frac{Q}{RC}e^{-t/RC}$     (7.60)

Similar to the discussion on the charging of a capacitor, from the equation above we conclude that the capacitor is fully discharged in $RC$ seconds after the start of the discharging process.


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