12.1 Important points in $R$-$C$ circuits

  Capacitors

• A capacitor gives zero resistance in a circuit when it is uncharged, i.e., it can be assumed as short circuited and it offers infinite resistance when it is fully charged. That means that no current will passes through that branch.

• Equivalent Time Constant: To find the equivalent time constant of a circuit, following steps are followed:
  
  (a) Short circuit the battery.
  
  (b) Find net resistance across the capacitor (say it${R_{net}}$)
  
  (c) ${\tau } = \left( {{R_{net}}} \right)C$

• Alternate method of finding current in the circuit and charge on the capacitor at any time $t$: In a complicated $R-C$ circuit it is easy to find current in the circuit and charge stored in the capacitor at time $t=0 $ and $t=\infty $ by the methods discussed. But to find the current and charge at time $t$ following steps may be followed.
  
  (a) Find the equivalent time constant ${\tau}$ of the circuit.
  
  (b) Find steady state charge $q_o$ (at time $t=\infty $ on the capacitor).
  
  (c) Charge on the capacitor at any time $t$ is,$$q = {q_o}(1 - {e^{ - \frac{t}{\tau }}})$$ By differentiating it with respect to time we can find current through the capacitor at time $t$. Then by using Kirchhoff's laws we can find current in other parts of the circuit also.

Example: In the circuit shown in the figure switch is closed at time $t=0$. Find the current through different branches and charge stored on the capacitor at any time $t$.

Solution: First we will calculate $\tau$ then the steady state charge $q_o$. So for $\tau$, we have to find the equivalent resistance across the capacitor after short circuiting the battery,

$$\begin{equation} \begin{aligned} {R_{net}} = R + \frac{{(6R)(3R)}}{{6R + 3R}} = 3R \\ \tau = {R_{net}}C = 3RC \\\end{aligned} \end{equation} $$At $t=\infty$, capacitor is fully charged and no current flows through it.

Then the potential difference across capacitor = potential difference across $3R$

$$\begin{equation} \begin{aligned} = \left( {\frac{V}{{9R}}} \right)(3R) \\ = \frac{V}{3} \\ {q_o} = \frac{{CV}}{3} \\\end{aligned} \end{equation} $$Now, let the charge on the capacitor at any time $t$ be $q$ and current through it is $i_1$. Then,$$\begin{equation} \begin{aligned} q = {q_o}(1 - {e^{ - \frac{t}{\tau }}}) \\ {i_1} = \frac{{dq}}{{dt}} = \frac{{{q_o}}}{\tau }{e^{ - \frac{t}{\tau }}} \\\end{aligned} \end{equation} $$

Applying Kirchhoff's second law in loop $ABCDFA$, we have

$$\begin{equation} \begin{aligned} - 6iR - 3{i_2}R + V = 0 \\ 2i + {i_2} = \frac{V}{{3R}} \\\end{aligned} \end{equation} $$ Applying Kirchhoff's junction law at $B$, we have $$i = {i_1} + {i_2}$$ Solving eqs.$$\begin{equation} \begin{aligned} {i_2} = \frac{V}{{9R}} - \frac{2}{3}{i_1} = \frac{V}{{9R}} - \frac{{2{q_ \circ }}}{{3\tau }}{e^{ - t/\tau }} \\ i = \frac{V}{{9R}} + \frac{{{q_ \circ }}}{{3\tau }}{e^{ - t/\tau }} \\\end{aligned} \end{equation} $$