RC-Circuit

RC-CIRCUIT

Charging: Let us assume that the capacitor in the shown network is uncharged for t < 0. The switch is connected to position 1 at t = 0.

Now, 'C' is getting charged.

If the charge on capacitor at time 't' is q.

writing the loop rule,

$\frac{q}{C}$ + IR - E = 0

$\Rightarrow R\frac{dq}{dt}=E-\frac{q}{c}$

$\Rightarrow RC\frac{dq}{dt}=EC-q$

$\Rightarrow \frac{dq}{EC-q}=\frac{1}{RC}dt$

Integrating

${\int }_o^q\frac{dq}{EC-q}=\frac{1}{RC}{\int }_o^{t}dt$

$-\ln |EC-q{|}_o^q=\frac{1}{RC}.t$

$\Rightarrow \ln \left|\frac{EC-q}{EC}\right|=\frac{-t}{RC}$

$\Rightarrow q=EC\left[1-e^{-t/RC}\right]$ $\Rightarrow$ At t = 0, q = 0

and at t = $\infty$, q = E C (the maximum charge.) = q_max

Thus,

\begin{align}q=\,\,{{q}_{\max }}\left[ 1-{{e}^{\dfrac{-t}{RC}}} \right] \\ i=\dfrac{dq}{dt}=\dfrac{{{q}_{\max }}}{RC}{{e}^{-t/RC}}=\dfrac{E}{R}{{e}^{-t/RC}} \\ i={{i}_{\max}}{{et/RC}}where{{i}_{\max }}=\dfrac{E}{R} \\\end{align}

Discharging

Consider the same arrangement as we had in the previous case with one difference that the capacitor has charge q_o for t<0 and the switch is connected to position 2 at t = 0. If the charge on capacitor is q at any later moment t then the loop equation is given as

Flip the switch to 2

\begin{align} \dfrac{q}{c}+IR=0\\ \Rightarrow R\dfrac{dq}{dt}=\dfrac{-q}{c}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Rightarrow\dfrac{dq}{q}=\dfrac{-1}{RC}dt\\\end{align}

Integrating, at t = 0, q = q₀

t = t, q = q

$\Rightarrow {\int }_{q_0}^q\frac{dq}{q}=-\frac{1}{RC}{\int }_o^{t}dt$ $\Rightarrow \ln \left|\frac{q}{q_o}\right|=-\frac{t}{RC}orq=q_0.e^{-t/RC}$

$i=\frac{q_0}{RC}{e}^{-t/RC}$

$i=\frac{-EC}{RC}{e}^{-t/RC}$

$i=-i_0{e}^{-t/RC}$

'-ve' sign indicates that the discharging current flows in a direction opposite to the charging current.

Example : Calculate the steady-state current in the 2$\Omega$ resistor shown. The internal resistance of the battery is negligible and the capacitance of the capacitor is 0.2 mF.

Solution : The resistance of the parallel combination of 2$\Omega$ and 3$\Omega$ resistors is given by

$\frac{1}{R}=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\Rightarrow R=1.2\Omega$

This resistance is in series with 2.8 $\Omega$ giving a total effective resistance

= 1.2 + 2.8 $\Omega$ = 4 $\Omega$.

In the steady state, charge on the capacitor C has stablised and hence no current passes through 4$\Omega$ resistor which is in series with the capacitor.

Thus the current through the circuit = 6/4 = 1.5 A,

V_AB = 1.5 x 1.2 = 1.8 V, I through 2$\Omega$ resistor = 1.8/2 = 0.9 A.