Charging a capacitor with a voltage source

Take 10 minutes to prepare this exercise.

Then, if you lack ideas to begin, look at the given clue and start searching for the solution.

A detailed solution is then proposed to you.

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For \(t < 0\), the circuit is at rest and \(e(t)\) is \(0\) for \(t<0\) and \(E\) for \(t>0\).

Question

We are interested in the circuit just after application of voltage \(E\) ; determine \(i_1 (0^+)\), \(i_2 (0^+)\), \(i (0^+)\) and \(v(0^+)\).

Solution

It is known that the voltage and the charge of a capacitor are continuous functions. Therefore :

\(v({0^ + }) = v(\left( {{0^ - }} \right) = 0\;\;\;\;\;;\;\;\;\;\;{i_2}({0^ + }) = \frac{{v({0^ + })}}{{{R_2}}} = 0\)

The Kirchhoff's Voltage Law (CVL) and Kirchhoff's Current Law (KCL) then gives :

\({i_1}({0^ + }) = i({0^ + }) = \frac{E}{{{R_1}}}\)

Question

We are interested in the permanent regime ; determine \(i_1(+\infty),\;i_2(+\infty),\;i(+\infty),\;and\;v(+\infty)\).

Solution

The permanent regime, \(i = 0\), then :

\({i_1}(\infty ) = {i_2}(\infty ) = \frac{E}{{{R_1} + {R_{_2}}}}\;\;\;\;\;and\;\;\;\;\;v(\infty ) = {R_2}{i_2}(\infty ) = \frac{{{R_2}}}{{{R_1} + {R_{_2}}}}E\)

Question

Determine expression of \(v(t)\).

Solution

The KCL and KVL allow you to write :

\(\left\{ \begin{array}{l}E = {R_1}{i_1} + v(t) \\{R_2}{i_2} = v(t) \\{i_1} = {i_2} + i\;\;\;\;\;(with\;:\;i = C\frac{{dv(t)}}{{dt}}) \\\end{array} \right.\)

We can deduce the differential equation checked by \(v(t)\) :

\(\frac{{dv(t)}}{{dt}} + \frac{1}{C}\left( {\frac{{{R_1} + {R_2}}}{{{R_1 R_2}}}} \right)v(t) = \frac{E}{R_1C}\)

The voltage across the capacitor is then given by taking into account the initial conditions :

\(v(t) = {E_{eq}}(1 - {e^{ - t/{R_{eq}}C}})\)

With :

\(E_{eq}=\frac {R_2}{R_1+R_2}E\)

And :

\(R_{eq}=\frac {R_1R_2}{R_1+R_2}\)

Question

Make an energy balance.

Solution

The energy balance is :

\(E_G=\int_0^\infty {E{i_1}(t)dt} = \frac{1}{2}Cv{(+\infty)^2} + \int_0^\infty {{R_1}i_1^2(t)dt} + \int_0^\infty {{R_2}i_2^2(t)dt}\)

It merely reflects that the energy supplied by the generator (\(E_G\)) during the charging of the capacitor is found only partially stored by the capacitor ; a part is in fact dissipated by Joule effect in the two resistors.