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Exercise E4 (Dis)Charging Capacities
(written test, approx. 14 % of a 60-minute written test, SS2023)

The circuit below has to be analyzed. The component values are:

  • $U = 10 ~\rm V$
  • $C_1 = 200 ~\rm nF$
  • $R_1 = 8.0 ~\rm k\Omega$
  • $R_2 = 17 ~\rm k\Omega$
  • $R_3 = 7.0 ~\rm k\Omega$
  • $I = 2.0 ~\rm mA$

electrical_engineering_1:p8yrdjr60k6bvc4ncircuit.svg

Before $t_0$ all switches are switched as shown and the capacitor is fully discharged.
At $t_0=0 ~\rm s$ the switch $S_1$ shall switch to the voltage source.

1. Calculate the time constant for charging the capacitor.

Solution

The time constant is generally given as: \begin{align*} \tau &= R\cdot C \end{align*}

Once $S_1$ is closed and $S_2$ is open at $t_0$, the source $U$ drives the current through the series circuit given by $S_1$, $C$, $R_1$ and $R_3$.
Therefore, $R= R_1 + R_3$ \begin{align*} \tau_1 &= (R_1+R_3)\cdot C \\ &= (8~\rm k\Omega + 7~\rm k\Omega) \cdot 0.2 ~\rm \mu F \\ &= 15\cdot 10^{3} \cdot 0.2 \cdot 10^{-6} ~\rm {{V}\over{A}}{{As}\over{V}}\\ \end{align*}

Result

\begin{align*} \tau_1 = 3.0 ~\rm ms \end{align*}

2. What is the voltage $u_C$ at $t_1=t_0+4 ~\rm ms$?

Solution

\begin{align*} u_C(t_1) &= U_0 \cdot(1-e^{-t/\tau} ) \\ &= 10~\rm V \cdot(1-e^{-4~ms/3~ ms} ) \end{align*}

Result

\begin{align*} u_C(t_1) &= 7.36~\rm V \end{align*}

3. The capacitors shall be charged to $U=10 ~\rm V$ at the time $t_2$.
At this point in time, the switch $S_1$ switches to the situation shown in the drawing. What is the new time constant?

Solution

Again, the time constant is generally given as: \begin{align*} \tau &= R\cdot C \end{align*}

Now, $\rm S_1$ is opened and $\rm S_2$ is closed. Then, the source $U$ drives the current through the series circuit given by $\rm S_1$, $C$, $R_1$ and $R_2$.
Therefore, $R= R_1 + R_2$ \begin{align*} \tau_2 &= (R_1+R_2)\cdot C \\ &= (8~\rm k\Omega + 17~\rm k\Omega) \cdot 0.2 ~\rm \mu F \\ &= 25\cdot 10^{3} \cdot 0.2 \cdot 10^{-6} ~\rm {{V}\over{A}}{{As}\over{V}}\\ \end{align*}

Result

\begin{align*} \tau_2 = 5.0 ~\rm ms \end{align*}

4. Draw the overall course of the voltage $u_C(t)$ over time in the diagram below.
Use an appropriate $x$-axis scale.

electrical_engineering_1:p8yrdjr60k6bvc4nutdiagramquestion.svg

Solution

Both courses of the voltage for charging and discharging are described with an exponential function. However, the curve for charging increases first steep and flattens out for longer time scales ($\propto (1-e^{-x})$).

Result

electrical_engineering_1:p8yrdjr60k6bvc4nutdiagramsolution.svg

5. The switch $S_2$ is then closed, with $S_1$ still as shown in the drawing. What will be the maximum voltage of $u_C$?

Solution

The current of the source flows through the circuit consisting of $C$ in parallel with $R_1+R_2$. Without the parallel resistors, the current source would charge the capacitor „to infinity“ ($u_C \rightarrow \infty$) . This is here limited by the parallel resistors $R_1+R_2$.

The maximum voltage on the branch with the resisors $R_1+R_2$ is

\begin{align*} U_{12} &= R \cdot I \\ &= (R_1+R_2) \cdot I \end{align*}

This is also the maximum voltage on the capacitor, since it is in parallel with the resisors.

Result

\begin{align*} U_C &= 50 ~\rm V \end{align*}