{{tag>charging capacities exam_ee1_SS2023}}
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#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ (Dis)Charging Capacities         \\ <fs medium>(written test, approx. 14 % of a 60-minute written test, SS2023)</fs>
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The circuit below has to be analyzed. 
The component values are:

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

{{drawio>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.
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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*}
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#@ResultBegin_HTML~1~@# 
\begin{align*}
\tau_1 = 3.0 ~\rm ms
\end{align*}
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2. What is the voltage $u_C$ at $t_1=t_0+4 ~\rm ms$?
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\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*}
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#@ResultBegin_HTML~2~@# 
\begin{align*}
u_C(t_1) &= 7.36~\rm V
\end{align*}
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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?

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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*}
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#@ResultBegin_HTML~3~@#
\begin{align*}
\tau_2 = 5.0 ~\rm ms
\end{align*}
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4. Draw the overall course of the voltage $u_C(t)$ over time in the diagram below. \\ Use an appropriate $x$-axis scale. 

{{drawio>electrical_engineering_1:p8yrdjr60k6bvc4nUtDiagramQuestion.svg}}

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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 ($\propto (1-e^{-x})$). 
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{{drawio>electrical_engineering_1:p8yrdjr60k6bvc4nUtDiagramSolution.svg}}

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

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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. 
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#@ResultBegin_HTML~5~@#
 
\begin{align*}
U_C   &= 50 ~\rm V
\end{align*}
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