Learning Objectives

Preparation at Home

Well, again

• read through the present chapter and write down anything you did not understand.
• Also here, there are some clips for more clarification under 'Embedded resources' (check the text above/below, sometimes only part of the clip is interesting).

For checking your understanding please do the following exercises:

• …

90-minute plan

1. Warm-up (x min):
   1. ….
2. Core concepts & derivations (x min):
   1. …
3. Practice (x min): …
4. Wrap-up (x min): Summary box; common pitfalls checklist.

Conceptual overview

Core content

Time Course of the Charging and Discharging Process

In the simulation below you can see the circuit mentioned above in a slightly modified form:

• The capacitance $C$ can be charged via the resistor $R$ if the toggle switch $S$ connects the DC voltage source $U_{\rm s}$ to the two.
• But it is also possible to short-circuit the series circuit of $R$ and $C$ via the switch $S$.
• Furthermore, the current $i_C$ and the voltage $u_C$ are displayed in the oscilloscope as data points over time and in the circuit as numerical values.
• Additionally it is possible to change the capacitance value $C$ and resistance value $R$ with the sliders Capacitance C and Resistance R.

Exercises:

1. Become familiar with how the capacitor current $i_C$ and capacitor voltage $u_C$ depend on the given capacitance $C$ and resistance $R$.
   To do this, use for $R=\{ 10~\Omega, 100~\Omega, 1~k\Omega\}$ and $C=\{ 1~\rm µF, 10 ~ µF\}$. How fast does the capacitor voltage $u_C$ increase in each case n?
2. Which quantity ($i_C$ or $u_C$) is continuous here? Why must this one be continuous? Why must the other quantity be discontinuous?

In the following, this circuit is divided into two separate circuits, which consider only charging and only discharging.

To understand the charging process of a capacitor, an initially uncharged capacitor with capacitance $C$ is to be charged by a DC voltage source $U_{\rm s}$ via a resistor $R$.

• In order that the voltage $U_{\rm s}$ acts at a certain time $t_0 = 0 ~s$ the switch $S$ is closed at this time.
• Directly after the time $t_0$ the maximum current (“charging current”) flows in the circuit. This is only limited by the resistor $R$. The uncharged capacitor has a voltage $u_C(t_0)=0~V$ at that time. The maximum voltage $u_R(t_0)=U_{\rm s}$ is applied to the resistor. The current is $i_C(t_0)={{U_{\rm s}}\over{R}}$.
• The current causes charge carriers to flow from one electrode to the other. Thus the capacitor is charged and its voltage increases $u_C$.
• Thus the voltage $u_R$ across the resistor is reduced and so is the current $i_R$.
• With the current thus reduced, less charge flows on the capacitor.
• Ideally, the capacitor is not fully charged to the specified voltage $U_{\rm s}$ until $t \rightarrow \infty$. It then carries the charge: $q(t \rightarrow \infty) = Q = C \cdot U_{\rm s}$

The process is now to be summarized in detail in formulas. Linear components are used in the circuit, i.e. the component values for the resistor $R$ and the capacitance $C$ are independent of the current or the voltage. Then definition equations for the resistor $R$ and the capacitance $C$ are also valid for time-varying or infinitesimal quantities:

\begin{align*} R = {{u_R(t)}\over{i_R(t)}} = {{{\rm d}u_R}\over{{\rm d}i_R}} = {\rm const.} \\ C = {{q(t)} \over{u_C(t)}} = {{{\rm d}q} \over{{\rm d}u_C}} = {\rm const.} \tag{5.1.1} \end{align*}

Charging a capacitor at time t=0

By considering the loop, the general result is: the voltage of the source is equal to the sum of the two voltages across the resistor and capacitor.

\begin{align*} U_{\rm s} =u_R + u_C = R \cdot i_C + u_C \tag{5.1.2} \end{align*}

At the first instant ${\rm d}t$, an infinitesimally small charge “chunk” ${\rm d}q$ flows through the circuit driven by the current $i_C$ from the voltage source. For this, $(5.1.1)$ gives:

\begin{align*} i_C = {{{\rm d}q}\over{{\rm d}t}} \quad \text{and} \quad {\rm d}q = C \cdot {\rm d}u_C \end{align*}

The charging current $i_C$ can be determined from the two formulas:

\begin{align*} i_C = C \cdot {{{\rm d}u_C}\over{{\rm d}t}} \tag{5.1.3} \end{align*}

Thus $(5.1.2)$ becomes:

\begin{align*} U_{\rm s} &= u_R + u_C \\ &= R \cdot C \cdot {{{\rm d}u_C}\over{{\rm d}t}} + u_C \end{align*}

here follows some mathematics:

This result represents a 1st order differential equation. This should generally be rewritten so that the part that depends (on the variable) is on one side and the rest is on the other. This is already present here. The appropriate approach to such a problem is:

\begin{align*} u_C(t) = \mathcal{A} \cdot {\rm e}^{\mathcal{B}\cdot t} + \mathcal{C} \end{align*}

\begin{align*} U_{\rm s} &= R \cdot C \cdot {{\rm d}\over{{\rm d}t}}(\mathcal{A} \cdot {\rm e}^{\mathcal{B}\cdot t} + \mathcal{C}) + \mathcal{A} \cdot {\rm e}^{\mathcal{B}\cdot t} + \mathcal{C} \\ &= R \cdot C \cdot \mathcal{AB} \cdot {\rm e}^{\mathcal{B}\cdot t} + \mathcal{A} \cdot {\rm e}^{\mathcal{B}\cdot t} + \mathcal{C} \\ U_{\rm s} - \mathcal{C} &= ( R \cdot C \cdot \mathcal{AB} + \mathcal{A})\cdot {\rm e}^{\mathcal{B}\cdot t} \\ \end{align*}

This equation must hold for every $t$. This is only possible if the left, as well as the right term, become equal to 0.
Thus:

\begin{align*} \mathcal{C} = U_{\rm s} \\ \\ R \cdot C \cdot \mathcal{AB} + \mathcal{A} &= 0 \quad \quad | : \mathcal{A} \quad | -1 \\ R \cdot C \cdot \mathcal{B} &= - 1 \\ \mathcal{B} &= - {{1}\over{R C}} \\ \end{align*}

So it follows:

\begin{align*} u_C(t) = \mathcal{A} \cdot {\rm e}^{\large{- {{t}\over{R C}} }} + U_{\rm s} \end{align*}

For the solution it must still hold that at time $t_0=0$ $u_C(t_0) = 0$ just holds:

\begin{align*} 0 &= \mathcal{A} \cdot {\rm e}^{\large{0}} + U_{\rm s} \\ 0 &= \mathcal{A} + U_{\rm s} \\ \mathcal{A} &= - U_{\rm s} \end{align*}

So the solution is:

\begin{align*} u_C(t) &= - U_{\rm s} \cdot {\rm e}^{\large{- {{t}\over{R C}}}} + U_{\rm s} \end{align*}

And this results in: \begin{align*} u_C(t) &= U_{\rm s} \cdot (1 - {\rm e}^{\large{- {{t}\over{R C}}}}) \end{align*}

And with $(5.1.3)$, $i_C$ becomes: \begin{align*} i_C(t) &= {{U_{\rm s}}\over{R}} \cdot {\rm e}^{\large{- {{t}\over{R C}} } } \end{align*}

In figure 4, the two time course diagrams for the charging voltage $u_C(t)$ and the charging current $i_C(t)$ of the capacitor are shown.

Discharging a capacitor at time t=0

The following situation is considered for the discharge:

• A capacitor charged to voltage $U_{\rm s}$ with capacitance $C$ is short-circuited across a resistor $R$ at time $t=t_0$.
• As a result, the full voltage $U_{\rm s}$ is initially applied to the resistor: $u_R(t_0)=U_{\rm s}$
• The initial discharge current is thus defined by the resistance: $i_C ={{u_R}\over{R}}$
• The discharging charges lower the voltage of the capacitor $u_C$, since: $u_C = {{q(t)}\over{C}}$
• Ideally, the capacitor is not fully discharged before $t \rightarrow \infty$.

Also, this process now is to put into a formula in detail. By looking at the loop, the general result is: the sum of the two voltages across the resistor and capacitor adds up to zero.

\begin{align*} 0 =u_R + u_C = R \cdot i_C + u_C \end{align*}

This gives $(5.1.3)$:

\begin{align*} 0 =u_R + u_C = R \cdot C \cdot {{{\rm d}u_C}\over{{\rm d}t}} + u_C \end{align*}

also here uses some mathematics:

This result again represents a 1st order differential equation. The appropriate approach to such a problem is:

\begin{align*} u_C(t) = \mathcal{A} \cdot {\rm e}^{\mathcal{B}\cdot t} + \mathcal{C} \end{align*}

\begin{align*} 0 &= R \cdot C \cdot {{\rm d}\over{{\rm d}t}}(\mathcal{A} \cdot {\rm e}^{\mathcal{B}\cdot t} + \mathcal{C}) + \mathcal{A} \cdot {\rm e}^{\mathcal{B}\cdot t} + \mathcal{C} \\ &= R \cdot C \cdot \mathcal{AB} \cdot {\rm e}^{\mathcal{B}\cdot t} + \mathcal{A} \cdot {\rm e}^{\mathcal{B}\cdot t} + \mathcal{C} \\ - \mathcal{C} &= ( R \cdot C \cdot \mathcal{AB} + \mathcal{A}) \cdot {\rm e}^{\mathcal{B}\cdot t} \\ \end{align*}

This equation must hold for every $t$. This is only possible if the left, as well as the right term, become equal to 0. Thus:

\begin{align*} \mathcal{C} = 0 \\ \\ R \cdot C \cdot \mathcal{AB} + \mathcal{A} &= 0 \quad \quad | : \mathcal{A} \quad | -1 \\ R \cdot C \cdot \mathcal{B} &= - 1 \\ \mathcal{B} &= - {{1}\over{R C}} \\ \end{align*}

So it follows:

\begin{align*} u_C(t) = \mathcal{A} \cdot {\rm e}^{\large{- {{t}\over{R C}} }} + 0 \end{align*}

For the solution it must still hold that at time $t_0=0$ $u_C(t_0) = U_{\rm s}$ just holds:

\begin{align*} U_{\rm s} &= \mathcal{A} \cdot {\rm e}^{\large{0}} \\ \mathcal{A} &= U_{\rm s} \end{align*}

Therefore, the result is:

\begin{align*} u_C(t) &= U_{\rm s} \cdot {\rm e}^{\large{- {{t}\over{R C}}}} \end{align*}

And this results in: \begin{align*} u_C(t) &= U_{\rm s} \cdot {\rm e}^{\large{- {{t}\over{\tau}}}} \quad \text{with} \quad \tau = R C \end{align*}

And with $(5.1.3)$, $i_C$ becomes: \begin{align*} i_C(t) &=- {{U_{\rm s}}\over{R}} \cdot {\rm e}^{\large{- {{t}\over{R C}} } } \end{align*}

In figure 6 the two time course diagrams are again shown; now for the discharge voltage $u_C(t)$ and the discharge current $i_C(t)$ of the capacitor. Since the current now flows out of the capacitor, the sign of $i_C$ is negative.

Periodic switching operations

In the simulation on the right, a periodic switching operation can be seen. The capacitor is periodically charged and discharged via the switch. Three sliders are given in the simulation to change the resistance $R$ (Resistance R), the capacity $C$ (Capacity C), and the frequency $f$ (Frequency f). In the simulation below, the voltage $u_C$ across the capacitor is shown in green and the current $i_C$ is shown in yellow.

Exercises:

1. Increase the the frequency to $f=10~{\rm kHz}$ using the appropriate slider. What is the change for $u_C$ and $i_C$?
2. Now increase the capacitance to $C=10 ~{\rm µF}$ using the corresponding slider. What is the change for $u_C$ and $i_C$?
3. Now increase the resistance to $R= 1 ~\rm k\Omega$ using the corresponding slider. What is the change for $u_C$ and $i_C$?

Energy stored in a Capacitor

Now the capacitor as energy storage is to be looked at more closely. For this, we consider again the circuit in figure 2 an. According to the chapter Preparation, Properties, and Proportions, the power for constant values (DC) is defined as:

\begin{align*} P={{\Delta W}\over{\Delta t}} = U \cdot I \end{align*}

For variable signals, the instantaneous power is given as:

\begin{align*} p={{{\rm d}w}\over{{\rm d}t}} = u \cdot i \end{align*}

Energy consideration of the capacitor

Charging the capacitor at time $t_0 = 0$ results in $\Delta W = \Delta W_C$ for the stored energy at a later time $t_1 =t$:

\begin{align*} \Delta W_C = \int_{t_0}^{t_1} {\rm d}w = \int_{0}^t u \cdot i \cdot {\rm d}t = \int_{0}^t u_C \cdot i_C {\rm d}t \tag{5.2.1} \end{align*}

During the charging process \begin{align*} u_C(t) = U_{\rm s} \cdot (1 - {\rm e}^{ -{{t}\over{\tau}} }) \\ i_C(t) = {{U_{\rm s}}\over{R}} \cdot {\rm e}^{ -{{t}\over{\tau}} } \tag{5.2.2} \end{align*}

In particular:

\begin{align*} C = {{q(t)}\over{u_C(t)}} \quad &\rightarrow \quad &q(t) &= {u_C(t)}\cdot{C} \\ i_C(t) = {{{\rm d} q(t)}\over{{\rm d}t}} \quad &\xrightarrow{C=\rm const.} \quad &i_C(t) &= C \cdot {{{\rm d} u_C(t)}\over{{\rm d}t}} \end{align*}

Thus, the stored energy from formula $(5.2.1)$:

\begin{align*} \Delta W_C &= \int_{0}^t u_C(t) \cdot C \cdot {{{\rm d} u_C(t)}\over{{\rm d}t}} {\rm d}t \quad & | \text{ substitution of integration variable: } t \rightarrow u_C\\ &= \int_{U_0}^{U_1} u_C(t) \cdot C \cdot {\rm d}u_C \quad & | \text{ Since the capacity is constant, it can be written ahead of the integral}\\ &= C \cdot \int_{U_0}^{U_1} u_C \, {\rm d} u_C \\ &= C \cdot \left[{{1}\over{2}} u_C^2 \right] _{U_0}^{U_1} \\ \end{align*} \begin{align*} \boxed{\Delta W_C= {{1}\over{2}} C \cdot (U_1^2-U_0^2)} \tag{5.2.3} \end{align*}

Thus, for a fully discharged capacitor ($U_{\rm s}=0~{\rm V}$), the energy stored when charging to voltage $U_{\rm s}$ is $\Delta W_C={{1}\over{2}} C \cdot U_{\rm s}^2$.

Energy Consideration on the Resistor

The converted energy can also be determined for the resistor:

\begin{align*} \Delta W_R = \int_{0}^t u_R \cdot i_R {\rm d}t = \int_{0}^t R \cdot i_R \cdot i_R {\rm d}t = R \cdot \int_{0}^t i_R^2 {\rm d}t \end{align*}

Since the current through the capacitor $i_C$ is equal to that through the resistor $i_R$, it follows via $(5.2.2)$:

\begin{align*} \Delta W_R &= R \cdot \int_{0}^t \left( { {U_{\rm s}}\over{R}} \cdot {\rm e}^ { -{{t}\over{\tau}}} \right)^2 {\rm d}t \\ &= { {U_{\rm s}^2}\over{R}} \cdot \int_{0}^t {\rm e}^ { -{{2 \cdot t}\over{\tau}}} {\rm d}t \\ &= { {U_{\rm s}^2}\over{R}} \cdot \left[ -{{\tau }\over{2}} \cdot {\rm e}^ { -{{2 \cdot t}\over{\tau}}} \right]_{0}^t \quad & | \text{with } \tau = R \cdot C \\ &= -{{1}\over{2}} \cdot { U_{\rm s}^2}\cdot{C} \cdot \left[ {\rm e}^ { -{{2 \cdot t}\over{\tau}}} \right]_{0}^t \\ \end{align*}

For $t \rightarrow \infty$ we get:

\begin{align*} \Delta W_R &= -{{1}\over{2}} \cdot {U_{\rm s}^2}\cdot{C} \cdot \left[ {\rm e}^ { -{{2 \cdot t}\over{\tau}}} \right]_{0}^{\infty} \\ &= -{{1}\over{2}} \cdot {U_{\rm s}^2}\cdot{C} \cdot \left[ 0 - 1 \right] \\ \end{align*} \begin{align*} \boxed{ \Delta W_R = {{1}\over{2}} \cdot {U_{\rm s}^2}\cdot{C}} \tag{5.2.4} \end{align*}

This means that the energy converted at the resistor is independent of the resistance value (for an ideal constant voltage source $U_{\rm s}$ and given capacitor $C$)! At first, this doesn't really sound comprehensible. No matter if there is a very large resistor $R_1$ or a tiny small resistor $R_2$: The same waste heat is always produced. Graphically, this apparent contradiction can be resolved like this: A higher resistor $R_2$ slows down the small charge packets $\Delta q_1$, $\Delta q_2$, … $\Delta q_n$ more strongly. But a considered single charge packet $\Delta q_k$ will nevertheless pass the same voltage across the resistor $R_1$ or $R_2$ since this is given only by the accumulated packets in the capacitor: $u_r = U_{\rm s} - u_C = U_{\rm s} - {{q}\over{C}}$.

In real applications, as mentioned in previous chapters, ideal voltage sources are not possible. Thus, without a real resistor, the waste heat is dissipated proportionally to the internal resistance of the source and the internal resistance of the capacitor. The internal resistance of the capacitor depends on the frequency but is usually smaller than the internal resistance of the source.

Consideration of total energy turnover

In the previous considerations, the energy conversion during the complete charging process was also considered. It was found that the capacitor stores the energy $W_C= {{1}\over{2}} \cdot {U_{\rm s}^2}\cdot{C} $ (see $(5.2.3)$) and at the resistor the energy $W_R= {{1}\over{2}} \cdot {U_{\rm s}^2}\cdot{C} $ (see $(5.2.4)$) into heat. So, in total, the voltage source injects the following energy:

\begin{align*} \Delta W_0 &=\Delta W_R + \Delta W_C = {U_{\rm s}^2}\cdot{C} \end{align*}

This also follows via $(5.2.1)$:

\begin{align*} \Delta W_0 &= \int_{0}^{\infty} u_0 \cdot i_0 \cdot {\rm d}t \quad | \quad u_0 = U_{\rm s} \text{ is constant because constant voltage source!} \\ &= U_{\rm s} \cdot \int_{0}^{\infty} i_C {\rm d}t \\ &= U_{\rm s} \cdot \int_{0}^{\infty} {{{\rm d}q}\over{{\rm d}t}} {\rm d}t \\ &= U_{\rm s} \cdot \int_{0}^Q {\rm d}q = U_{\rm s} \cdot Q \quad | \quad \text{where } Q= C \cdot U_{\rm s} \\ &= U_{\rm s}^2 \cdot C \\ \end{align*}

This means that only half of the energy emitted by the source is stored in the capacitor! Again, This doesn't really sound comprehensible at first. And again, it helps to look at small packets of charge that have to be transferred from the ideal source to the capacitor. figure 8 shows current and voltage waveforms across the capacitor and the stored energy for different resistance values. There, too, it can be seen that the maximum stored energy (dashed line in the figure at right) is given by $\Delta W= {{1}\over{2}} {U_{\rm s}^2}\cdot{C} $ alone. $U_{\rm s}^2 \cdot C = {{1}\over{2}} \cdot (5~{\rm V})^2 \cdot 1 ~{\rm µF} = 12.5 ~{\rm µWs}$ is given.

This can also be tested in the following simulation. In addition to the RC element shown so far, a power meter and an integrator are also drawn here. It is possible to display the instantaneous power and the stored energy. Via the slider Resistance R the resistance value can be varied. The following values are shown in the oscilloscopes:

• left: Current $u_C$ and voltage $i_C$ at the capacitor.
• middle: Instantaneous power $p_C = u_C \cdot i_C$ of the capacitor.
• right: stored energy $w_C = \int u_C \cdot i_C \; {\rm d}t$ of the capacitor

Common pitfalls

• …

Exercises

Worked examples

Exercise E 1.2 Capacitor charging/discharging

The following circuit shows a charging/discharging circuit for a capacitor.

The values of the components shall be the following:

• $R_1 = 1.0 \rm k\Omega$
• $R_2 = 2.0 \rm k\Omega$
• $R_3 = 3.0 \rm k\Omega$
• $C = 1 \rm \mu F$
• $S_1$ and $S_2$ are opened in the beginning (open-circuit)

1. For the first tasks, the switch $S_1$ gets closed at $t=t_0 = 0s$.

1.1 What is the value of the time constant $\tau_1$?

Solution

The time constant $\tau$ is generally given as: $\tau= R\cdot C$.
Now, we try to determine which $R$ and $C$ must be used here.
To find this out, we have to look at the circuit when $S_1$ gets closed.

We see that for the time constant, we need to use $R=R_1 + R_2$.

Result

\begin{align*} \tau_1 &= R\cdot C \\ &= (R_1 + R_2) \cdot C \\ &= 3~\rm ms \\ \end{align*}

1.2 What is the formula for the voltage $u_{R2}$ over the resistor $R_2$? Derive a general formula without using component values!

Solution

To get a general formula, we again look at the circuit, but this time with the voltage arrows.

We see, that: $U_1 = u_C + u_{R2}$ and there is only one current in the loop: $i = i_C = i_{R2}$
The current is generally given with the exponential function: $i_c = {{U}\over{R}}\cdot e^{-t/\tau}$, with $R$ given here as $R = R_1 + R_2$. Therefore, $u_{R2}$ can be written as:

\begin{align*} u_{R2} &= R_2 \cdot i_{R2} \\ &= U_1 \cdot {{R_2}\over{R_1 + R_2}} \cdot e^{-t/ \tau} \end{align*}

Result

\begin{align*} u_{R2} = U_1 \cdot {{R_2}\over{R_1 + R_2}} \cdot e^{t/ \tau} \end{align*}

2. At a distinct time $t_1$, the voltage $u_C$ is charged up to $4/5 \cdot U_1$. At this point, the switch $S_1$ will be opened.
Calculate $t_1$!

Solution

We can derive $u_{C}$ based on the exponential function: $u_C(t) = U_1 \cdot (1-e^{-t/\tau})$.
Therefore, we get $t_1$ by:

\begin{align*} u_C = 4/5 \cdot U_1 &= U_1 \cdot (1-e^{-t/\tau}) \\ 4/5 &= 1-e^{-t/\tau} \\ e^{-t/\tau} &= 1-4/5 = 1/5\\ -t/\tau &= \rm ln (1/5) \\ t &= -\tau \cdot \rm ln (1/5) \\ \end{align*}

Result

\begin{align*} t &= 3~{\rm ms} \cdot 1.61 \approx 4.8~\rm ms \\ \end{align*}

3. The switch $S_2$ will get closed at the moment $t_2 = 10 ~\rm ms$. The values of the voltage sources are now: $U_1 = 5.0 ~\rm V$ and $U_2 = 10 ~\rm V$.

3.1 What is the new time constant $\tau_2$?

Solution

Again, the time constant $\tau$ is given as: $\tau= R\cdot C$.
Again, we try to determine which $R$ and $C$ must be used here.
To find this out, we have to look at the circuit when $S_1$ is open and $S_2$ is closed.

We see that for the time constant, we now need to use $R=R_3 + R_2$.

\begin{align*} \tau_2 &= R\cdot C \\ &= (R_3 + R_2) \cdot C \\ \end{align*}

Result

\begin{align*} \tau_2 &= 5~\rm ms \\ \end{align*}

3.2 Calculate the moment $t_3$ when $u_{R2}$ is smaller than $1/10 \cdot U_2$.

Solution

To calculate the moment $t_3$ when $u_{R2}$ is smaller than $1/10 \cdot U_2$, we first have to find out the value of $u_{R2}(t_2 = 10 ~\rm ms)$, when $S_2$ just got closed.

• Starting from $t_2 = 10 ~\rm ms$, the voltage source $U_2$ charges up the capacitor $C$ further.
• Before at $t_1$, when $S_1$ got opened, the value of $u_c$ was: $u_c(t_1) = 4/5 \cdot U_1 = 4 ~\rm V$.
• This is also true for $t_2$, since between $t_1$ and $t_2$ the charge on $C$ does not change: $u_c(t_2) = 4 ~\rm V$.
• In the first moment after closing $S_2$ at $t_2$, the voltage drop on $R_3 + R_2$ is: $U_{R3+R2} = U_2 - u_c(t_2) = 6 ~\rm V$.
• So the voltage divider of $R_3 + R_2$ lead to $ \boldsymbol{u_{R2}(t_2 = 10 ~\rm ms)} = {{R_2}\over{R_3 + R2}} \cdot U_{R3+R2} = {{2 {~\rm k\Omega}}\over{3 {~\rm k\Omega} + 2 {~\rm k\Omega} }} \cdot 6 ~\rm V = \boldsymbol{2.4 ~\rm V} $

We see that the voltage on $R_2$ has to decrease from $2.4 ~\rm V $ to $1/10 \cdot U_2 = 1 ~\rm V$.
To calculate this, there are multiple ways. In the following, one shall be retraced:

• We know, that the current $i_C = i_{R2}$ subsides exponentially: $i_{R2}(t) = I_{R2~ 0} \cdot {\rm e}^{-t/\tau}$
• So we can rearrange the task to focus on the change in current instead of the voltage.
• The exponential decay is true regardless of where it starts.

So from ${{i_{R2}(t)}\over{I_{R2~ 0}}} = {\rm e}^{-t/\tau}$ we get \begin{align*} {{i_{R2}(t_3)}\over{i_{R2}(t_2)}} &= {\rm exp} \left( -{{t_3 - t_2}\over{\tau_2}} \right) \\ -{{t_3 - t_2}\over{\tau_2}} &= {\rm ln } \left( {{i_{R2}(t_3)}\over{i_{R2}(t_2)}} \right) \\ t_3 &= t_2 - \tau_2 \cdot {\rm ln } \left( {{i_{R2}(t_3)}\over{i_{R2}(t_2)}} \right) \\ t_3 &= 10 ~{\rm ms} - 5~{\rm ms} \cdot {\rm ln } \left( {{1 ~\rm V }\over{2.4 ~\rm V }} \right) \\ \end{align*}

Result

\begin{align*} t_3 &= 14.4~\rm ms \\ \end{align*}

3.3 Draw the course of time of the voltage $u_C(t)$ over the capacitor.

Result

Embedded resources