Unterschiede
Hier werden die Unterschiede zwischen zwei Versionen angezeigt.
Beide Seiten der vorigen Revision Vorhergehende Überarbeitung Nächste Überarbeitung | Vorhergehende Überarbeitung | ||
electrical_engineering_1:dc_circuit_transients [2023/12/02 00:38] mexleadmin [Exercises] |
electrical_engineering_1:dc_circuit_transients [2023/12/03 16:53] (aktuell) mexleadmin [Exercises] |
||
---|---|---|---|
Zeile 479: | Zeile 479: | ||
The values of the components shall be the following: | The values of the components shall be the following: | ||
- | * $R_1 = 1 \rm k\Omega$ | + | * $R_1 = 1.0 \rm k\Omega$ |
- | * $R_2 = 2 \rm k\Omega$ | + | * $R_2 = 2.0 \rm k\Omega$ |
- | * $R_3 = 3 \rm k\Omega$ | + | * $R_3 = 3.0 \rm k\Omega$ |
* $C = 1 \rm \mu F$ | * $C = 1 \rm \mu F$ | ||
- | * $S_1$ and $S_2$ are opened (open-circuit) | + | * $S_1$ and $S_2$ are opened |
{{drawio> | {{drawio> | ||
Zeile 538: | Zeile 538: | ||
2. At a distinct time $t_1$, the voltage $u_C$ is charged up to $4/5 \cdot U_1$. | 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 $S_2$ will be closed. \\ Calculate $t_1$! | + | At this point, the switch $S_1$ will be opened. \\ Calculate $t_1$! |
# | # | ||
We can derive $u_{C}$ based on the exponential function: $u_C(t) = U_1 \cdot (1-e^{-t/ | We can derive $u_{C}$ based on the exponential function: $u_C(t) = U_1 \cdot (1-e^{-t/ | ||
- | Therefore we get $t_1$ by: | + | Therefore, we get $t_1$ by: |
\begin{align*} | \begin{align*} | ||
Zeile 557: | Zeile 557: | ||
# | # | ||
\begin{align*} | \begin{align*} | ||
- | t & | + | t & |
\end{align*} | \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: $U_1 = 5 ~\rm V$ and $U_2 = 10 ~\rm V$. | + | 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 $\tau_2$? | ||
# | # | ||
- | Again the time constant $\tau$ is given as: $\tau= R\cdot C$. \\ | + | 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. \\ | 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 both $S_1$ and $S_2$ are closed. | + | To find this out, we have to look at the circuit when $S_1$ is open and $S_2$ is closed. |
{{drawio> | {{drawio> | ||
- | We see that for the time constant, we need to use $R=R_1 + R_2$. | + | We see that for the time constant, we now need to use $R=R_3 + R_2$. |
+ | |||
+ | \begin{align*} | ||
+ | \tau_2 &= R\cdot C \\ | ||
+ | & | ||
+ | \end{align*} | ||
# | # | ||
Zeile 577: | Zeile 584: | ||
# | # | ||
\begin{align*} | \begin{align*} | ||
- | \tau_1 & | + | \tau_2 & |
- | & | + | |
- | & | + | |
\end{align*} | \end{align*} | ||
- | |||
# | # | ||
- | |||
- | 3.1 What is the new time constant $\tau_2$? | ||
3.2 Calculate the moment $t_3$ when $u_{R2}$ is smaller than $1/10 \cdot U_2$. | 3.2 Calculate the moment $t_3$ when $u_{R2}$ is smaller than $1/10 \cdot U_2$. | ||
+ | |||
+ | # | ||
+ | |||
+ | 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: | ||
+ | * 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/ | ||
+ | \begin{align*} | ||
+ | {{i_{R2}(t_3)}\over{i_{R2}(t_2)}} & | ||
+ | -{{t_3 - t_2}\over{\tau_2}} | ||
+ | | ||
+ | | ||
+ | \end{align*} | ||
+ | |||
+ | # | ||
+ | |||
+ | # | ||
+ | \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. | 3.3 Draw the course of time of the voltage $u_C(t)$ over the capacitor. | ||
+ | |||
+ | {{drawio> | ||
+ | |||
+ | |||
+ | # | ||
+ | {{drawio> | ||
+ | # | ||
+ | |||
# | # | ||