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electrical_engineering_1:dc_circuit_transients [2023/12/02 00:55]
mexleadmin [Exercises] 		electrical_engineering_1:dc_circuit_transients [2023/12/03 16:53]
mexleadmin [Exercises]
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 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 in the beginning (open-circuit)   * $S_1$ and $S_2$ are opened in the beginning (open-circuit)
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 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 $t_1$!
  
 #@HiddenBegin_HTML~Solution3,Solution~@# #@HiddenBegin_HTML~Solution3,Solution~@#
  
 We can derive $u_{C}$ based on the exponential function: $u_C(t) = U_1 \cdot (1-e^{-t/\tau})$. \\ 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:+Therefore, we get $t_1$ by:
  
 \begin{align*} \begin{align*}
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 #@HiddenBegin_HTML~Result3,Result~@# #@HiddenBegin_HTML~Result3,Result~@#
 \begin{align*} \begin{align*}
-          t                      &=  3~{\rm ms} \cdot 1.61 = 4.8~\rm ms \\+          t                      &=  3~{\rm ms} \cdot 1.61 \approx 4.8~\rm ms \\
 \end{align*} \end{align*}
 #@HiddenEnd_HTML~Result3,Result~@# #@HiddenEnd_HTML~Result3,Result~@#
  
-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 $\tau_2$? 3.1 What is the new time constant $\tau_2$?
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 #@HiddenBegin_HTML~Solution4,Solution~@# #@HiddenBegin_HTML~Solution4,Solution~@#
  
-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.
-In this case, we can "fold" over the resistor $R_3$ to the left-side of the capacitor $C$.+
  
 {{drawio>electrical_engineering_1:Exercise522sol4.svg}} {{drawio>electrical_engineering_1:Exercise522sol4.svg}}
  
-We see that for the time constant, we now need to use $R=R_1 || R_3 + R_2$.+We see that for the time constant, we now need to use $R=R_3 + R_2$.
  
 \begin{align*} \begin{align*}
 \tau_2 &= R\cdot C \\ \tau_2 &= R\cdot C \\
-       &= (R_1 || R_3 + R_2) \cdot C \\ +       &= (R_3 + R_2) \cdot C \\
-       &= \left({{\rm  1 ~k\Omega \cdot 3 ~k\Omega }\over{\rm  1 ~k\Omega + 3 ~k\Omega}} + 2 ~{\rm k\Omega}\right) \cdot 1 ~{\rm \mu F} \\+
 \end{align*} \end{align*}
  
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 #@HiddenBegin_HTML~Result4,Result~@# #@HiddenBegin_HTML~Result4,Result~@#
 \begin{align*} \begin{align*}
-\tau_2 &= 2.75~\rm ms \\+\tau_2 &= 5~\rm ms \\
 \end{align*} \end{align*}
- 
 #@HiddenEnd_HTML~Result4,Result~@# #@HiddenEnd_HTML~Result4,Result~@#
 +
 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 $1/10 \cdot U_2$.
 +
 +#@HiddenBegin_HTML~Solution5,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*}
 +
 +#@HiddenEnd_HTML~Solution5,Solution ~@#
 +
 +#@HiddenBegin_HTML~Result5,Result~@#
 +\begin{align*}
 +t_3 &= 14.4~\rm ms \\
 +\end{align*}
 +#@HiddenEnd_HTML~Result5,Result~@#
  
 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>electrical_engineering_1:Exercise522task6.svg}}
 +
 +
 +#@HiddenBegin_HTML~Result6,Result~@#
 +{{drawio>electrical_engineering_1:Exercise522sol6.svg}}
 +#@HiddenEnd_HTML~Result6,Result~@#
 +
 #@TaskEnd_HTML@# #@TaskEnd_HTML@#