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electrical_engineering_1:dc_circuit_transients [2023/12/02 00:38]
mexleadmin [Exercises] 		electrical_engineering_1:dc_circuit_transients [2023/12/02 01:08]
mexleadmin [Exercises]
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   * $R_3 = 3 \rm k\Omega$   * $R_3 = 3 \rm k\Omega$
   * $C   = 1 \rm \mu F$   * $C   = 1 \rm \mu F$
-  * $S_1$ and $S_2$ are opened (open-circuit)+  * $S_1$ and $S_2$ are opened in the beginning (open-circuit)
  
 {{drawio>electrical_engineering_1:Exercise522setup.svg}} {{drawio>electrical_engineering_1:Exercise522setup.svg}}
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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*}
Zeile 562: Zeile 562:
  
 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: $U_1 = 5 ~\rm V$ and $U_2 = 10 ~\rm V$.
 +
 +3.1 What is the new time constant $\tau_2$?
  
 #@HiddenBegin_HTML~Solution4,Solution~@# #@HiddenBegin_HTML~Solution4,Solution~@#
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 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>electrical_engineering_1:Exercise522sol4.svg}} {{drawio>electrical_engineering_1:Exercise522sol4.svg}}
  
-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 \\ 
 +       &= (R_3 + R_2) \cdot C \\ 
 +\end{align*}
  
 #@HiddenEnd_HTML~Solution4,Solution ~@# #@HiddenEnd_HTML~Solution4,Solution ~@#
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 #@HiddenBegin_HTML~Result4,Result~@# #@HiddenBegin_HTML~Result4,Result~@#
 \begin{align*} \begin{align*}
-\tau_1 &= R\cdot C \\ +\tau_2 &= 5~\rm ms \\
-       &= (R_1 + R_2) \cdot C \\ +
-       &= 3~\rm ms \\+
 \end{align*} \end{align*}
  
 #@HiddenEnd_HTML~Result4,Result~@# #@HiddenEnd_HTML~Result4,Result~@#
- 
-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 $1/10 \cdot U_2$.