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electrical_engineering_1:dc_circuit_transients [2023/12/02 00:52]
mexleadmin [Exercises] 		electrical_engineering_1:dc_circuit_transients [2023/12/02 01:08]
mexleadmin [Exercises]
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 $t_1$!
  
 #@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 569: Zeile 569:
 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 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 \\
-       &= {{\rm  1 ~k\Omega \cdot 3 ~k\Omega }\over{\rm  1 ~k\Omega + 3 ~k\Omega}} + 2 ~{\rm k\Omega}) \cdot 1 ~{\rm \mu F} \\+
 \end{align*} \end{align*}
  
Zeile 585: Zeile 584:
 #@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*}