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electrical_engineering_1:dc_circuit_transients [2023/12/01 02:05]
mexleadmin [Bearbeiten - Panel] 		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}}
  
-1. For the first tasks the switch $S_1$ gets closed at $t=t_0 = 0s$. \\+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 $\tau_1$? 1.1 What is the value of the time constant $\tau_1$?
  
-1.2 What is the formula of the voltage $u_{R2}$ over the resistor $R_2$? Derive a general formula without using component values!+#@HiddenBegin_HTML~Solution1,Solution~@#
  
-2. At a distinct time $t_1$ the voltage $u_C$ is charged up to $4/5 \cdot U_1$. At this point of time, the switch $S_1$ will be closed. \\ Calculate $t_1$ !+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.
  
-3. The switch $S_2$ will get closed at the moment $t_2 = 10 ~\rm ms$.+{{drawio>electrical_engineering_1:Exercise522sol1.svg}} 
 + 
 +We see that for the time constant, we need to use $R=R_1 + R_2$. 
 + 
 +#@HiddenEnd_HTML~Solution1,Solution ~@# 
 + 
 +#@HiddenBegin_HTML~Result1,Result~@# 
 +\begin{align*} 
 +\tau_1 &= R\cdot C \\ 
 +       &= (R_1 + R_2) \cdot C \\ 
 +       &= 3~\rm ms \\ 
 +\end{align*} 
 + 
 +#@HiddenEnd_HTML~Result1,Result~@# 
 + 
 +1.2 What is the formula for the voltage $u_{R2}$ over the resistor $R_2$? Derive a general formula without using component values! 
 + 
 +#@HiddenBegin_HTML~Solution2,Solution~@# 
 + 
 +To get a general formula, we again take a look at the circuit, but this time with the voltage arrows. 
 + 
 +{{drawio>electrical_engineering_1:Exercise522sol2.svg}} 
 + 
 +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*} 
 + 
 +#@HiddenEnd_HTML~Solution2,Solution ~@# 
 + 
 +#@HiddenBegin_HTML~Result2,Result~@# 
 +\begin{align*} 
 +u_{R2} = U_1 \cdot {{R_2}\over{R_1 + R_2}} \cdot e^{t/ \tau} 
 +\end{align*} 
 +#@HiddenEnd_HTML~Result2,Result~@# 
 + 
 +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 $t_1$! 
 + 
 +#@HiddenBegin_HTML~Solution3,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*} 
 + 
 +#@HiddenEnd_HTML~Solution3,Solution ~@# 
 + 
 +#@HiddenBegin_HTML~Result3,Result~@# 
 +\begin{align*} 
 +          t                      &=  3~{\rm ms} \cdot 1.61 = 4.8~\rm ms \\ 
 +\end{align*} 
 +#@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 $U_2 = 10 ~\rm V$.
  
 3.1 What is the new time constant $\tau_2$? 3.1 What is the new time constant $\tau_2$?
  
-3.2 Calculate the moment $t_3$, when $u_{R2}$ is smaller than $1/10 \cdot U_2$+#@HiddenBegin_HTML~Solution4,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. 
 + 
 +{{drawio>electrical_engineering_1:Exercise522sol4.svg}} 
 + 
 +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 ~@# 
 + 
 +#@HiddenBegin_HTML~Result4,Result~@# 
 +\begin{align*} 
 +\tau_2 &= 5~\rm ms \\ 
 +\end{align*} 
 + 
 +#@HiddenEnd_HTML~Result4,Result~@# 
 +3.2 Calculate the moment $t_3$ when $u_{R2}$ is smaller than $1/10 \cdot U_2$.
  
 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.
 #@TaskEnd_HTML@# #@TaskEnd_HTML@#
  
 +{{page>aufgabe_7.2.6_mit_rechnung&nofooter}}
  
-#@TaskTitle_HTML@#5.2.2 Charge balance of two capacitors \\ <fs medium>(educational exercise, not part of an exam)</fs>#@TaskText_HTML@#+#@TaskTitle_HTML@#5.2.4 Charge balance of two capacitors \\ <fs medium>(educational exercise, not part of an exam)</fs>#@TaskText_HTML@#
  
  
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