Beide Seiten der vorigen Revision
Vorhergehende Überarbeitung
Nächste Überarbeitung
|
Vorhergehende Überarbeitung
Nächste Überarbeitung
Beide Seiten der Revision
|
electrical_engineering_1:dc_circuit_transients [2023/11/29 23:59] mexleadmin |
electrical_engineering_1:dc_circuit_transients [2023/12/02 00:55] mexleadmin [Exercises] |
</WRAP></WRAP></panel> | </WRAP></WRAP></panel> |
| |
<panel type="info" title="Exercise 5.2.2 Charge balance of two capacitors"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> | #@TaskTitle_HTML@# Exercise 5.2.2 Capacitor charging/discharging #@TaskText_HTML@# |
| |
| The following circuit shows a charging/discharging circuit for a capacitor. |
| |
| The values of the components shall be the following: |
| * $R_1 = 1 \rm k\Omega$ |
| * $R_2 = 2 \rm k\Omega$ |
| * $R_3 = 3 \rm k\Omega$ |
| * $C = 1 \rm \mu F$ |
| * $S_1$ and $S_2$ are opened in the beginning (open-circuit) |
| |
| {{drawio>electrical_engineering_1:Exercise522setup.svg}} |
| |
| 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 $\tau_1$? |
| |
| #@HiddenBegin_HTML~Solution1,Solution~@# |
| |
| 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. |
| |
| {{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 $S_2$ will be closed. \\ 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 $\tau_2$? |
| |
| #@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 both $S_1$ and $S_2$ are 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}} |
| |
| We see that for the time constant, we now need to use $R=R_1 || R_3 + R_2$. |
| |
| \begin{align*} |
| \tau_2 &= R\cdot C \\ |
| &= (R_1 || 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*} |
| |
| #@HiddenEnd_HTML~Solution4,Solution ~@# |
| |
| #@HiddenBegin_HTML~Result4,Result~@# |
| \begin{align*} |
| \tau_2 &= 2.75~\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. |
| #@TaskEnd_HTML@# |
| |
| {{page>aufgabe_7.2.6_mit_rechnung&nofooter}} |
| |
| #@TaskTitle_HTML@#5.2.4 Charge balance of two capacitors \\ <fs medium>(educational exercise, not part of an exam)</fs>#@TaskText_HTML@# |
| |
<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=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-KmvXR-oW9gADyoXz05JI6b09RQ7SPVDA5LwfBIp40EHu2UKd+QKTfeln3wfpid4aABIRsAB8izgId7YJYeD3EQDgpEBsawTg5K0BgqDknw0G3seGAkCEOLVOSQFbLBKjtFhIhCPhHDHigWE4pAEBEaSlCUceuB8MwDxPHWID8c2mQgJwoycDkIwSUiACWgoQkygoDEySIABaMEyjDzn4cmMC0ILgrpsItDk8KIn66K9CARIQJsxLGGY6KAIhEJLZCmlLUrSCrMqyKotFyUq8gKwpihKKoymF8qSrQloqkyIZ6sqyoAA7bmaFpWllB5jh24QMOWSQCMZXqmeZllKi0AZJeaDLpTuJR7nuNZPJAz4fJeXTZJMjRthUbUCPWl6dj1GBVmUA2ugIBV3P0BXZC5kytROYRJs4a2kr1raTYWq0dF2NxbeNTRlMadBrMwayoKgKQYEh+jXIIsyhJ2-YSBw1y1LscQ9lYH3sIKpKbIeyBvmYGYOe0QZphiMAqHFgMMASlig+DegKIS0Nxcj5Lw-dkRA6+oho2xFjiIg0DsdimYEymnn44jRP2STYNk-9lMQOS5PwHT95w7AdNI10INsz2FOY-c-0I2SNmM5EJDzOi7GWAAkhCcnyaMTIUoKIxIkMfgTEAA noborder}} </WRAP> | <WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=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-KmvXR-oW9gADyoXz05JI6b09RQ7SPVDA5LwfBIp40EHu2UKd+QKTfeln3wfpid4aABIRsAB8izgId7YJYeD3EQDgpEBsawTg5K0BgqDknw0G3seGAkCEOLVOSQFbLBKjtFhIhCPhHDHigWE4pAEBEaSlCUceuB8MwDxPHWID8c2mQgJwoycDkIwSUiACWgoQkygoDEySIABaMEyjDzn4cmMC0ILgrpsItDk8KIn66K9CARIQJsxLGGY6KAIhEJLZCmlLUrSCrMqyKotFyUq8gKwpihKKoymF8qSrQloqkyIZ6sqyoAA7bmaFpWllB5jh24QMOWSQCMZXqmeZllKi0AZJeaDLpTuJR7nuNZPJAz4fJeXTZJMjRthUbUCPWl6dj1GBVmUA2ugIBV3P0BXZC5kytROYRJs4a2kr1raTYWq0dF2NxbeNTRlMadBrMwayoKgKQYEh+jXIIsyhJ2-YSBw1y1LscQ9lYH3sIKpKbIeyBvmYGYOe0QZphiMAqHFgMMASlig+DegKIS0Nxcj5Lw-dkRA6+oho2xFjiIg0DsdimYEymnn44jRP2STYNk-9lMQOS5PwHT95w7AdNI10INsz2FOY-c-0I2SNmM5EJDzOi7GWAAkhCcnyaMTIUoKIxIkMfgTEAA noborder}} </WRAP> |
- What are the energies and the total energy? \\ How is this understandable with the previous total energy? | - What are the energies and the total energy? \\ How is this understandable with the previous total energy? |
| |
</WRAP></WRAP></panel> | #@TaskEnd_HTML@# |
| |
{{page>aufgabe_7.2.6_mit_rechnung&nofooter}} | |
| |
| |