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electrical_engineering_and_electronics_2:block01 [2026/03/05 01:58] – mexleadminelectrical_engineering_and_electronics_2:block01 [2026/04/06 23:36] (current) – mexleadmin
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 ===== Learning Objectives ===== ===== Learning Objectives =====
 <callout> <callout>
-After this 90-minute block, you can +After this 90-minute block, you  
-  * ...+  * know the time constant $\tau$ and in particularly calculate it. 
 +  * determine the time characteristic of the currents and voltages at the RC element for a given resistance and capacitance. 
 +  * know the continuity conditions of electrical quantities. 
 +  * know when (=according to which measure) the capacitor is considered to be fully charged/discharged, i.e. a steady state can be considered to have been reached. 
 +  * can calculate the energy content in a capacitor. 
 +  * can calculate the change in energy of a capacitor resulting from a change in voltage between the capacitor terminals. 
 +  * can calculate (initial) current, (final) voltage, and charge when balancing the charge of several capacitors (also via resistors).
 </callout> </callout>
 +
  
 ===== Preparation at Home ===== ===== Preparation at Home =====
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 </callout> </callout>
  
-===== 5.1 Time Course of the Charging and Discharging Process =====+====Time Course of the Charging and Discharging Process ====
  
-<callout> 
- 
-=== Learning Objectives === 
- 
-By the end of this section, you will be able to: 
-  - know the time constant $\tau$ and in particularly calculate it. 
-  - determine the time characteristic of the currents and voltages at the RC element for a given resistance and capacitance. 
-  - know the continuity conditions of electrical quantities. 
-  - know when (=according to which measure) the capacitor is considered to be fully charged/discharged, i.e. a steady state can be considered to have been reached. 
- 
-</callout> 
  
 In the simulation below you can see the circuit mentioned above in a slightly modified form: In the simulation below you can see the circuit mentioned above in a slightly modified form:
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 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
  
-===== 5.2 Energy stored in a Capacitor ===== +==== Energy stored in a Capacitor ====
- +
-<callout> +
- +
-=== Learning Objectives === +
- +
-By the end of this section, you will be able to: +
-  - calculate the energy content in a capacitor. +
-  - calculate the change in energy of a capacitor resulting from a change in voltage between the capacitor terminals. +
-  - calculate (initial) current, (final) voltage, and charge when balancing the charge of several capacitors (also via resistors). +
- +
-</callout>+
  
 <WRAP> <imgcaption imageNo02 | circuit for viewing the charge curve> </imgcaption> {{drawio>SchaltungEntladekurve2.svg}} </WRAP> <WRAP> <imgcaption imageNo02 | circuit for viewing the charge curve> </imgcaption> {{drawio>SchaltungEntladekurve2.svg}} </WRAP>
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-<panel type="info" title="Exercise 5.2.1 Capacitor charging/discharging practice Exercise "> +<panel type="info" title="Exercise 1.1 Capacitor charging/discharging practice Exercise "> 
 <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
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 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-#@TaskTitle_HTML@# 5.2.2 Capacitor charging/discharging #@TaskText_HTML@#+#@TaskTitle_HTML@# 1.2 Capacitor charging/discharging #@TaskText_HTML@#
  
 The following circuit shows a charging/discharging circuit for a capacitor. The following circuit shows a charging/discharging circuit for a capacitor.
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 #@TaskEnd_HTML@# #@TaskEnd_HTML@#
  
-{{page>aufgabe_7.2.6_mit_rechnung&nofooter}} +#@TaskTitle_HTML@#1.3 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@#+
  
  
-<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=CQAgjCAMB0l3EZgBwDZkCYAsBWAnDgOxYDMG52IJEOIpIOApgLRhgBQAxuBsnX2F7hCqOlFjxIeaTNly84sNMKRsYQmFWFCyNOEjjJkDtxI4M-KueGishyfMcyQeaMjgY8hHJmR5UFgYwRhwA5jwCIhHgeIFQ7OFmFuqiSTFxkOwYkITRWAJCYLYgAK5gABQAwmAAlFk5Vhb5jeDFZVUYdQDOIOR2ZL2EFgMGEABmAIYANl2M7ADueZG2fJkkKlSQds1pzYjxAMo2ln2WQSCTM4xQvewATr2UqHYYQyDPNxyLrxYfJFtnBaDYbDAHNTLQEAAUQAdow7qEAJ4AW0YABd4TcLMMQJVeliQABVcAkiwAdTEBhIIAAapUaQcoVNkQAdLpgNkYNkkNlYNk4Fl3ADidwA9iUYQATJmsroC4ViiXS5ncqAs7nqrowHAAKnKE2YnBqAD0MHqAEaGmpsyCa21au3sLCaBiCKgWHCoAwfCCw+FI1EYh72nla6C6-VW00Wq3gG1xrXsABKrpB-CpGToQQk8CcThu2Zw9VyODdYCw23c4ArYnalU6xdT70r3u2pQq1TqKdLyTd+VGbvOWGzRjzjgL4iLzoMPfAbs9oxrvrhCJR6MxIZt4b1BqN0fKlqNCdGNvYAA8qP99NSSMP9MUULcL9RqXg+CQr5oIC87JVzyBkFyL99DsTQ20oP8LzA+851AqQxEfDgL2wAw8BeIgXFyH8QCTf8MBwalaAwVBqT4UC+AwPCSFEAx8JxbCDjw1ReiIt0hHI8B-xQIjaOMXpqQY-9cD4ZhXneUsQFEtsihAThpk4EopjktEAEtRRhNlRTGNk0QAC0YNlGBXUIVMYLp2EhP1Vy6EpkVRYN8RxMkIH2clsVJEBAEQiCk7CpWl6UZFV2U5DUuj5OVBRFcUpRlDUFWi5VZVoW0NTZSMjXVdUAAcDytG07Xy09pybJQPRrbIBGhYzERsuyN1C9LrRZHLDxqY9j0bL131+O8BjseYO1qTrIAEN1ngHJoQHmDAOjqbJci69Jm1GWIxE8+ZG1nUr0yW-rBrmhpZ1OftHjbabZvYc06GaZhmlQVBcnw3JMlFBCbjPXgv2SbNXIA3pwxufpDHw+JXtofYPuQL79EUDyKIBmcqGB5L2Fez93s+4xAgkRBoAgdzVhzEGqT84JidRqgXIxqGsfsXGIGpCxCfgYnKaRsmUbR5Jqeh4I+KQMQmfsVRkvddnYHJkhVhAAAxfHSQMVgQAASRhFTVOmNkaVFKY0QmUI5iAA noborder}} </WRAP>
  
 In the simulation, you see the two capacitors $C_1$ and $C_2$ (The two small resistors with $1 ~\rm µ\Omega$ have to be there for the simulation to run). At the beginning, $C_1$ is charged to $10~{\rm V}$ and $C_2$ to $0~{\rm V}$. With the switches $S_1$ and $S_2$ you can choose whether In the simulation, you see the two capacitors $C_1$ and $C_2$ (The two small resistors with $1 ~\rm µ\Omega$ have to be there for the simulation to run). At the beginning, $C_1$ is charged to $10~{\rm V}$ and $C_2$ to $0~{\rm V}$. With the switches $S_1$ and $S_2$ you can choose whether
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 #@TaskEnd_HTML@# #@TaskEnd_HTML@#
 +
 +
 +#@TaskTitle_HTML@#1.4  Machine-Vision Strobe Unit: Charging and Safe Discharge of a Flash Capacitor                    #@TaskText_HTML@#
 +
 +A machine-vision inspection system on a production line uses a short high-voltage flash pulse.
 +For this purpose, an energy-storage capacitor is charged from a DC source and must be safely discharged before maintenance.
 +
 +Data:
 +\begin{align*}
 +C &= 1~{\rm \mu F} \\
 +W_e &= 0.1~{\rm J} \\
 +I_{\rm max} &= 100~{\rm mA} \\
 +R_i &= 10~{\rm M\Omega}
 +\end{align*}
 +
 +1. What voltage must the capacitor have so that it stores the required energy?
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~101~@#
 +<WRAP leftalign>
 +\begin{align*}
 +W_e &= \frac{1}{2} C U^2 \\
 +U &= \sqrt{\frac{2W_e}{C}}
 +   = \sqrt{\frac{2 \cdot 0.1~{\rm J}}{1 \cdot 10^{-6}~{\rm F}}} \\
 +  &= \sqrt{200000}~{\rm V}
 +   \approx 447.2~{\rm V}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~101~@#
 +\begin{align*}
 +U = 447.2~{\rm V}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +2. The charging current must not exceed $100~\rm mA$ at the start of charging. What charging resistor is required?
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~102~@#
 +<WRAP leftalign>
 +At the beginning of charging, the capacitor behaves like a short circuit, so
 +\begin{align*}
 +i_{C{\rm max}} = i_C(t=0) = \frac{U}{R}
 +\end{align*}
 +Thus,
 +\begin{align*}
 +R &\ge \frac{U}{I_{\rm max}}
 +   = \frac{447.2~{\rm V}}{0.1~{\rm A}} \\
 +  &\approx 4472~{\rm \Omega}
 +   = 4.47~{\rm k\Omega}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~102~@#
 +\begin{align*}
 +R \ge 4.47~{\rm k\Omega}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +3. How long does the charging process take until the capacitor is practically fully charged?
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~103~@#
 +<WRAP leftalign>
 +The time constant is
 +\begin{align*}
 +T = RC = 4.47~{\rm k\Omega} \cdot 1~{\rm \mu F} = 4.47~{\rm ms}
 +\end{align*}
 +In engineering practice, a capacitor is considered practically fully charged after about $5T$:
 +\begin{align*}
 +t \approx 5T = 5 \cdot 4.47~{\rm ms} = 22.35~{\rm ms}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~103~@#
 +\begin{align*}
 +t \approx 22.35~{\rm ms}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +4. Give the time-dependent capacitor voltage and the voltage across the charging resistor.
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~104~@#
 +<WRAP leftalign>
 +For the charging process:
 +\begin{align*}
 +u_C(t) &= U\left(1-e^{-t/T}\right) \\
 +u_R(t) &= Ue^{-t/T}
 +\end{align*}
 +with
 +\begin{align*}
 +U &= 447.2~{\rm V} \\
 +T &= 4.47~{\rm ms}
 +\end{align*}
 +So the capacitor voltage rises exponentially from $0$ to $447.2~\rm V$, while the resistor voltage falls exponentially from $447.2~\rm V$ to $0$.
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~104~@#
 +\begin{align*}
 +u_C(t) &= 447.2\left(1-e^{-t/4.47{\rm ms}}\right)~{\rm V} \\
 +u_R(t) &= 447.2\,e^{-t/4.47{\rm ms}}~{\rm V}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +5. After charging, the capacitor is disconnected from the source. Its leakage can be modeled by an internal resistance of $10~\rm M\Omega$.
 +After what time has the stored energy dropped to one half, and what is the capacitor voltage then?
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~105~@#
 +<WRAP leftalign>
 +Half the energy means
 +\begin{align*}
 +W_e' = 0.5W_e
 +\end{align*}
 +Since
 +\begin{align*}
 +W_e = \frac{1}{2}CU^2
 +\end{align*}
 +the voltage at half energy is
 +\begin{align*}
 +U' = \frac{U}{\sqrt{2}}
 +    = \frac{447.2~{\rm V}}{\sqrt{2}}
 +    = 316.2~{\rm V}
 +\end{align*}
 +For the discharge through the internal resistance:
 +\begin{align*}
 +u_C(t) = Ue^{-t/T_2}
 +\end{align*}
 +with
 +\begin{align*}
 +T_2 = R_iC = 10~{\rm M\Omega} \cdot 1~{\rm \mu F} = 10~{\rm s}
 +\end{align*}
 +Set $u_C(t)=U'$:
 +\begin{align*}
 +Ue^{-t/T_2} &= U' \\
 +t &= T_2 \ln\left(\frac{U}{U'}\right) \\
 +  &= 10~{\rm s}\cdot\ln\left(\frac{447.2}{316.2}\right) \\
 +  &\approx 3.47~{\rm s}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~105~@#
 +\begin{align*}
 +U' = 316.2~{\rm V} \\
 +t = 3.47~{\rm s}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +6. The fully charged capacitor is discharged through the charging resistor before maintenance.
 +How long does the discharge take, and how much energy is converted into heat in the resistor?
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~106~@#
 +<WRAP leftalign>
 +The discharge time constant through the same resistor is again
 +\begin{align*}
 +T = RC = 4.47~{\rm ms}
 +\end{align*}
 +Thus the practical discharge time is
 +\begin{align*}
 +t \approx 5T = 22.35~{\rm ms}
 +\end{align*}
 +The complete stored capacitor energy is converted into heat in the resistor:
 +\begin{align*}
 +W_R = W_e = 0.1~{\rm Ws}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~106~@#
 +\begin{align*}
 +t \approx 22.35~{\rm ms} \\
 +W_R = 0.1~{\rm Ws}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +#@TaskEnd_HTML@#
 +
 +
 +#@TaskTitle_HTML@#1.5 Sensor Input Buffer: Source, T-Network and Capacitor                    #@TaskText_HTML@#
 +
 +<wrap right>{{drawio>electrical_engineering_and_electronics_2:ExTNetworkWithCapV01.svg}}</wrap>
 +A 12 V industrial sensor electronics unit feeds a buffered measurement node through a resistor T-network.
 +A capacitor smooths the node voltage. At first, the load is disconnected.
 +After the capacitor is fully charged, a measurement load is connected by a switch.
 +
 +Data:
 +\begin{align*}
 +U   &= 12~{\rm V} \\
 +R_1 &= 2~{\rm k\Omega} \\
 +R_2 &= 10~{\rm k\Omega} \\
 +R_3 &= 3.33~{\rm k\Omega} \\
 +C   &= 2~{\rm \mu F} \\
 +R_L &= 5~{\rm k\Omega}
 +\end{align*}
 +
 +Initially, the capacitor is uncharged and the switch is open.
 +
 +1. What is the capacitor voltage after it is fully charged?
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~201~@#
 +<WRAP leftalign>
 +Using the equivalent voltage source of the network on the left-hand side, the open-circuit voltage is
 +\begin{align*}
 +U_{0e} &= \frac{R_2}{R_1+R_2}\,U \\
 +       &= \frac{10~{\rm k\Omega}}{2~{\rm k\Omega}+10~{\rm k\Omega}}\cdot 12~{\rm V} \\
 +       &= 10~{\rm V}
 +\end{align*}
 +After full charging, the capacitor voltage equals this voltage.
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~201~@#
 +\begin{align*}
 +U_C = U_{0e} = 10~{\rm V}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +2. How long does the charging process take?
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~202~@#
 +<WRAP leftalign>
 +The internal resistance seen by the capacitor is
 +\begin{align*}
 +R_{ie} &= R_3 + (R_1 \parallel R_2) \\
 +       &= 3.33~{\rm k\Omega} + \frac{2~{\rm k\Omega}\cdot 10~{\rm k\Omega}}{2~{\rm k\Omega}+10~{\rm k\Omega}} \\
 +       &= 3.33~{\rm k\Omega} + 1.67~{\rm k\Omega} \\
 +       &= 5.00~{\rm k\Omega}
 +\end{align*}
 +So the time constant is
 +\begin{align*}
 +T &= R_{ie}C
 +   = 5.00~{\rm k\Omega}\cdot 2~{\rm \mu F}
 +   = 10~{\rm ms}
 +\end{align*}
 +Practical charging time:
 +\begin{align*}
 +t \approx 5T = 50~{\rm ms}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~202~@#
 +\begin{align*}
 +R_{ie} = 5.00~{\rm k\Omega} \\
 +t \approx 50~{\rm ms}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +3. Give the time-dependent capacitor voltage.
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~203~@#
 +<WRAP leftalign>
 +The charging law is
 +\begin{align*}
 +u_C(t) &= U_{0e}\left(1-e^{-t/T}\right) \\
 +       &= 10\left(1-e^{-t/10{\rm ms}}\right)~{\rm V}
 +\end{align*}
 +So the capacitor voltage rises exponentially from $0~\rm V$ to $10~\rm V$.
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~203~@#
 +\begin{align*}
 +u_C(t) = 10\left(1-e^{-t/10{\rm ms}}\right)~{\rm V}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +4. After the capacitor is fully charged, the switch is closed and the load resistor is connected.
 +What is the stationary load voltage?
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~204~@#
 +<WRAP leftalign>
 +Now use a second equivalent voltage-source step.
 +The Thevenin source seen by the load has
 +\begin{align*}
 +U_{0e} &= 10~{\rm V} \\
 +R_{ie} &= 5.00~{\rm k\Omega}
 +\end{align*}
 +Thus, the stationary load voltage is
 +\begin{align*}
 +U_C' = U_{0e}' &= \frac{R_L}{R_{ie}+R_L}\,U_{0e} \\
 +               &= \frac{5~{\rm k\Omega}}{5~{\rm k\Omega}+5~{\rm k\Omega}}\cdot 10~{\rm V} \\
 +               &= 5~{\rm V}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~204~@#
 +\begin{align*}
 +U_L = 5~{\rm V}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +5. How long does it take until this new stationary state is practically reached?
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~205~@#
 +<WRAP leftalign>
 +The new internal resistance is
 +\begin{align*}
 +R_{ie}' &= R_{ie}\parallel R_L \\
 +        &= 5.00~{\rm k\Omega}\parallel 5.00~{\rm k\Omega} \\
 +        &= 2.50~{\rm k\Omega}
 +\end{align*}
 +Hence the new time constant is
 +\begin{align*}
 +T' &= R_{ie}'C
 +    = 2.50~{\rm k\Omega}\cdot 2~{\rm \mu F}
 +    = 5~{\rm ms}
 +\end{align*}
 +Practical settling time:
 +\begin{align*}
 +t \approx 5T' = 25~{\rm ms}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~205~@#
 +\begin{align*}
 +R_{ie}' = 2.50~{\rm k\Omega} \\
 +t \approx 25~{\rm ms}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +6. Give the time-dependent load voltage after the switch is closed.
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~206~@#
 +<WRAP leftalign>
 +At the switching instant, the capacitor voltage cannot jump.
 +Therefore:
 +\begin{align*}
 +u_L(0^+) &= 10~{\rm V} \\
 +u_L(\infty) &= 5~{\rm V}
 +\end{align*}
 +The voltage therefore decays exponentially toward the new final value:
 +\begin{align*}
 +u_L(t) &= u_L(\infty) + \left(u_L(0^+)-u_L(\infty)\right)e^{-t/T'} \\
 +       &= 5 + 5e^{-t/5{\rm ms}}~{\rm V}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~206~@#
 +\begin{align*}
 +u_L(t) = 5 + 5e^{-t/5{\rm ms}}~{\rm V}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +#@TaskEnd_HTML@#
 +
 +
 +{{tag>inductors air_core_coil magnetic_field hall_sensor transient_response current_density chapter1_1}}
 +{{include_n>300}}
 +
 +#@TaskTitle_HTML@#1.6 Hall-Sensor Calibration Coil: Short Air-Core Coil                    #@TaskText_HTML@#
 +
 +A Hall-sensor calibration bench uses a short air-core coil to create a defined magnetic field.
 +An air-core coil is chosen because it avoids hysteresis and remanence effects.
 +The coil is wound as a short cylindrical coil.
 +
 +Data:
 +\begin{align*}
 +l &= 22~{\rm mm} \\
 +d &= 20~{\rm mm} \\
 +d_{\rm Cu} &= 0.8~{\rm mm} \\
 +N &= 25 \\
 +\rho_{\rm Cu,20^\circ C} &= 0.0178~{\rm \Omega\,mm^2/m}
 +\end{align*}
 +
 +A DC current of $1~\rm A$ shall flow through the coil.
 +
 +1. Calculate the coil resistance $R$ at room temperature.
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~301~@#
 +<WRAP leftalign>
 +The wire cross section is
 +\begin{align*}
 +A_{\rm Cu} &= \pi\left(\frac{d_{\rm Cu}}{2}\right)^2
 +           = \pi(0.4~{\rm mm})^2 \\
 +           &= 0.503~{\rm mm^2}
 +\end{align*}
 +The total wire length is approximated by the number of turns times the circumference:
 +\begin{align*}
 +l_{\rm Cu} &= N\pi d \\
 +           &= 25\pi \cdot 20~{\rm mm} \\
 +           &= 1570.8~{\rm mm}
 +           = 1.571~{\rm m}
 +\end{align*}
 +Thus,
 +\begin{align*}
 +R &= \rho_{\rm Cu}\frac{l_{\rm Cu}}{A_{\rm Cu}} \\
 +  &= 0.0178~{\rm \Omega\,mm^2/m}\cdot\frac{1.571~{\rm m}}{0.503~{\rm mm^2}} \\
 +  &\approx 0.0556~{\rm \Omega}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~301~@#
 +\begin{align*}
 +R = 55.6~{\rm m\Omega}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +2. Calculate the coil inductance $L$.
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~302~@#
 +<WRAP leftalign>
 +For this short air-core coil, use
 +\begin{align*}
 +L = N^2 \cdot \frac{\mu_0 A}{l}\cdot\frac{1}{1+\frac{d}{2l}}
 +\end{align*}
 +with
 +\begin{align*}
 +A &= \pi\left(\frac{d}{2}\right)^2
 +   = \pi(10~{\rm mm})^2
 +   = 314.16~{\rm mm^2}
 +   = 3.1416\cdot 10^{-4}~{\rm m^2} \\
 +\mu_0 &= 4\pi\cdot 10^{-7}~{\rm Vs/(Am)}
 +\end{align*}
 +Therefore,
 +\begin{align*}
 +L &= 25^2 \cdot \frac{4\pi\cdot 10^{-7}\,{\rm Vs/(Am)} \cdot 3.1416\cdot 10^{-4}~{\rm m^2}}{22\cdot 10^{-3}~{\rm m}}
 +   \cdot \frac{1}{1+\frac{20}{2\cdot 22}} \\
 +  &\approx 7.71\cdot 10^{-6}~{\rm H}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~302~@#
 +\begin{align*}
 +L = 7.71~{\rm \mu H}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +3. Which DC voltage must be applied so that the stationary current becomes $I=1~\rm A$?
 +How large is the current density $j$ in the copper wire?
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~303~@#
 +<WRAP leftalign>
 +In the stationary DC state, the coil behaves like its ohmic resistance:
 +\begin{align*}
 +U &= RI \\
 +  &= 55.6~{\rm m\Omega}\cdot 1~{\rm A} \\
 +  &= 55.6~{\rm mV}
 +\end{align*}
 +The current density is
 +\begin{align*}
 +j &= \frac{I}{A_{\rm Cu}} \\
 +  &= \frac{1~{\rm A}}{0.503~{\rm mm^2}} \\
 +  &\approx 1.99~{\rm A/mm^2}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~303~@#
 +\begin{align*}
 +U = 55.6~{\rm mV} \\
 +j = 1.99~{\rm A/mm^2}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +4. How much magnetic energy is stored in the coil in the stationary state?
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~304~@#
 +<WRAP leftalign>
 +\begin{align*}
 +W_m &= \frac{1}{2}LI^2 \\
 +    &= \frac{1}{2}\cdot 7.71\cdot 10^{-6}~{\rm H}\cdot (1~{\rm A})^2 \\
 +    &= 3.86\cdot 10^{-6}~{\rm Ws}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~304~@#
 +\begin{align*}
 +W_m = 3.86\cdot 10^{-6}~{\rm Ws}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +5. Give the time-dependent coil current $i(t)$ when the coil is switched on.
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~305~@#
 +<WRAP leftalign>
 +A coil current cannot jump instantly.
 +It starts at $0$ and approaches the final value $I=1~\rm A$ exponentially:
 +\begin{align*}
 +i(t) = I\left(1-e^{-t/T}\right)
 +\end{align*}
 +So the sketch starts at $0~\rm A$, rises quickly, and then slowly approaches $1~\rm A$.
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~305~@#
 +\begin{align*}
 +i(t) = 1\left(1-e^{-t/T}\right)~{\rm A}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +6. How long does it take until the current has practically reached its stationary value?
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~306~@#
 +<WRAP leftalign>
 +The time constant is
 +\begin{align*}
 +T &= \frac{L}{R}
 +   = \frac{7.71~{\rm \mu H}}{55.6~{\rm m\Omega}} \\
 +  &\approx 138.9~{\rm \mu s}
 +\end{align*}
 +A practical final value is reached after about $5T$:
 +\begin{align*}
 +t \approx 5T = 5\cdot 138.9~{\rm \mu s} \approx 695~{\rm \mu s}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~306~@#
 +\begin{align*}
 +t \approx 695~{\rm \mu s}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +7. How much energy is dissipated as heat in the coil resistance during the current build-up?
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~307~@#
 +<WRAP leftalign>
 +Using the current from task 5,
 +\begin{align*}
 +i(t) = I\left(1-e^{-t/T}\right)
 +\end{align*}
 +the heat dissipated in the winding resistance up to the practical final time $5T$ is
 +\begin{align*}
 +W_R &= \int_0^{5T} R\,i^2(t)\,dt \\
 +    &= R I^2 \int_0^{5T} \left(1-e^{-t/T}\right)^2 dt
 +\end{align*}
 +For this interval, the integral is approximately
 +\begin{align*}
 +\int_0^{5T} \left(1-e^{-t/T}\right)^2 dt \approx \frac{7}{2}T
 +\end{align*}
 +Thus,
 +\begin{align*}
 +W_R &\approx RI^2\cdot \frac{7}{2}T \\
 +    &= 0.0556~{\rm \Omega}\cdot (1~{\rm A})^2 \cdot \frac{7}{2}\cdot 138.9~{\rm \mu s} \\
 +    &\approx 27.05\cdot 10^{-6}~{\rm Ws}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~307~@#
 +\begin{align*}
 +W_R \approx 27.05\cdot 10^{-6}~{\rm Ws}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +#@TaskEnd_HTML@#
 +
 +
 +{{page>aufgabe_7.2.6_mit_rechnung&nofooter}}