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This shows you the differences between two versions of the page.

--- electrical_engineering_and_electronics_2:block01 [2026/03/05 02:45] – mexleadmin
+++ electrical_engineering_and_electronics_2:block01 [2026/04/06 23:36] (current) – mexleadmin
@@ Line 475: / Line 475: @@
-<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%>
@@ Line 482: / Line 482: @@
 </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.
@@ Line 640: / Line 640: @@
 #@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@#
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+<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
@@ Line 677: / Line 675: @@
 #@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*}
+. 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>
+. 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>
+. 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>
+. 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>
+. 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>
+. 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.
+. 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>
+. 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>
+. 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>
+. 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>
+. 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>
+. 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.
+. 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>
+. 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>
+. 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>
+. 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>
+. 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>
+. 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>
+. 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}}

inductors air core coil magnetic field hall sensor transient response current density chapter1 1