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| {{tag>capacitors rc_circuit transient_response energy_storage industrial_safety chapter1_1}} | {{drawio>diagram99.png}} |
| {{include_n>300}} | |
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| #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Machine-Vision Strobe: Capacitor Charging and Safe Discharge #@TaskText_HTML@# | #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Machine-Vision Strobe Unit: Charging and Safe Discharge of a Flash Capacitor #@TaskText_HTML@# |
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| A high-speed machine-vision inspection system on a production line uses a short high-voltage flash pulse. | 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. |
| For this purpose, an energy-storage capacitor is charged by a regulated DC high-voltage source and must be safely discharged before maintenance. | |
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| Data: | 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*} |
| \begin{align*} | |
| C &= 1~{\rm \mu F} \\ | |
| W &= 0.1~{\rm J} \\ | |
| I_{\rm max} &= 100~{\rm mA} \\ | |
| R_i &= 10~{\rm M\Omega} | |
| \end{align*} | |
| |
| Assume the DC charging source is adjusted to the required final capacitor voltage. | 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> |
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| 1. What voltage must be present across the capacitor so that it stores the required energy? | 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> |
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| #@PathBegin_HTML~1~@# | 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> |
| \begin{align*} | |
| W &= \frac{1}{2} C U^2 \\ | |
| U &= \sqrt{\frac{2W}{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*} | |
| #@PathEnd_HTML@# | |
| |
| #@ResultBegin_HTML~1~@# | 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> |
| \begin{align*} | |
| U_C \approx 447~{\rm V} | |
| \end{align*} | |
| #@ResultEnd_HTML@# | |
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| 2. The charging current must not exceed $100~\rm mA$ at the start of charging. What series charging resistor is required? | 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> |
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| #@PathBegin_HTML~2~@# | 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> |
| \begin{align*} | |
| I_{\rm max} &= \frac{U}{R} \\ | |
| R &= \frac{U}{I_{\rm max}} | |
| = \frac{447.2~{\rm V}}{0.1~{\rm A}} | |
| \approx 4472~{\rm \Omega} | |
| \end{align*} | |
| \end{align*} | |
| A practical standard value would be about $4.7~{\rm k\Omega}$. | |
| #@PathEnd_HTML@# | |
| |
| #@ResultBegin_HTML~2~@# | #@TaskEnd_HTML@# |
| \begin{align*} | |
| R \approx 4.47~{\rm k\Omega} | |
| \end{align*} | |
| #@ResultEnd_HTML@# | |
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| 3. How long does the charging process take until the capacitor is considered practically fully charged? | {{tag>rc_circuit thevenin_equivalent transient_response sensor_interface industrial_electronics chapter1_1}}{{include_n>300}} |
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| #@PathBegin_HTML~3~@# | #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Sensor Input Buffer: Source, T-Network and Capacitor #@TaskText_HTML@# |
| \begin{align*} | |
| \tau &= RC | |
| = 4.472~{\rm k\Omega} \cdot 1~{\rm \mu F} | |
| \approx 4.47~{\rm ms} | |
| \end{align*} | |
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| In engineering practice, a capacitor is usually considered fully charged after about $5\tau$. | 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. |
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| \begin{align*} | 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*} |
| t_{\rm charge} &\approx 5\tau | |
| \approx 5 \cdot 4.47~{\rm ms} | |
| \approx 22.4~{\rm ms} | |
| \end{align*} | |
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| Mathematically, the charging time to exactly $100\%$ is infinite. | Initially, the capacitor is uncharged and the switch is open. |
| #@PathEnd_HTML@# | |
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| #@ResultBegin_HTML~3~@# | 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> |
| \begin{align*} | |
| t_{\rm charge} \approx 22.4~{\rm ms} \qquad \text{(practically, }5\tau\text{)} | |
| \end{align*} | |
| #@ResultEnd_HTML@# | |
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| 4. Sketch the capacitor voltage and the voltage across the charging resistor during charging. | 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> |
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| #@PathBegin_HTML~4~@# | 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> |
| \begin{align*} | |
| u_C(t) &= U_0 \left(1 - e^{-t/(RC)}\right) \\ | |
| u_R(t) &= U_0 e^{-t/(RC)} | |
| \end{align*} | |
| |
| with | 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> |
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| \begin{align*} | 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> |
| U_0 &= 447.2~{\rm V} \\ | |
| RC &= 4.47~{\rm ms} | |
| \end{align*} | |
| |
| Initial and final values: | 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> |
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| \begin{align*} | #@TaskEnd_HTML@# |
| u_C(0) &= 0, & u_C(\infty) &= U_0 \\ | |
| u_R(0) &= U_0, & u_R(\infty) &= 0 | |
| \end{align*} | |
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| Thus, the capacitor voltage rises exponentially, while the resistor voltage falls exponentially. | {{tag>inductors air_core_coil magnetic_field hall_sensor transient_response current_density chapter1_1}}{{include_n>300}} |
| #@PathEnd_HTML@# | |
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| #@ResultBegin_HTML~4~@# | #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Hall-Sensor Calibration Coil: Short Air-Core Coil #@TaskText_HTML@# |
| \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@# | |
| |
| 5. After charging, the capacitor is disconnected from the source. Its leakage can be modeled by an internal resistance of $10~\rm M\Omega$. | 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. |
| After what time has the stored energy dropped to one half, and what is the capacitor voltage then? | |
| |
| #@PathBegin_HTML~5~@# | 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*} |
| \begin{align*} | |
| u_C(t) &= U_0 e^{-t/(R_i C)} | |
| \end{align*} | |
| |
| Since capacitor energy is | A DC current of $1~\rm A$ shall flow through the coil. |
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| \begin{align*} | 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> |
| W(t) &= \frac{1}{2} C u_C^2(t) | |
| \end{align*} | |
| |
| it follows that | 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> |
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| \begin{align*} | 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> |
| \frac{W(t)}{W_0} &= e^{-2t/(R_i C)} | |
| \end{align*} | |
| |
| For half the initial energy: | 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> |
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| \begin{align*} | 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> |
| \frac{1}{2} &= e^{-2t/(R_i C)} \\ | |
| t &= \frac{R_i C}{2}\ln(2) | |
| \end{align*} | |
| |
| With | 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> |
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| \begin{align*} | 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> |
| R_i C = 10~{\rm M\Omega} \cdot 1~{\rm \mu F} = 10~{\rm s} | |
| \end{align*} | |
| |
| we get | #@TaskEnd_HTML@# |
| |
| \begin{align*} | |
| t &= \frac{10~{\rm s}}{2}\ln(2) | |
| \approx 3.47~{\rm s} | |
| \end{align*} | |
| |
| The voltage at this instant is | |
| |
| \begin{align*} | |
| u_C &= U_0 \cdot \frac{1}{\sqrt{2}} | |
| = \frac{447.2~{\rm V}}{\sqrt{2}} | |
| \approx 316.2~{\rm V} | |
| \end{align*} | |
| #@PathEnd_HTML@# | |
| |
| #@ResultBegin_HTML~5~@# | |
| \begin{align*} | |
| t_{\frac{W_0}{2}} \approx 3.47~{\rm s} \\ | |
| u_C \approx 316~{\rm V} | |
| \end{align*} | |
| #@ResultEnd_HTML@# | |
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| 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? | |
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| #@PathBegin_HTML~6~@# | |
| For discharge through the same resistor: | |
| |
| \begin{align*} | |
| u_C(t) &= U_0 e^{-t/(RC)} | |
| \end{align*} | |
| |
| with | |
| |
| \begin{align*} | |
| \tau &= RC \approx 4.47~{\rm ms} | |
| \end{align*} | |
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| Again, practical discharge time is about | |
| |
| \begin{align*} | |
| t_{\rm discharge} &\approx 5\tau | |
| \approx 22.4~{\rm ms} | |
| \end{align*} | |
| |
| The energy stored initially in the capacitor is | |
| |
| \begin{align*} | |
| W_0 &= 0.1~{\rm J} | |
| \end{align*} | |
| |
| After full discharge, the capacitor energy is zero, so the entire initial energy is converted into heat in the resistor: | |
| |
| \begin{align*} | |
| W_R &= W_0 = 0.1~{\rm J} | |
| \end{align*} | |
| |
| In a real design, the resistor must therefore be checked for pulse-load capability. | |
| #@PathEnd_HTML@# | |
| |
| #@ResultBegin_HTML~6~@# | |
| \begin{align*} | |
| t_{\rm discharge} \approx 22.4~{\rm ms} \qquad \text{(practically, }5\tau\text{)} \\ | |
| W_R = 0.1~{\rm J} | |
| \end{align*} | |
| #@ResultEnd_HTML@# | |
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| #@TaskEnd_HTML@# | |