Differences

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--- dummy [2026/03/27 01:24] – mexleadmin
+++ dummy [2026/05/26 13:48] (current) – removed mexleadmin
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-{{tag>capacitors rc_circuit transient_response energy_storage industrial_safety chapter1_1}}
-{{include_n>300}}
-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  Machine-Vision Strobe: Capacitor Charging and Safe Discharge                    #@TaskText_HTML@#
-A high-speed machine-vision inspection system on a production line uses a short high-voltage flash pulse.
-For this purpose, an energy-storage capacitor is charged by a regulated DC high-voltage source and must be safely discharged before maintenance.
-Data:
-\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.
-. What voltage must be present across the capacitor so that it stores the required energy?
-#@PathBegin_HTML~1~@#
-\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~@#
-\begin{align*}
-U_C \approx 447~{\rm V}
-\end{align*}
-#@ResultEnd_HTML@#
-. The charging current must not exceed $100~\rm mA$ at the start of charging. What series charging resistor is required?
-#@PathBegin_HTML~2~@#
-\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~@#
-\begin{align*}
-R \approx 4.47~{\rm k\Omega}
-\end{align*}
-#@ResultEnd_HTML@#
-. How long does the charging process take until the capacitor is considered practically fully charged?
-#@PathBegin_HTML~3~@#
-\begin{align*}
-\tau &= RC
-     = 4.472~{\rm k\Omega} \cdot 1~{\rm \mu F}
-     \approx 4.47~{\rm ms}
-\end{align*}
-In engineering practice, a capacitor is usually considered fully charged after about $5\tau$.
-\begin{align*}
-t_{\rm charge} &\approx 5\tau
-              \approx 5 \cdot 4.47~{\rm ms}
-              \approx 22.4~{\rm ms}
-\end{align*}
-Mathematically, the charging time to exactly $100\%$ is infinite.
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~3~@#
-\begin{align*}
-t_{\rm charge} \approx 22.4~{\rm ms} \qquad \text{(practically, }5\tau\text{)}
-\end{align*}
-#@ResultEnd_HTML@#
-. Sketch the capacitor voltage and the voltage across the charging resistor during charging.
-#@PathBegin_HTML~4~@#
-\begin{align*}
-u_C(t) &= U_0 \left(1 - e^{-t/(RC)}\right) \\
-u_R(t) &= U_0 e^{-t/(RC)}
-\end{align*}
-with
-\begin{align*}
-U_0 &= 447.2~{\rm V} \\
-RC  &= 4.47~{\rm ms}
-\end{align*}
-Initial and final values:
-\begin{align*}
-u_C(0) &= 0,                 & u_C(\infty) &= U_0 \\
-u_R(0) &= U_0,              & u_R(\infty) &= 0
-\end{align*}
-Thus, the capacitor voltage rises exponentially, while the resistor voltage falls exponentially.
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~4~@#
-\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@#
-. 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?
-#@PathBegin_HTML~5~@#
-\begin{align*}
-u_C(t) &= U_0 e^{-t/(R_i C)}
-\end{align*}
-Since capacitor energy is
-\begin{align*}
-W(t) &= \frac{1}{2} C u_C^2(t)
-\end{align*}
-it follows that
-\begin{align*}
-\frac{W(t)}{W_0} &= e^{-2t/(R_i C)}
-\end{align*}
-For half the initial energy:
-\begin{align*}
-\frac{1}{2} &= e^{-2t/(R_i C)} \\
-t &= \frac{R_i C}{2}\ln(2)
-\end{align*}
-With
-\begin{align*}
-R_i C = 10~{\rm M\Omega} \cdot 1~{\rm \mu F} = 10~{\rm s}
-\end{align*}
-we get
-\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@#
-. 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?
-#@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*}
-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@#
-#@TaskEnd_HTML@#