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electrical_engineering_and_electronics_1:block13 [2025/10/31 21:20] – angelegt mexleadminelectrical_engineering_and_electronics_1:block13 [2026/01/10 12:51] (aktuell) – mexleadmin
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-====== Block xx - xxx ======+====== Block 13 - Capacitor Circuits and Energy ======
  
-===== Learning objectives =====+===== 13.0 Intro ===== 
 + 
 +==== 13.0.1 Learning Objectives ====
 <callout> <callout>
 After this 90-minute block, you can After this 90-minute block, you can
-  * Recognize a series connection of capacitors and distinguish it from a parallel connection, +  * identify series vs. parallel connections of capacitors from a circuit diagram, 
-  * Calculate the resulting total capacitance of a series or parallel circuit, +  * compute equivalent capacitance $C_{\rm eq}$ for series and parallelnetworks, 
-  * Know how the total charge is distributed among the individual capacitors in a parallel circuit, +  * use the key sharing rules: in **series** $Q_k=\text{const.}$ and voltages divide; in **parallel** $U_k=\text{const.}$ and charges divide, 
-  * Determine the voltage across a single capacitor in a series circuit.+  * apply the capacitor divider relation (two series capacitors), 
 +  * determine stored energy, including a dimensional check to $\rm J$.
 </callout> </callout>
  
-===== Preparation at Home =====+==== 13.0.2 Preparation at Home ====
  
 Well, again  Well, again 
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 For checking your understanding please do the following exercises: For checking your understanding please do the following exercises:
-  * ...+  * 5.9.5
  
-===== 90-minute plan ===== +==== 13.0.3 90-minute plan ==== 
-  - Warm-up (x min):  +  - Warm-up (10 min): 
-    - ....  +    - Quick quiz (2–3 items): series or parallel? which rule applies (constant $U$ or constant $Q$)? 
-  - Core concepts & derivations (x min): +    - Recall $Q=C\,U$ and energy $W=\tfrac12 C U^2$ (units). 
-    - ... +  - Core concepts & derivations (35 min): 
-  - Practice (x min): ... +    - Derive $C_{\rm eq}$ for **series** from Kirchhoff’s voltage law and $Q=\text{const.}$; derive voltage division $U_k=\dfrac{Q}{C_k}$. 
-  - Wrap-up (x min): Summary box; common pitfalls checklist.+    - Derive $C_{\rm eq}$ for **parallel** from Kirchhoff’s current/charge balance and $U=\text{const.}$; obtain $Q_k=C_k U$. 
 +    - Energy in the electric field: integrate $dW=U\,dq$ → $W=\tfrac12 C U^2$; short dimensional check. 
 +  - Practice (35 min): 
 +    - Two short worked examples: mixed series/parallel network; two-capacitor divider with given $U$ (find $U_1$, $U_2$, $W$ on each). 
 +    - Short simulation tasks (use the two embedded Falstad circuits in this page): observe $U_k$, $Q_k$ when toggling the switch or changing values. 
 +    - Mini-problems: “double a plate area / halve distance” reasoning on $C$ and $W$. 
 +  - Wrap-up (10 min): 
 +    - Common-pitfalls checklist and one exit-ticket calculation.
  
-===== Conceptual overview =====+==== 13.0.4 Conceptual overview ====
 <callout icon="fa fa-lightbulb-o" color="blue"> <callout icon="fa fa-lightbulb-o" color="blue">
-  - ...+  - **What stays the same?** In **series** all capacitors carry the **same charge** $Q$; in **parallel** all capacitors see the **same voltage** $U$. 
 +  - **How do totals form?** Capacitances **add inversely** in series and **add directly** in parallel. This mirrors resistors but with the roles swapped. 
 +  - **Voltage/charge sharing:** In series, the **smaller** $C_k$ takes the **larger** $U_k$ ($U_k=Q/C_k$). In parallel, the **larger** $C_k$ takes the **larger** $Q_k$ ($Q_k=C_k U$). 
 +  - **Energy viewpoint:** Charging needs work against the field; $W=\tfrac12 C U^2=\tfrac12 Q U=\dfrac{Q^2}{2C}$. Dimensional check: $[C]=\rm F=\dfrac{A\,s}{V}$, so $[C U^2]=\dfrac{A\,s}{V}\,V^2=A\,s\,V=J$. 
 +  - **Design intuition:** Increasing plate area $A$ or dielectric $\varepsilon_r$ raises $C$ and thus stored $W$ at the same $U$; increasing gap $d$ lowers $C$.
 </callout> </callout>
  
-===== Core content =====+===== 13.1 Core content =====
  
-==== Series Circuit of Capacitor ====+==== 13.1.1 Series Circuit of Capacitor ====
  
 If capacitors are connected in series, the charging current $I$ into the individual capacitors $C_1 ... C_n$ is equal. If capacitors are connected in series, the charging current $I$ into the individual capacitors $C_1 ... C_n$ is equal.
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   * The capacitors can be discharged again via the lamp.   * The capacitors can be discharged again via the lamp.
  
-This derivation is also well explained, for example, in [[https://www.youtube.com/watch?v=9-Bp9Cvr7Jg|this video]]. +<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0lwrFWAmZBmMbJgGzYCwAc+yAnIciGmiIvjfAKYC0YYAUAMYjIDslYXjh79wpSviTZw0ZPEjJ8pZSuW8p0yOzCEIRSCFLD9VNMINYYcazZvCs4A1c2xs7AMoiB+SX0pGoQIAzAEMAGwBnRkDkdgB3EBNqe3hKZKh4nnhhAMEc83YAN3AhU2E8w3NA-ANEJyR4dgBzEvszVqocQkCtBJMA5EJugN6qVLKvCdG-cB9Jio4EmeRIbuWa8C5JxQNl7MSpCDBZeTlbGxpnCGnRZdEsSQ5uGYfJtBWD50RkaCVSeXOcEGGmumRmYDmaHGEMkN0oK26UPhqwyAAcxt5JEieCjrj12OjsQiMfMemiSS9xuk8U5MkTbgzKHCssI7kywaIwOJtijme9dvd2loAB48MCICroAS8UgHRQgADC7FFUvAuh4JDVkke3Q4KrQvg2ikNNB1PGVYvU6G+Bp4aG6hsVFuQXJ4pCt7UscskSv19gUGv9dDVVAtEKtAfFrMgwZ0IAiYYcXPU4ZxBkdAFULWYBPB1GgZeA8N6QFnRQXZbg0oRKzhY7rs4XUIh7bLUA6Q7Fyzh4R8Kzx0CWy514YR803SOmNaXG23ua23fXQyqIeBgegDJg9NP3By0u0KuktEA 600,600 noborder}}
- +
-<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0lwrFWAmZBmMbJgGzYCwAc+yAnIciGmiIvjfAKYC0YYAUAMYjIDslYXjh79wpSpCTZw0ZPEjJ8pZSuW8p2dmEIQiksPnwg9VNMMlYYcazZvCs4SVenOOAZREDDnkKXNQQADMAQwAbAGdGAOR2AHcqYl9hEz8oOONCSVSTAyNIdJNqexxCU3N0uWFUwSrygDdwITLG2oDJfElEJyR4dgAnFua+SiK26F6Ac0HRmqoStvS0RNTkQlLU-Pi0eBGzH1HNn1yjpo544Z5IUovFfS4fW4f4fysIMFl5OVsbGlcK0RuoiwRg43AuwP2yG6zkQyGgSlI8m+cFWGjOR282y8eX+lChpSxl1K+QADlQduBMRT8eAFmTCTSGaIIE52PTqUCKTMFlsuXszAS9ocmXiAaJDjdng9xbixKK8Vc0ucgVD9kL2AAPHhgRCzdACXikYw8IwAYU1PDQAh0JutRhBpQ4WvQRkeild9GN2h4FuQgktsLQrqWxsUIHNzrARrI6nQ9jgobNvv5ChN8bo4AJFoMsdTOuEUIz3vC2YcUfUOcu7RNIAAqhazAJ4Oo0IbwHhE3WG23cCNCEbcEXHd3o3IEqPSg6fVrGzxVa3RzQPV2Zzg8YQW22yNWw-WZ1vxOOeIivVnIyDUeh9BhO25ff6zAWg3MjcuAErsIA 600,600 noborder}}+
 </WRAP> </WRAP>
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
-==== Parallel Circuit of Capacitors ====+==== 13.1.2 Parallel Circuit of Capacitors ====
  
 If capacitors are connected in parallel, the voltage $U$ across the individual capacitors $C_1 ... C_n$ is equal. If capacitors are connected in parallel, the voltage $U$ across the individual capacitors $C_1 ... C_n$ is equal.
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 In the simulation below, again, besides the parallel connected capacitors $C_1$, $C_2$,$C_3$, an ideal voltage source $U_q$, a resistor $R$, a switch $S$, and a lamp are installed. In the simulation below, again, besides the parallel connected capacitors $C_1$, $C_2$,$C_3$, an ideal voltage source $U_q$, a resistor $R$, a switch $S$, and a lamp are installed.
  
-This derivation is also well explained, for example, in [[https://www.youtube.com/watch?v=fH-9pUeEpZU|this video]]. +<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3EaQMxjAFhQJi8gnBgKx4DsJ45hI6yIhApgLRoBQAxiFpABwUBsnHuDxZqUWJFHpoeWXPkLxCKCzDcIyPrzwDNvLIQGRwOQtDVbCk5N0LqQyOOCXwFbxUlfvvLAMqDeDHQAkB0ocIAzAEMAGwBnenCsFgB3By1OQ05ufSzIVOFRMMZ0bSMWADcQEtyBGtCjcMw6cOMYQhYAc2rSzLrerBzWgr0GzixgsPy0rlqTOemQoKWSATACsEHwdGDNwNXwdnn+Y7ARMU8IaW93F0QC3uXHg-WOZ4FH8+DPSCprm48EnuMyEy0GgR2KhBc3BfRUAAdshDgrCDGtWlDqHM9nDFjizqIcWjMbNCiFiYtPlIFgVSct8ZC8VsnsjMfViQM8iwAB7VbjkMCEciMEjGMB8Ywo4IAVR51T4eHAfAgjAlSuCu14cTlWxwGn0EzEakOvLAxiwYHIWCtXDEhoAwnK8LQLQKzeNJZxgo7TWhxhBNhAcBrwPo5ZaXZbhObbVKQD7hEGLYIkyHjcgWEA 700,500 noborder}}
- +
-<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3EaQMxjAFhQJi8gnBgKx4DsJ45hI6yIhApgLRoBQAxiFpABwUBsnHuDxYoUWJFHpoeWXPkLxCFmG4RkfXhnQgNvLIQGRwuacnQHuyQujBZu3Hcjjgl8BR8VJ3n3ywDKglroOly8eEZiAGYAhgA2AM70YlgsAO66XCARuprZRum54QJ64CFQhaUGAvb6hhUZ6MUgjE35FQBuLW3V3c3Gxph0YgPQhCwA5n2c9a36DiOVWTk4OjmQhWEzAnZ1BRlb2kH84IW7ZTrnYCQ77CZaN-fCoqMI1DK+Hm5ghW1Hf48fhwAQI-iIRhJIFRpJ8vt4fgchEdahcGsdeijehsAA6cBbIha9RBonr1c5Ys72Z5PLYbRFacFbWm-BlSfRCOnUZrndaU7ng3n01FXcqckU6P6ilgAJ3RQiZ9VekB+svJSKpWLcqVlPSEpMi3nQhTm2y5po2AA8WtxyGBCORGCRjGA+INODoAKosK2MPh4cB8CC+518CXgXgJb14ii0ZAPUHh05WsAh-1YW0kKTukAAYSjeFoLv9aGdmeo2bzybQMZMEGuCdUnCj10LfuEpazWB0leEdZItC4fYbvGQLCAA 700,500 noborder}}+
 </WRAP> </WRAP>
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
 +==== 13.1.3 Energy in the electric Field ====
  
-===== Common pitfalls ===== +Now we want to see how much energy is stored in a capacitor during charging. 
-  * ...+When we want to charge a capacior charges have be separated (see <imgref ImgNr616>). This gets more and more difficult as more charges were moved, since these already moved charges create an electric field. 
 + 
 +<WRAP> 
 +<imgcaption ImgNr616 | summary of electrostatics> 
 +</imgcaption> 
 +{{drawio>electrical_engineering_and_electronics_1:chargingACapacitor.svg}} 
 +</WRAP> 
 + 
 +We already had a first look onto the energy in the electric field in [[https://wiki.mexle.org/electrical_engineering_and_electronics_1/block09#energy_required_to_displace_a_charge_in_the_electric_field|block09]]. \\ 
 +There, we got: 
 + 
 +\begin{align*} 
 +\Delta W &= \int    \vec{F} d\vec{r} \\ 
 +         &= q  \int \vec{E} d\vec{r} \\ 
 +         &= q  \cdot U \\ 
 +dW       &= dq \cdot U  
 +\end{align*} 
 + 
 +Now, For a capacitor we include the formula for the capacitor $C = {{q}\over{U}}$, or better its rearranged version $U = {{q}\over{C}}$: 
 + 
 +\begin{align*} 
 +     dW &= dq \cdot {{q}\over{C}}  \\ 
 +\int dW &= \int {{q}\over{C}} dq  
 +\end{align*} 
 + 
 +Here we again see, that the needed energy portion $dW$ to move a portion $dq$ is also related to the already moved charges $q$. \\ 
 +To get the energy $Delta W$ needed to move all of the charges $$Q = \int dq$$ we have to integrate from $0$ to $Q$: 
 + 
 +\begin{align*} 
 +\Delta W &= \int_0^Q dW \\  
 +         &= \int_0^Q {{q}\over{C}} dq \\ 
 +         &= {{1}\over{2}}{{Q^2}\over{C}} \\  
 +\end{align*} 
 +\begin{align*} 
 +\boxed{ \Delta W =  {{1}\over{2}}{{Q^2}\over{C}} = {{1}\over{2}}QU = {{1}\over{2}}CQ^2 } 
 +\end{align*} 
 + 
 +~~PAGEBREAK~~ ~~CLEARFIX~~ 
 +===== 13.2 Common pitfalls ===== 
 +  * Mixing up the rules: writing $C_{\rm eq}=C_1+C_2$ for **series** (wrong) or $\dfrac{1}{C_{\rm eq}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}$ for **parallel** (wrong). 
 +  * Forgetting which quantity is equal: **series $\Rightarrow Q_k=\text{const.}$**, **parallel $\Rightarrow U_k=\text{const.}$**. 
 +  * Applying the **resistive** voltage divider $U_1=\dfrac{R_1}{R_1+R_2}U$ to capacitors. For capacitors in series it inverts: $U_1=\dfrac{C_2}{C_1+C_2}U$. 
 +  * Ignoring **initial charge states**: pre-charged capacitors reconnected will redistribute charge; use charge conservation on isolated nodes before using $Q=C\,U$. 
 +  * Dropping units or mixing forms of energy: always keep $W=\tfrac12 C U^2=\tfrac12 Q U=\dfrac{Q^2}{2C}$ and check $\rm J$.
  
-===== Exercises =====+===== 13.3 Exercises =====
  
 <panel type="info" title="Task 5.8.1 Calculating a circuit of different capacitors"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Task 5.8.1 Calculating a circuit of different capacitors"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
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 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-{{page>electrical_engineering_and_electronics_2:task_5.9.3_with_calculation&nofooter}} +{{page>electrical_engineering_and_electronics:task_5.9.3_with_calculation&nofooter}} 
- +{{page>electrical_engineering_and_electronics:task_k4wrrhf8v46gct49_with_calculation&nofooter}} 
 +{{page>electrical_engineering_and_electronics:task_y7dozgdsljqvnqge_with_calculation&nofooter}}
  
 ===== Embedded resources ===== ===== Embedded resources =====
 <WRAP column half> <WRAP column half>
-Explanation (video): ...+The equivalent capacitor for series of parallel configuration is well explained here 
 +{{youtube>9-Bp9Cvr7Jg}}
 </WRAP> </WRAP>