electrical_engineering_and_electronics_1:block13 [MEXLE Wiki]

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====== Block 13 - Capacitor Circuits and Energy ======

===== Learning objectives =====
<callout>
After this 90-minute block, you can
  * identify series vs. parallel connections of capacitors from a circuit diagram,
  * compute equivalent capacitance $C_{\rm eq}$ for series and parallelnetworks,
  * use the key sharing rules: in **series** $Q_k=\text{const.}$ and voltages divide; in **parallel** $U_k=\text{const.}$ and charges divide,
  * apply the capacitor divider relation (two series capacitors),
  * determine stored energy, including a dimensional check to $\rm J$.
</callout>

===== Preparation at Home =====

Well, again 
  * read through the present chapter and write down anything you did not understand.
  * Also here, there are some clips for more clarification under 'Embedded resources' (check the text above/below, sometimes only part of the clip is interesting). 

For checking your understanding please do the following exercises:
  * ...

===== 90-minute plan =====
  - Warm-up (10 min):
    - Quick quiz (2–3 items): series or parallel? which rule applies (constant $U$ or constant $Q$)?
    - Recall $Q=C\,U$ and energy $W=\tfrac12 C U^2$ (units).
  - Core concepts & derivations (35 min):
    - Derive $C_{\rm eq}$ for **series** from Kirchhoff’s voltage law and $Q=\text{const.}$; derive voltage division $U_k=\dfrac{Q}{C_k}$.
    - 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 =====
<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>

===== Core content =====

==== Series Circuit of Capacitor ====

If capacitors are connected in series, the charging current $I$ into the individual capacitors $C_1 ... C_n$ is equal.
Thus, the charges absorbed $\Delta Q$ are also equal:
\begin{align*}
\Delta Q = \Delta Q_1 = \Delta Q_2 = ... = \Delta Q_n
\end{align*}

Furthermore, after charging, a voltage is formed across the series circuit, which corresponds to the source voltage $U_q$. This results from the addition of partial voltages across the individual capacitors.
\begin{align*}
U_q = U_1 + U_2 + ... + U_n = \sum_{k=1}^n U_k
\end{align*}

It holds for the voltage $U_k = \Large{{Q_k}\over{C_k}}$. \\
If all capacitors are initially discharged, then $U_k = \Large{{\Delta Q}\over{C_k}}$ holds.
Thus
\begin{align*}
U_q &= &U_1 &+ &U_2 &+ &... &+ &U_n &= \sum_{k=1}^n U_k \\
U_q &= &{{\Delta Q}\over{C_1}} &+ &{{\Delta Q}\over{C_2}} &+ &... &+ &{{\Delta Q}\over{C_3}} &= \sum_{k=1}^n {{1}\over{C_k}}\cdot \Delta Q \\
{{1}\over{C_{ \rm eq}}}\cdot \Delta Q &= &&&&&&&&\sum_{k=1}^n {{1}\over{C_k}}\cdot \Delta Q
\end{align*}

Thus, for the series connection of capacitors $C_1 ... C_n$ :

\begin{align*}
\boxed{ {{1}\over{C_{ \rm eq}}} = \sum_{k=1}^n {{1}\over{C_k}} } 
\end{align*}
\begin{align*}
\boxed{ \Delta Q_k = {\rm const.}} 
\end{align*}

For initially uncharged capacitors, (voltage divider for capacitors) holds:
\begin{align*}
\boxed{Q = Q_k}
\end{align*}
\begin{align*}
\boxed{U_{ \rm eq} \cdot C_{ \rm eq} = U_{k} \cdot C_{k} }
\end{align*}

In the simulation below, 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.
  * The switch $S$ allows the voltage source to charge the capacitors.
  * The resistor $R$ is necessary because the simulation cannot represent instantaneous charging. The resistor limits the charging current to a maximum value. 
  * The capacitors can be discharged again via the lamp.

<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0lwrFWAmZBmMbJgGzYCwAc+yAnIciGmiIvjfAKYC0YYAUAMYjIDslYXjh79wpSpCTZw0ZPEjJ8pZSuW8p2dmEIQiksPnwg9VNMMlYYcazZvCs4SVenOOAZREDDnkKXNQQADMAQwAbAGdGAOR2AHcqYl9hEz8oOONCSVSTAyNIdJNqexxCU3N0uWFUwSrygDdwITLG2oDJfElEJyR4dgAnFua+SiK26F6Ac0HRmqoStvS0RNTkQlLU-Pi0eBGzH1HNn1yjpo544Z5IUovFfS4fW4f4fysIMFl5OVsbGlcK0RuoiwRg43AuwP2yG6zkQyGgSlI8m+cFWGjOR282y8eX+lChpSxl1K+QADlQduBMRT8eAFmTCTSGaIIE52PTqUCKTMFlsuXszAS9ocmXiAaJDjdng9xbixKK8Vc0ucgVD9kL2AAPHhgRCzdACXikYw8IwAYU1PDQAh0JutRhBpQ4WvQRkeild9GN2h4FuQgktsLQrqWxsUIHNzrARrI6nQ9jgobNvv5ChN8bo4AJFoMsdTOuEUIz3vC2YcUfUOcu7RNIAAqhazAJ4Oo0IbwHhE3WG23cCNCEbcEXHd3o3IEqPSg6fVrGzxVa3RzQPV2Zzg8YQW22yNWw-WZ1vxOOeIivVnIyDUeh9BhO25ff6zAWg3MjcuAErsIA 600,600 noborder}}
</WRAP>

~~PAGEBREAK~~ ~~CLEARFIX~~
==== Parallel Circuit of Capacitors ====

If capacitors are connected in parallel, the voltage $U$ across the individual capacitors $C_1 ... C_n$ is equal.
It is therefore valid:

\begin{align*}
U_q = U_1 = U_2 = ... = U_n
\end{align*}

Furthermore, during charging, the total charge $\Delta Q$ from the source is distributed to the individual capacitors.
This gives the following for the individual charges absorbed:
\begin{align*}
\Delta Q = \Delta Q_1 + \Delta Q_2 + ... + \Delta Q_n = \sum_{k=1}^n \Delta Q_k
\end{align*}

If all capacitors are initially discharged, then $Q_k  = \Delta Q_k = C_k \cdot U$ \\
Thus
\begin{align*}
\Delta Q &= & Q_1 &+ & Q_2 &+ &... &+ & Q_n &=  \sum_{k=1}^n Q_k \\
\Delta Q &= &C_1 \cdot U &+ &C_2 \cdot U &+ &... &+ &C_n \cdot U &=  \sum_{k=1}^n C_k \cdot U \\
C_{ \rm eq} \cdot U &= &&&&&&&& \sum_{k=1}^n C_k \cdot U \\
\end{align*}

Thus, for the parallel connection of capacitors $C_1 ... C_n$ :
<WRAP>
\begin{align*}
\boxed{ C_{ \rm eq} = \sum_{k=1}^n C_k }
\end{align*}
\begin{align*}
\boxed{ U_k = {\rm const.}}
\end{align*}
</WRAP>

For initially uncharged capacitors, (charge divider for capacitors) holds:
\begin{align*}
\boxed{\Delta Q = \sum_{k=1}^n Q_k} 
\end{align*}

\begin{align*}
\boxed{ {{Q_k}\over{C_k}} = {{\Delta Q}\over{C_{ \rm eq}}} }
\end{align*}

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.

<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>

~~PAGEBREAK~~ ~~CLEARFIX~~
==== Energy in the electric Field ====

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~~
===== 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 =====

<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%>

See https://www.youtube.com/watch?v=vSeSHAmpd4Y
</WRAP></WRAP></panel>


<panel type="info" title="Task 5.9.1 Layered Capacitor Task"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>ATXnPRXXDi4}}

</WRAP></WRAP></panel>

<panel type="info" title="Exercise 5.9.2 Further capacitor charging/discharging practice Exercise "> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>a-gPuw6JsxQ}}

</WRAP></WRAP></panel>

<panel type="info" title="Exercise 5.9.3 Further practice charging the capacitor"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>L0S_Aw8pBto}}

</WRAP></WRAP></panel>

<panel type="info" title="Exercise 5.9.4 Charge balance of two capacitors"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

{{youtube>EMdpkDoMXXE}}

</WRAP></WRAP></panel>

<panel type="info" title="Exercise 5.9.5 Capacitor with glass plate"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>

<WRAP>
<imgcaption ImgNr15 | Structure of a capacitor with glass plate>
</imgcaption>
{{drawio>CapacitorWithGlassPlate.svg}}
</WRAP>

Two parallel capacitor plates face each other with a distance $d_{ \rm K} = 10~{ \rm mm}$. A voltage of $U = 3'000~{ \rm V}$ is applied to the capacitor. 
Parallel to the capacitor plates, there is a glass plate ($\varepsilon_{ \rm r, G}=8$) with a thickness $d_{ \rm G} = 3~{ \rm mm}$ in the capacitor.

  - Calculate the partial voltages $U_{ \rm G}$ in the glass and $U_{ \rm A}$ in the air gap.
  - What is the maximum thickness of the glass pane if the electric field $E_{ \rm 0, G} =12 ~{ \rm kV/cm}$ must not exceed?

~~PAGEBREAK~~ ~~CLEARFIX~~
</WRAP></WRAP></panel>

{{page>electrical_engineering_and_electronics_2:task_5.9.3_with_calculation&nofooter}}



===== Embedded resources =====
<WRAP column half>
The equivalent capacitor for series of parallel configuration is well explained here
{{youtube>9-Bp9Cvr7Jg}}
</WRAP>

~~PAGEBREAK~~ ~~CLEARFIX~~