Block 13 - Capacitor Circuits and Energy

Learning objectives

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:

• 5.9.5

90-minute plan

1. Warm-up (10 min):
   1. Quick quiz (2–3 items): series or parallel? which rule applies (constant $U$ or constant $Q$)?
   2. Recall $Q=C\,U$ and energy $W=\tfrac12 C U^2$ (units).
2. Core concepts & derivations (35 min):
   1. Derive $C_{\rm eq}$ for series from Kirchhoff’s voltage law and $Q=\text{const.}$; derive voltage division $U_k=\dfrac{Q}{C_k}$.
   2. Derive $C_{\rm eq}$ for parallel from Kirchhoff’s current/charge balance and $U=\text{const.}$; obtain $Q_k=C_k U$.
   3. Energy in the electric field: integrate $dW=U\,dq$ → $W=\tfrac12 C U^2$; short dimensional check.
3. Practice (35 min):
   1. Two short worked examples: mixed series/parallel network; two-capacitor divider with given $U$ (find $U_1$, $U_2$, $W$ on each).
   2. Short simulation tasks (use the two embedded Falstad circuits in this page): observe $U_k$, $Q_k$ when toggling the switch or changing values.
   3. Mini-problems: “double a plate area / halve distance” reasoning on $C$ and $W$.
4. Wrap-up (10 min):
   1. Common-pitfalls checklist and one exit-ticket calculation.

Conceptual overview

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.

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 $C_1 ... C_n$ :

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.

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 Abbildung 1). This gets more and more difficult as more charges were moved, since these already moved charges create an electric field.

We already had a first look onto the energy in the electric field in 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*}

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

Embedded resources