Exercises — Series and parallel capacitors — derivations

WHAT: parallel → capacitances add. WHY: both capacitors sit across the same two nodes (same voltage), so charges just pile up together. $C_{eq}=C_1+C_2=4+12=16\mu F$ Sanity check: larger than the biggest member ($12$) — correct for parallel.

WHAT: series → reciprocals add. Use the two-capacitor shortcut $C_{eq}=\frac{C_1{C}_2}{C_1+C_2}$. WHY: the same charge marches through both, voltages add, so it is the $1/C$ values that add. $C_{eq}=\frac{4\times 12}{4+12}=\frac{48}{16}=3\mu F$ Sanity check: smaller than the smallest member ($4$) — correct for series.

(a) Series shortcut $C_{eq}=\frac{C_1{C}_2}{C_1+C_2}$: $C_{eq}=\frac{2\times 6}{2+6}=\frac{12}{8}=1.5\mu F$ Why this formula? Because these are in series, the same charge $Q$ passes through both while their voltages add ($V=V_1+V_2$). Feeding $V_1=Q/C_1$ and $V_2=Q/C_2$ into that sum gives $\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}$, which for two capacitors rearranges to product-over-sum. The shortcut is the same-charge idea in disguise. (b) Same charge everywhere in series; get it from the equivalent capacitor: $Q=C_{eq}V=1.5\times 16=24\mu C$ Why the equivalent? $C_{eq}$ by definition draws the same total charge at the same voltage, so $Q=C_{eq}V$ hands us the single charge that then sits on both capacitors. (Units: $\mu F\times V=\mu C$.) (c) Use $V=Q/C$ on each (from Capacitance and the relation Q = CV): $V_1=\frac{24}{2}=12V,V_2=\frac{24}{6}=4V$ Check with Kirchhoff's voltage law: $12+4=16V$ ✓. The smaller capacitor ($C_1$) takes the larger voltage, since $V\propto 1/C$.

(a) Parallel → add: $C_{eq}=3+5=8\mu F$. Why add? Both capacitors are wired across the same two nodes, so they feel the same voltage $V$; their charges $Q_1=C_{1}V$ and $Q_2=C_{2}V$ add, giving $Q_{total}=(C_1+C_2)V$ — i.e. capacitances add. (b) $Q_{total}=C_{eq}V=8\times 20=160\mu C$. Why $C_{eq}$? By definition the equivalent capacitor stores the whole network's charge at the applied voltage. (Units: $\mu F\times V=\mu C$.) (c) Same voltage on each; apply $Q=CV$ per branch: $Q_1=3\times 20=60\mu C,Q_2=5\times 20=100\mu C$ Check (Charge conservation): $60+100=160\mu C$ ✓. Bigger capacitor grabs more charge ($Q\propto C$). (d) Energy from Energy stored in a capacitor using the equivalent. Convert $C$ to farads so the answer lands in joules: $U=\frac{1}{2}{C}_{eq}{V}^2=\frac{1}{2}(8\times 10^{-6})(20{)}^2=1.6\times 10^{-3}J=1.6mJ$

Step 1 — reduce the innermost group (parallel $C_2\parallel C_3$): $C_{23}=C_2+C_3=2+4=6\mu F$ Why first, and why add? $C_2$ and $C_3$ hang between the same nodes A and B (see figure), so they share a voltage and their capacitances add (parallel rule). Collapsing this group first turns the whole thing into a clean series pair. Step 2 — now $C_1$ in series with $C_{23}$: $C_{eq}=\frac{C_1{C}_{23}}{C_1+C_{23}}=\frac{4\times 6}{4+6}=\frac{24}{10}=2.4\mu F$ Why product-over-sum? $C_1$ and the block $C_{23}$ now carry the same charge with adding voltages — the two-capacitor series shortcut applies. Step 3 — total charge from the battery: $Q=C_{eq}V=2.4\times 24=57.6\mu C$ Why? The equivalent capacitor draws the network's total charge at $24V$; and because $C_1$ and the block are in series, this same $Q$ sits on $C_1$ and on the $C_{23}$ block. (Units: $\mu F\times V=\mu C$.) Step 4 — voltages using $V=Q/C$: $V_1=\frac{57.6}{4}=14.4V,V_{23}=\frac{57.6}{6}=9.6V$ Check (Kirchhoff's voltage law): $14.4+9.6=24V$ ✓. Step 5 — split charge inside the parallel block (both feel $V_{23}=9.6V$): $Q_2=2\times 9.6=19.2\mu C,Q_3=4\times 9.6=38.4\mu C$ Why the same $V_{23}$ for both? They are parallel — same two nodes A, B — so each obeys $Q=CV$ with the common $V_{23}$. Check: $Q_2+Q_3=57.6\mu C=Q$ ✓ (the block's charge equals the series charge).

Battery voltage is the same in both cases, so use $U=\frac{1}{2}{C}_{eq}{V}^2$ — energy scales directly with $C_{eq}$. Parallel: $C_{eq}^{\text{par}}=10+10=20\mu F$. Why add? Same two nodes → same voltage → capacitances add. $U_{\text{par}}=\frac{1}{2}(20\times 10^{-6})(100{)}^2=0.10J$ Series: $C_{eq}^{\text{ser}}=\frac{10\times 10}{10+10}=5\mu F$. Why product-over-sum? Same charge, adding voltages → reciprocals add. $U_{\text{ser}}=\frac{1}{2}(5\times 10^{-6})(100{)}^2=0.025J$ Ratio: $\frac{U_{\text{par}}}{U_{\text{ser}}}=\frac{20}{5}=4$ Parallel stores 4× more energy. Makes sense: same voltage, four times bigger equivalent capacitance ($20$ vs $5$). (Both energies used $C$ in farads so the results are in joules.)

Step 1 — reciprocals add: $\frac{1}{C_{eq}}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}=\frac{6+3+2}{6}=\frac{11}{6}(\mu F{)}^{-1}\Rightarrow C_{eq}=\frac{6}{11}\mu F\approx 0.545\mu F$ Why reciprocals? All three are in series → same charge through each, voltages add; substituting $V_i=Q/C_i$ into $V=\sum V_i$ gives $\frac{1}{C_{eq}}=\sum \frac{1}{C_i}$. Step 2 — common charge: $Q=C_{eq}V=\frac{6}{11}\times 11=6\mu C$ Why? The equivalent capacitor sets the single charge that, in series, is identical on every member. (Units: $\mu F\times V=\mu C$.) Step 3 — individual voltages via $V=Q/C$: $V_1=\frac{6}{1}=6V,V_2=\frac{6}{2}=3V,V_3=\frac{6}{3}=2V$ Check: $6+3+2=11V$ ✓. The smallest capacitor $C_1=1\mu F$ takes the most voltage ($6V$) — exactly because $V\propto 1/C$.

Use $C_{eq}=\frac{C_1{C}_2}{C_1+C_2}$ (the series shortcut — valid because the pair shares one charge with adding voltages). As $C_2\to 0$: numerator $\to 0$ faster than denominator, so $C_{eq}\to 0$. A vanishing capacitor is a huge "difficulty" ($1/C_2\to \infty$); series difficulties add, so the whole chain becomes impossible to charge — $C_{eq}\to 0$. As $C_2\to \infty$: divide top and bottom by $C_2$: $C_{eq}=\frac{C_1}{C_1/C_2+1}\stackrel{C_2\to \infty }{\to }\frac{C_1}{0+1}=C_1=5\mu F$ A giant $C_2$ contributes zero difficulty ($1/C_2\to 0$); it behaves like a plain wire, so only $C_1$ limits the chain — $C_{eq}\to C_1$. Lesson: in series, the smallest capacitor dominates $C_{eq}$.

(a) Voltage rating forces series. A single $2\mu F$ survives only $50V$, so to hold $150V$ the voltage must split, i.e. series. With identical capacitors in series the voltage splits equally, so we need $n=\frac{150V}{50V}=3\text{capacitors in series per string,}$ and each then sees exactly $150/3=50V$ (at rating, safe). The string's capacitance: $\frac{1}{C_{\text{string}}}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{3}{2}\Rightarrow C_{\text{string}}=\frac{2}{3}\mu F\approx 0.667\mu F.$ Why 3 and not fewer? $n=2$ would put $75V$ on each capacitor — over the $50V$ rating. $n=3$ is the minimum that keeps every capacitor at or below $50V$. (b) Parallel raises capacitance, and it is unbounded. Adding strings in parallel does not change the voltage any capacitor sees (each string still splits $150V$ into $50/50/50$), so it is always safe, and parallel capacitances add: $\boxed{C_{\text{block}}(m)=m\times \frac{2}{3}\mu F.}$ This grows linearly with $m$ without any upper bound — with unlimited capacitors you can make the block capacitance as large as you please. "Largest possible" has no finite answer; the design instead trades capacitors for capacitance ($m$ strings of 3 = $3m$ capacitors give $\frac{2m}{3}\mu F$). (c) Smallest count meeting only the voltage rating: a single string of $n=3$ capacitors, giving $C_{\text{string}}=\frac{2}{3}\mu F$ — 3 capacitors total.

1. Series of $4\mu F$ and $12\mu F$? ::: $3\mu F$ (product over sum).
2. In series across $16V$ with $C_1=2,C_2=6\mu F$, which takes more voltage? ::: $C_1$, the smaller one, $12V$ vs $4V$.
3. Fixed battery: parallel or series stores more energy? ::: Parallel — bigger $C_{eq}$, and $U=\frac{1}{2}{C}_{eq}{V}^2$.
4. Series with $C_2\to \infty$, $C_1=5\mu F$ fixed: $C_{eq}\to$? ::: $C_1=5\mu F$ (huge $C_2$ acts like a wire).