Capacitors in Series & Parallel - Physics 2

1. Switch $S$ is initially open. The network reaches equilibrium. Then $S$ is closed and the network reaches a new equilibrium. By how much does the charge delivered by the battery increase when $S$ is closed?
   Answer: $72\,\mu\text{C}$, because when $S$ closes, $C_3$ is added in parallel with $C_1$ and $C_2$. The equivalent capacitance increases from $5\,\mu\text{F}$ to $11\,\mu\text{F}$, and the additional charge delivered by the battery is $\Delta Q = C_3 \times V = 6 \times 12 = 72\,\mu\text{C}$. When analyzing a capacitor network that changes configuration, your first job is to clearly identify how each capacitor is connected before and after the switch closes — specifically, whether elements are in series or parallel. Before closing $S$, only $C_1$ and $C_2$ are active, both connected directly between nodes A and B — that's a parallel combination: $C_{eq,i} = 2 + 3 = 5\,\mu\text{F}$. The battery delivers $Q_i = 5 \times 12 = 60\,\mu\text{C}$. After closing $S$, capacitor $C_3$ is also connected directly between A and B — still in parallel with $C_1$ and $C_2$. The new equivalent capacitance is $C_{eq,f} = 2 + 3 + 6 = 11\,\mu\text{F}$, giving $Q_f = 11 \times 12 = 132\,\mu\text{C}$. The increase in charge delivered by the battery is $\Delta Q = 132 - 60 = 72\,\mu\text{C}$, which equals $C_3 \times V = 6 \times 12$. Answer D is correct. Choice A incorrectly subtracts the original 12 µC from $C_3$'s charge — but $C_1$ and $C_2$ are still fully charged by the battery, so no subtraction is needed; the battery simply supplies additional charge. Choice B misreads the circuit topology entirely, treating $C_3$ as series with the parallel pair — but since all three capacitors share the same two nodes A and B, they are always in parallel, never in series. Choice C repeats B's topology error and additionally concludes the charge decreases, which contradicts the physics of adding a capacitor in parallel at fixed voltage. Key strategy: At fixed voltage, adding a capacitor in parallel always increases total charge. The incremental charge is simply $\Delta Q = \Delta C_{eq} \times V$ — no subtraction needed.
2. What is the charge stored on capacitor $C_1$?
   Answer: $48\,\mu\text{C}$, because $C_1$ is in series with the parallel combination, so it carries the same charge as the equivalent series capacitor, which has capacitance $4\,\mu\text{F}$ at 12 V. When a circuit mixes series and parallel connections, your first move should always be to simplify from the inside out — reduce the parallel group first, then handle the series connection. Here, $C_2$ and $C_3$ are in parallel, so their equivalent capacitance is $C_{23} = 4 + 8 = 12\,\mu\text{F}$. Now you have $C_1 = 6\,\mu\text{F}$ in series with $C_{23} = 12\,\mu\text{F}$. The total equivalent capacitance is: $\frac{1}{C_{eq}} = \frac{1}{6} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} \implies C_{eq} = 4\,\mu\text{F}$ The charge stored by the entire series combination — and therefore the charge on $C_1$ — is: $Q = C_{eq} \times V = 4\,\mu\text{F} \times 12\,\text{V} = 48\,\mu\text{C}$ This confirms answer A is correct. Answer B uses $6\,\mu\text{F}$ (the value of $C_1$ alone) instead of the true equivalent capacitance, incorrectly treating $C_1$'s individual capacitance as if it were the series equivalent. Answer C arrives at $3\,\mu\text{F}$, which would result from adding the reciprocals of all three capacitors individually as if they were all in series — but $C_2$ and $C_3$ are in parallel, not series. Answer D makes two errors: it invents a $12\,\mu\text{F}$ equivalent by adding all three values directly, then incorrectly claims charge splits equally among branches. Study tip: On mixed-circuit problems, always redraw the circuit step by step. Label which components are parallel and which are series before touching any formula — rushing past this identification step is the source of nearly every wrong answer here.
3. A capacitor of capacitance $C_0$ is charged to voltage $V_0$ and then disconnected from the battery. An uncharged capacitor of capacitance $2C_0$ is then connected in parallel with the first capacitor. Which of the following correctly gives the final voltage across the combination and the ratio of final total energy to initial energy?
   Answer: Final voltage $= \frac{V_0}{3}$; ratio of final to initial energy $= \frac{1}{3}$. When a charged capacitor is disconnected from its battery and then connected to an uncharged capacitor, charge redistributes — but the total charge is conserved. This is the key principle to anchor your thinking. The initial charge on $C_0$ is $Q_0 = C_0 V_0$. When $2C_0$ is connected in parallel, the total capacitance becomes $C_0 + 2C_0 = 3C_0$. Since charge is conserved and the final voltage must be equal across both capacitors (they share terminals), the final voltage is: $V_f = \frac{Q_0}{3C_0} = \frac{C_0 V_0}{3C_0} = \frac{V_0}{3}$ Now for the energy ratio. Initial energy: $U_i = \frac{1}{2}C_0 V_0^2$. Final energy stored in the combination: $U_f = \frac{1}{2}(3C_0)V_f^2 = \frac{1}{2}(3C_0)\left(\frac{V_0}{3}\right)^2 = \frac{1}{2}(3C_0)\frac{V_0^2}{9} = \frac{C_0 V_0^2}{6}$. The ratio is $\frac{U_f}{U_i} = \frac{C_0 V_0^2/6}{C_0 V_0^2/2} = \frac{1}{3}$. This confirms answer C. Answer A gets the voltage right but incorrectly squares the voltage factor again, arriving at $\frac{1}{9}$ — a double-penalty error. Answer B assumes the two capacitors split the voltage evenly in half, forgetting that capacitance values determine the charge distribution, not a simple 50/50 split. Answer D confuses the redistribution direction, treating the larger capacitor as if it adds voltage rather than diluting it. A key study tip: energy is not conserved when capacitors share charge — the "missing" energy is lost to heat or radiation in the connecting wires. Always recalculate energy from $U = \frac{1}{2}CV^2$ after finding the new voltage.
4. After the plates of the second capacitor are pulled apart, what is the new voltage across the first capacitor?
   Answer: $5\,\text{V}$, because the first capacitor was charged to 5 V in the original series circuit, and since both capacitors are fully isolated after the battery is removed, the charge on each capacitor is fixed and the voltage across the first capacitor remains unchanged. When a capacitor is isolated — meaning no conducting path exists for charge to flow — its charge is permanently fixed. That principle is the key to unlocking this entire problem. In the original series circuit, the two capacitors share the same charge $Q = C_{series} \times V_{battery}$. Since $C_{series} = \frac{C_0}{2}$, we get $Q = \frac{C_0}{2} \times 10 = 5C_0$. Each capacitor holds charge $Q = 5C_0$, and each capacitor's voltage is $V = \frac{Q}{C_0} = 5\,\text{V}$. Once the battery is removed and the wires are detached, both capacitors become completely isolated — no charge can move anywhere. When you then pull apart the plates of the second capacitor, its capacitance drops to $C_0/2$, but its charge is still locked at $5C_0$. The first capacitor's charge is also still $5C_0$. Therefore, the voltage across the first capacitor remains $V_1 = \frac{5C_0}{C_0} = 5\,\text{V}$, confirming B is correct. Choice A misapplies the voltage-divider formula for series capacitors, which only applies when capacitors are connected and free to redistribute charge — that's not the case here. Choice C imagines charge flowing between the capacitors after isolation, which is physically impossible with no conducting path. Choice D similarly assumes charge redistribution, incorrectly concluding the first capacitor absorbs all the charge. A powerful study tip: whenever a problem says the battery is removed and wires are detached, immediately recognize that charge on each capacitor is frozen — voltage can only change if capacitance changes on that same capacitor.
5. How does the voltage across $C_1$ compare to the voltage across $C_3$, and what is the total charge delivered by the battery?
   Answer: The voltage across $C_1$ equals the voltage across $C_3$ (both are 10 V), and the total charge delivered by the battery is $40\,\mu\text{C}$. When a circuit mixes series and parallel connections, tackle each branch independently before combining them. Here, Branch A ($C_1$ and $C_2$ in series) and Branch B ($C_3$ and $C_4$ in series) are connected in parallel across 20 V. Because they're in parallel, both branches see the full 20 V. Within each branch, the two equal capacitors split that voltage equally — so $C_1$, $C_2$, $C_3$, and $C_4$ each carry 10 V. The voltage across $C_1$ equals the voltage across $C_3$: both are 10 V. ✓ Now for total charge. First find the equivalent capacitance of each series branch: $\frac{1}{C_{series}} = \frac{1}{2} + \frac{1}{2}$, giving $C_{series} = 1\,\mu\text{F}$. The two branches in parallel add: $C_{total} = 1 + 1 = 2\,\mu\text{F}$. Total charge from the battery: $Q = C_{total} \times V = 2\,\mu\text{F} \times 20\,\text{V} = 40\,\mu\text{C}$. This confirms answer A. Answer B incorrectly halves the charge (using only one branch), ignoring that the battery supplies both branches. Answer C falsely claims $C_1$ sees 20 V — that would only be true if $C_1$ were alone across the source, not in series with $C_2$. Answer D doubles the charge to 80 µC, likely by mistakenly treating all four capacitors as being in parallel rather than properly accounting for the series combinations. Study tip: Always reduce series groups first, then combine parallel branches — and remember that parallel branches each receive the full source voltage, not a divided share.
6. Three capacitors with capacitances $C$, $2C$, and $3C$ are connected in series across a voltage source $V$. A student claims that the capacitor with capacitance $3C$ stores the most energy because it has the greatest capacitance. A second student claims that the capacitor with capacitance $C$ stores the most energy. Which student, if either, is correct, and why?
   Answer: The second student is correct. In a series circuit, all capacitors carry the same charge $Q$, and since energy stored is $U = Q^2/(2C)$, the capacitor with the smallest capacitance $C$ stores the most energy. Whenever you see capacitors in series, the single most important fact to lock in is this: every capacitor in a series circuit carries the same charge $Q$. The source pushes charge onto the first plate, which induces the same charge on each subsequent capacitor — there's no branching path for charge to go elsewhere. With that in mind, the right formula for comparing stored energy isn't $U = \frac{1}{2}CV^2$ — it's the equivalent form $U = \frac{Q^2}{2C}$. Because $Q$ is identical for all three capacitors, the one with the smallest capacitance $C$ produces the largest value of $\frac{Q^2}{2C}$, meaning it stores the most energy. The second student is correct, making B the right answer. Here's where each wrong answer breaks down. A is doubly flawed: in series, larger capacitance does not mean more charge (charge is equal everywhere), and the reasoning about energy is backwards. C sounds balanced but is simply false — the capacitors do not store equal energy; energy scales inversely with capacitance when charge is fixed, so smaller capacitors store more. D starts with a true statement (voltage divides inversely with capacitance, so $3C$ sees the smallest voltage drop), but then makes a fatal error: when you substitute $V = Q/C$ into $U = \frac{1}{2}CV^2$, you recover $U = \frac{Q^2}{2C}$, and the larger capacitance loses — it does not compensate. Study tip: On series capacitor problems, always default to $U = \frac{Q^2}{2C}$ rather than $U = \frac{1}{2}CV^2$. Equal charge is the defining feature of series circuits, and picking the wrong energy formula is the most common trap on this type of question.
7. What is the magnitude of the final voltage across the parallel combination after the system reaches equilibrium?
   Answer: $0\,\text{V}$, because the initial charge on $C_A$ is $Q_A = 24\,\mu\text{C}$ and on $C_B$ is $Q_B = 24\,\mu\text{C}$; the opposite-polarity connection causes these charges to cancel, leaving zero net charge and zero final voltage. When capacitors are reconnected after being charged, conservation of charge governs the outcome — but you must be careful about sign conventions. Each capacitor's charge carries a polarity, and opposite-polarity reconnection means those charges work against each other. Here's the key setup: $C_A = 3\,\mu\text{F}$ charged to 8 V holds $Q_A = C_A V_A = 24\,\mu\text{C}$, and $C_B = 6\,\mu\text{F}$ charged to 4 V holds $Q_B = C_B V_B = 24\,\mu\text{C}$. When reconnected with opposite polarity, the positive plate of one connects to the negative plate of the other, so the net charge on the combined system is $Q_{net} = Q_A - Q_B = 24 - 24 = 0\,\mu\text{C}$. With zero net charge distributed over the total capacitance, the final voltage is $V_f = Q_{net}/(C_A + C_B) = 0/9\,\mu\text{F} = 0\,\text{V}$. Answer C is correct. Answer A incorrectly assumes $C_A$'s charge is fully neutralized first and only the "leftover" charge from $C_B$ remains — charge doesn't get consumed sequentially like this; it redistributes simultaneously based on net total. Answer B makes the opposite sign error, treating opposite-polarity connection as if it adds charges rather than subtracts, giving an inflated $Q_{net} = 48\,\mu\text{C}$. Answer D ignores both polarity and the capacitance weighting; a simple voltage average is never the correct formula here. Your strategy: always assign signed charges before combining. If polarities oppose, subtract. If they align, add. Then apply $V_f = Q_{net}/C_{total}$.

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