Physics 2 Quiz: Capacitors In Series And Parallel

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Capacitors In Series And ParallelQuestion 1 of 7

A parallel-plate capacitor with capacitance $C_0$ is connected in series with an identical capacitor (also capacitance $C_0$ ) and a 10 V battery. After the system reaches equilibrium, the battery is removed and the connecting wires are detached. The plates of the second capacitor are then pulled apart until its capacitance is reduced to $C_0/2$ .

After the plates of the second capacitor are pulled apart, what is the new voltage across the first capacitor?

\frac{10}{3}\,\text{V}

, because the reduced capacitance of the second capacitor changes the series voltage division ratio, and the first capacitor's voltage adjusts to

V_1 = \frac{C_0/2}{C_0 + C_0/2} \times 10\,\text{V}

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.

\frac{20}{3}\,\text{V}

, because the reduced capacitance of the second capacitor means it now holds less charge, so charge flows from the second capacitor to the first, increasing the first capacitor's voltage.

10\,\text{V}

, because once the second capacitor's capacitance decreases, the first capacitor absorbs all the charge originally shared between the two, resulting in the full battery voltage appearing across it.

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Physics 2 Quiz

Physics 2 Quiz: Capacitors In Series And Parallel

Practice Capacitors In Series And Parallel in Physics 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Capacitors In Series And Parallel, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

After the plates of the second capacitor are pulled apart, what is the new voltage across the first capacitor?

$\frac{10}{3}\,\text{V}$ , because the reduced capacitance of the second capacitor changes the series voltage division ratio, and the first capacitor's voltage adjusts to $V_1 = \frac{C_0/2}{C_0 + C_0/2} \times 10\,\text{V}$ .
$5\,\text{V}$ , 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. (correct answer)
$\frac{20}{3}\,\text{V}$ , because the reduced capacitance of the second capacitor means it now holds less charge, so charge flows from the second capacitor to the first, increasing the first capacitor's voltage.
$10\,\text{V}$ , because once the second capacitor's capacitance decreases, the first capacitor absorbs all the charge originally shared between the two, resulting in the full battery voltage appearing across it.

Explanation: 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.

Question 2

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?

The first student is correct. In a series circuit, greater capacitance leads to greater charge storage, and since energy depends on both charge and capacitance, the largest capacitor stores the most energy.
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. (correct answer)
Neither student is correct. In a series circuit, all capacitors store the same energy because they carry the same charge and the voltage differences between them average out, dividing energy equally regardless of individual capacitances.
The first student is correct. In a series circuit, voltage divides inversely with capacitance, so the largest capacitor has the smallest voltage drop. However, since energy is $U = \frac{1}{2}CV^2$ , the larger capacitance more than compensates for the smaller voltage, and the capacitor with capacitance $3C$ stores the greatest energy.

Explanation: 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.

Question 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?

Final voltage $= \frac{V_0}{3}$ ; ratio of final to initial energy $= \frac{1}{9}$ .
Final voltage $= \frac{V_0}{2}$ ; ratio of final to initial energy $= \frac{1}{4}$ .
Final voltage $= \frac{V_0}{3}$ ; ratio of final to initial energy $= \frac{1}{3}$ . (correct answer)
Final voltage $= \frac{2V_0}{3}$ ; ratio of final to initial energy $= \frac{2}{3}$ .

Explanation: 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

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.

Question 4

A student constructs a circuit with three capacitors. Capacitor $C_1 = 6\,\mu\text{F}$ is connected in series with a parallel combination of $C_2 = 4\,\mu\text{F}$ and $C_3 = 8\,\mu\text{F}$ . The entire network is connected to a 12 V battery.

What is the charge stored on capacitor $C_1$ ?

$48\,\mu\text{C}$ , 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. (correct answer)
$72\,\mu\text{C}$ , because $C_1$ is in series with the parallel combination, so it carries the same charge as the equivalent series capacitor, which has capacitance $6\,\mu\text{F}$ at 12 V.
$36\,\mu\text{C}$ , because $C_1$ is in series with the parallel combination, so it carries the same charge as the equivalent series capacitor, which has capacitance $3\,\mu\text{F}$ at 12 V.
$144\,\mu\text{C}$ , because all three capacitors together have an equivalent capacitance of $12\,\mu\text{F}$ , and at 12 V the total charge is split equally among the three branches.

Explanation: 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.

Question 5

A student constructs the following network: capacitor $C_1$ is connected from node A to node B. Capacitor $C_2$ is connected from node A to node B. Capacitor $C_3$ is connected from node A to node B. However, capacitor $C_3$ has a switch $S$ in series with it. The values are $C_1 = 2\,\mu\text{F}$ , $C_2 = 3\,\mu\text{F}$ , $C_3 = 6\,\mu\text{F}$ , and the battery maintains 12 V between A and B.

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?

$60\,\mu\text{C}$ , because closing $S$ adds $C_3 = 6\,\mu\text{F}$ in parallel, and the total charge after closing equals $C_3 \times V = 6 \times 12 = 72\,\mu\text{C}$ , but the pre-existing charge on $C_1$ and $C_2$ must be subtracted, yielding an increase of $72 - 12 = 60\,\mu\text{C}$ .
$48\,\mu\text{C}$ , because the switch connects $C_3$ in series with the parallel combination of $C_1$ and $C_2$ , giving a new equivalent capacitance of $\frac{30}{11}\,\mu\text{F}$ . The charge after closing is $\frac{30}{11} \times 12 \approx 32.7\,\mu\text{C}$ , a decrease, but the magnitude of the change in charge delivered to $C_3$ alone is $6 \times 8 = 48\,\mu\text{C}$ .
$36\,\mu\text{C}$ , because closing $S$ places $C_3$ in series with the parallel combination of $C_1$ and $C_2$ . The new equivalent capacitance is $\frac{5 \times 6}{5 + 6} = \frac{30}{11}\,\mu\text{F}$ , which is less than the original $5\,\mu\text{F}$ , so the total charge actually decreases by approximately $36\,\mu\text{C}$ .
$72\,\mu\text{C}$ , 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 $\Delta Q = C_3 \times V = 6 \times 12 = 72\,\mu\text{C}$ . (correct answer)

Explanation: 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.

Question 6

A network consists of four capacitors: $C_1 = 2\,\mu\text{F}$ , $C_2 = 2\,\mu\text{F}$ , $C_3 = 2\,\mu\text{F}$ , and $C_4 = 2\,\mu\text{F}$ . The network is arranged as follows: $C_1$ and $C_2$ are in series (call this branch A), and $C_3$ and $C_4$ are in series (call this branch B). Branch A and Branch B are then connected in parallel across a 20 V source.

How does the voltage across $C_1$ compare to the voltage across $C_3$ , and what is the total charge delivered by the battery?

The voltage across $C_1$ equals the voltage across $C_3$ (both are 10 V), and the total charge delivered by the battery is $40\,\mu\text{C}$ . (correct 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 $20\,\mu\text{C}$ .
The voltage across $C_1$ is 20 V and the voltage across $C_3$ is 10 V because they are in different branches, and the total charge delivered by the battery is $40\,\mu\text{C}$ .
The voltage across $C_1$ equals the voltage across $C_3$ (both are 10 V), and the total charge delivered by the battery is $80\,\mu\text{C}$ .

Explanation: 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.

Question 7

Two capacitors, $C_A = 3\,\mu\text{F}$ and $C_B = 6\,\mu\text{F}$ , are each independently charged: $C_A$ is charged to 8 V and $C_B$ is charged to 4 V. Both capacitors are then disconnected from their batteries. They are reconnected in parallel but with opposite polarity (positive plate of $C_A$ connected to negative plate of $C_B$ , and vice versa).

What is the magnitude of the final voltage across the parallel combination after the system reaches equilibrium?

$\frac{8}{3}\,\text{V}$ , because the final voltage is found by distributing the larger charge $Q_B = 24\,\mu\text{C}$ over the total capacitance $9\,\mu\text{F}$ , since the smaller capacitor's charge is fully neutralized first.
$\frac{16}{3}\,\text{V}$ , because opposite-polarity reconnection means the charges add rather than cancel, giving a net charge of $Q_A + Q_B = 48\,\mu\text{C}$ distributed over the total capacitance $C_A + C_B = 9\,\mu\text{F}$ .
$0\,\text{V}$ , 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. (correct answer)
$6\,\text{V}$ , because the final voltage is the simple average of the two initial voltages: $V_f = (V_A + V_B)/2 = (8 + 4)/2 = 6\,\text{V}$ , with the opposite polarity having no effect on the magnitude of the result.

Explanation: 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}