Q.In a circuit a cell of emf $E$ is connected through a key $K_1$ to a capacitor $C_1$; a second key $K_2$ then connects $C_1$ to another capacitor $C_2$. The top plate of $C_1$ is joined to the cell's positive terminal through $K_1$ and, through $K_2$, to the top plate of $C_2$, while the bottom plates of both capacitors share a common return wire to the cell. Initially $K_1$ is closed and $K_2$ is open, so $C_1$ charges up to the cell's emf while $C_2$ stays uncharged. Then $K_1$ is opened and $K_2$ is closed (the order is important). Taking $Q_1'$ and $Q_2'$ as the final charges on $C_1$ and $C_2$, $V_1$ and $V_2$ as their final voltages, and $Q$ as the charge originally stored on $C_1$, then

Showing the 12 most recent of 36 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.Four identical capacitors are connected in series with a battery of 16V between A and B as shown in the figure. If the point P is earthened, then the potentials at A and B are (Figure: a battery of 16V is connected across points A and B via the top wire. Between A and B along the main line, four identical capacitors of capacitance C are connected in series in the order: A — C — C — C — P — C — B, where P is the node between the 3rd and 4th capacitor (counting from A), and P is connected to earth/ground.) (A) 16V, 0 (B) 12V, -12V (C) 12V, -4V (D) 8V, -8V

   ›Reveal solutionSolution

   Four identical series capacitors share the 16 V equally (4 V each); earthing the node after 3 of them (P) fixes that node at 0 V, giving $V_A=12\text{ V}$ and $V_B=-4\text{ V}$.

   Concept and Intuition

   In a chain of capacitors in series (with nothing else branching off them), the same charge $Q$ must appear on every capacitor's plates — charge cannot pile up differently at an interior node with capacitors only. Since $V=Q/C$ and all four capacitors have equal $C$, they all have equal voltage across them. Earthing an interior point of a pure capacitor chain does not add any new current path in steady state (a single ground wire is a dead end), so it only shifts the reference — it fixes that node's potential at 0 V without changing how the 16 V splits among the capacitors.

   Step-by-Step Solution

   1. Four identical capacitors in series share the battery's 16 V equally: each drops $16/4 = 4$ V.
   2. Label the chain $A - C_1 - C_2 - C_3 - P - C_4 - B$. From A to P there are 3 capacitors, so before earthing, $V_A - V_P = 3\times4 = 12$ V.
   3. Earthing P sets $V_P = 0$. Hence $V_A = 12$ V. …

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2. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.In the arrangement shown (diagram: capacitor $C_1 = 2\mu F$ is connected in series with a parallel combination of capacitor $C_2 = 2\mu F$ and capacitor $C_3 = 4\mu F$; this series combination is connected across a $4\ V$ battery, forming a single closed loop), the charge on the capacitor $C_1$ is (A) $6\mu c$ (B) $4\mu c$ (C) $8\mu c$ (D) $2\mu c$

   ›Reveal solutionSolution

   This tests reducing a series-parallel capacitor network and using the fact that series elements carry the same charge. Answer: $6\ \mu C$.

   Concept and Intuition

   $C_1$ is in series with the parallel combination of $C_2$ and $C_3$. In a purely series path, the charge supplied by the battery is the same at every point along that path — so the charge on $C_1$ equals the total charge delivered by the battery to the whole network (which then splits between $C_2$ and $C_3$ according to their own values).

   Step-by-Step Solution

   1. Parallel combination: $C_{23} = C_2 + C_3 = 2 + 4 = 6\ \mu F$.
   2. This is in series with $C_1 = 2\ \mu F$: $\dfrac{1}{C_{eq}} = \dfrac{1}{2}+\dfrac{1}{6} = \dfrac{3+1}{6}=\dfrac{4}{6}$, so $C_{eq} = \dfrac{6}{4}=1.5\ \mu F$.
   3. Total charge supplied by the 4 V battery: $Q = C_{eq}V = 1.5 \times 4 = 6\ \mu C$. …

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3. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.When three parallel plate capacitors A, B and C are connected in series, the effective capacitance is 10 $\mu$F. If the capacitor C of capacitance 30 $\mu$F is removed from the combination, the effective capacitance becomes 15 $\mu$F. If the capacitances of A and B are in the ratio 1 : 3, then the energy stored when the three capacitors A, B and C are connected in parallel to a dc supply of 100 V is (A) 550 mJ (B) 275 mJ (C) 650 mJ (D) 325 mJ

   ›Reveal solutionSolution

   This tests reconstructing individual capacitances from series-combination data, then computing stored energy for the same capacitors reconnected in parallel.

   Concept and Intuition

   For capacitors in series, reciprocal capacitances add: $\dfrac{1}{C_{series}} = \sum \dfrac1{C_i}$. We're given two series measurements (all three, and then A+B after removing C) which, together with the ratio of A to B, is exactly enough information to solve for all three individual capacitances. Once known, reconnecting them in parallel is simple: parallel capacitances just add, $C_{parallel}=C_A+C_B+C_C$, and the energy stored when connected to a voltage $V$ is $U=\tfrac12 C_{parallel}V^2$.

   Step-by-Step Solution

   1. Let $C_A = x$ and $C_B = 3x$ (since $C_A:C_B = 1:3$), and $C_C = 30\ \mu\text{F}$ (given).
   2. With A, B, C in series (all three), effective capacitance is $10\ \mu\text{F}$:

   $\frac{1}{10} = \frac1x+\frac1{3x}+\frac1{30}.$

   3. With only A, B in series (C removed), effective capacitance is $15\ \mu\text{F}$:

   $\frac{1}{15} = \frac1x + \frac1{3x} = \frac{3+1}{3x} = \frac{4}{3x}.$

   Solve: $3x = 4\times15 = 60 \Rightarrow x = 20$.

   4. So $C_A = 20\ \mu\text{F}$, $C_B = 60\ \mu\text{F}$.

   5. Verify with the first (three-capacitor) equation: $\dfrac1{20}+\dfrac1{60}+\dfrac1{30} = \dfrac{3}{60}+\dfrac{1}{60}+\dfrac{2}{60}=\dfrac{6}{60}=\dfrac1{10}$ ✓ — matches the given $10\ \mu\text{F}$, confirming the values.

   6. Now connect A, B, C in parallel to a $100$ V DC supply: …

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4. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.A capacitor of 10 µF charged upto 200 V is connected in parallel with another capacitor of 20 µF charged upto 50 V. The common potential is (A) 400 V (B) 300 V (C) 200 V (D) 100 V

   ›Reveal solutionSolution

   When two charged capacitors are connected in parallel, total charge is conserved and redistributes to give a single common potential $V=\dfrac{C_1V_1+C_2V_2}{C_1+C_2}=100$ V.

   Concept and Intuition

   Connecting two charged capacitors in parallel lets charge flow between them until both plates reach the same potential (they're now electrically one node). No charge is lost in this process — it's simply redistributed — so the total charge before equals the total charge after. Since $Q=CV$ for each capacitor, this conservation law directly gives the common final potential as a charge-weighted average.

   Step-by-Step Solution

   1. Initial charge on capacitor 1: $Q_1=C_1V_1=10\ \mu F\times200\ V=2000\ \mu C$.
   2. Initial charge on capacitor 2: $Q_2=C_2V_2=20\ \mu F\times50\ V=1000\ \mu C$.
   3. Total charge (conserved): $Q_{total}=2000+1000=3000\ \mu C$. …

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5. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.Three capacitors each of capacitance 10 μF are to be connected in such a way that the effective capacitance becomes 15 μF. This can be done by connecting (A) All of them in series (B) All of them are in parallel (C) Two on series and the 3rd parallel to the combination (D) Two on parallel and the 3rd series to the combination

   ›Reveal solutionSolution

   This tests capacitor series/parallel combination arithmetic to hit a target effective capacitance of 15 μF from three identical 10 μF capacitors.

   Concept and Intuition

   Capacitors in series combine like resistors in parallel (reciprocal addition), and capacitors in parallel simply add. To build 15 μF — a value between 10 μF (one capacitor) and 30 μF (all parallel) — we need a mixed series-parallel arrangement.

   Step-by-Step Solution

   1. All three in series: $C=\dfrac{10}{3}\approx3.33\,\mu F$ — too small.
   2. All three in parallel: $C=30\,\mu F$ — too large.
   3. Two in series (giving $\dfrac{10\times10}{20}=5\,\mu F$) then this combination in parallel with the third 10 μF: $C=5+10=15\,\mu F$ — matches! …

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6. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
   Q.The surface charge density of an isolated sphere A of radius 2 cm having a charge of +10 μC is twice the surface charge density of another sphere B of radius 3 cm. If the two spheres are joined and then separated, the charges on the sphere A and B after separation are respectively (A) 12.75 μC, 8.5 μC (B) 1.5 μC, 8.5 μC (C) 8.5 μC, 12.75 μC (D) 8.5 μC, 1.5 μC

   ›Reveal solutionSolution

   Using $\sigma_A=2\sigma_B$ to find $Q_B$, then redistributing the total charge so both spheres reach a common potential (proportional to their radii), gives final charges 8.5 μC on A and 12.75 μC on B. Answer: (C).

   Concept and Intuition

   Surface charge density is charge spread over the sphere's surface area, $\sigma=Q/(4\pi r^2)$, so two spheres can hold very different charges yet have a specified density ratio depending on their radii. When two isolated conducting spheres are connected by a wire (however briefly) and then separated, the wire forces them to a common electric potential while the total charge is conserved. Since the potential of an isolated sphere is $V=kQ/r$, equal potentials mean each sphere's final charge is proportional to its own radius — bigger sphere ends up holding proportionally more charge.

   Step-by-Step Solution

   1. $\sigma = \dfrac{Q}{4\pi r^2}$. Given $\sigma_A = 2\sigma_B$: $\dfrac{Q_A}{r_A^2} = 2\dfrac{Q_B}{r_B^2}$.
   2. With $Q_A=10\ \mu C$, $r_A=2$ cm, $r_B=3$ cm: $\dfrac{10}{4} = 2\dfrac{Q_B}{9} \Rightarrow Q_B = \dfrac{10\times9}{8} = 11.25\ \mu C$.
   3. Total charge (conserved through joining and separating): $Q_{total} = 10+11.25 = 21.25\ \mu C$.
   4. After joining, common potential: $\dfrac{kQ_A'}{r_A}=\dfrac{kQ_B'}{r_B} \Rightarrow \dfrac{Q_A'}{2}=\dfrac{Q_B'}{3} \Rightarrow Q_A'=\dfrac{2}{3}Q_B'$. …

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7. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
   Q.If five parallel plate capacitors are connected to a battery of 240 V as shown in the figure, the charge on $1\mu F$ capacitor is [FIGURE] (a capacitor network: a $2\mu F$ capacitor forms a small loop with a $3\mu F$ capacitor below it on the left side; a $1\mu F$ capacitor is below the $3\mu F$ one; a $4\mu F$ capacitor sits to the right, in line with the $3\mu F$ capacitor, connecting to the right rail; a $12\mu F$ capacitor and the $240 \mathrm{~V}$ battery are in series along the bottom wire, completing the circuit loop) (A) 480 μC (B) 80 μC (C) 160 μC (D) 240 μC

   ›Reveal solutionSolution

   Reduce the left trio ($2+3+1=6\ \mu\text{F}$ in parallel) in series with $4\ \mu\text{F}$ to $2.4\ \mu\text{F}$; this is in series with $12\ \mu\text{F}$ across $240\text{ V}$; the series charge is $480\ \mu\text{C}$, giving $80\text{ V}$ across the $6\ \mu\text{F}$ group, so $Q_{1\mu F}=1\times80=80\ \mu\text{C}$ -> option (B).

   Step 1 - Left parallel group. The $2,\ 3,\ 1\ \mu\text{F}$ capacitors share the same two nodes:

   $C_{\text{left}}=2+3+1=6\ \mu\text{F}.$

   Step 2 - Series with $4\ \mu\text{F}$.

   $\frac{1}{C'}=\frac{1}{6}+\frac{1}{4}=\frac{5}{12}\ \Rightarrow\ C'=2.4\ \mu\text{F}.$

   Step 3 - Series with $12\ \mu\text{F}$ across the battery.

   $C_{eq}=\frac{2.4\times12}{2.4+12}=\frac{28.8}{14.4}=2\ \mu\text{F}.$

   The charge delivered through this series chain is

   $Q=C_{eq}V=2\ \mu\text{F}\times240\text{ V}=480\ \mu\text{C},$

   and the same $480\ \mu\text{C}$ sits on the $2.4\ \mu\text{F}$ branch. …

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8. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.In the given circuit, the potential difference across the plates of the capacitor C in steady state is [FIGURE] (a rectangular circuit: the top-left branch has capacitor C (3 μF); the top-right branch (in parallel, joining the same two top nodes) has a 3 Ω resistor; a middle vertical branch of 1 Ω connects the midpoint between C and the 3 Ω resistor down to the midpoint of the bottom rail; the bottom-left branch has a 6 Ω resistor; the bottom-right branch has a 4 Ω resistor; below the bottom rail is a 9 V battery in series with a 1 Ω resistor completing the loop back to the left side) (A) 6.5 V (B) 6 V (C) 9 V (D) 7.5 V

   ›Reveal solutionSolution

   In steady state a capacitor carries no current, so it can be treated as an open branch; solving the remaining bridge network of resistors gives the potential difference across it as $6.5$ V.

   Concept and Intuition

   Once a capacitor is fully charged in a DC circuit, no more charge flows onto it, so its branch current is exactly zero — it behaves like an open circuit for the purpose of finding currents everywhere else. The voltage that then appears across it is simply the potential difference between its two connection nodes, computed from the rest of the resistor network exactly as if the capacitor weren't there at all.

   Step-by-Step Solution

   1. Label the capacitor's top node as $T$ and bottom node as $B$. Since no current flows in the $T$–$B$ (capacitor) branch, this branch can be deleted for current analysis; the capacitor's voltage will simply be $V_T-V_B$ from the rest of the network.
   2. The remaining network is a 4-node bridge: $T$ connects to the right node $R$ via 3 Ω and to a middle node $Q$ via 1 Ω; $Q$ connects to $B$ via 6 Ω and to $R$ via 4 Ω; and $B$ connects to $R$ through the 9 V battery in series with its 1 Ω internal resistance. …

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9. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.A capacitor of capacitance $2\ \mu\mathrm{F}$ is charged with the help of a 60 V battery. After disconnecting the battery, if this capacitor is connected in parallel with another uncharged capacitor of capacitance $1\ \mu\mathrm{F}$, then the potential difference across the plates of $2\ \mu\mathrm{F}$ capacitor is (A) 30 V (B) 60 V (C) 40 V (D) 20 V

   ›Reveal solutionSolution

   Charge is conserved when isolated capacitors are connected in parallel; redistributing the original $120\,\mu\mathrm C$ over the combined $3\,\mu\mathrm F$ gives a common voltage of $40\ \mathrm V$.

   Concept and Intuition

   Once a charged capacitor is disconnected from its battery, its charge is fixed (isolated system). Connecting it to another (initially uncharged) capacitor in parallel does not create or destroy charge — it only lets charge redistribute between the two capacitors until they reach a common potential difference (since parallel connection forces equal voltage across both). Total charge before = total charge after.

   Step-by-Step Solution

   1. Initial charge on the $2\,\mu\mathrm F$ capacitor: $Q = CV = 2\,\mu\mathrm F \times 60\,\mathrm V = 120\,\mu\mathrm C$.
   2. The $1\,\mu\mathrm F$ capacitor starts uncharged, so total charge in the system is still $120\,\mu\mathrm C$.
   3. Combined capacitance in parallel: $C_{\text{total}} = 2 + 1 = 3\,\mu\mathrm F$. …

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10. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
    Q.Four capacitors are connected as shown in the figure. If $C_1, C_2, C_3$ and $C_4$ are in the ratio of 1 : 2 : 3 : 4, then the ratio of the charges on the capacitors $C_2$ and $C_4$ is [FIGURE] (a network of four capacitors: $C_3$ (left outer branch) and $C_1$ (right outer branch) are the two outer vertical branches connecting a top node to a bottom node; $C_2$ and $C_4$ are both connected in parallel with each other between the same top and bottom nodes, between the $C_3$ and $C_1$ branches; a battery V is connected across the bottom of the circuit) (A) 1 : 4 (B) 2 : 3 (C) 6 : 11 (D) 3 : 22

    ›Reveal solutionSolution

    $Q_2:Q_4=3:22$ — option (D).

    Take $C_1:C_2:C_3:C_4=1:2:3:4$ (in arbitrary units). In the network $C_1,C_2,C_3$ form a series branch across the source, while $C_4$ is connected directly across the source voltage $V$.

    Series combination $C_1,C_2,C_3$ carries a single charge — this is $Q_2$:

    $\frac{1}{C_s}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}=\frac{11}{6}\ \Rightarrow\ C_s=\frac{6}{11},$ …

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11. AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
    Q.The resultant capacitance of parallel combination of two capacitors C1 and C2 is 20 $\mu\mathrm{F}$. When these capacitors are individually connected to a voltage source of 1V, then the energy stored in C2 is 9 times that of C1. If these two capacitors are connected in series, the resultant capacitance value (A) $1.4\ \mu\mathrm{F}$ (B) $8\ \mu\mathrm{F}$ (C) $18\ \mu\mathrm{F}$ (D) $1.8\ \mu\mathrm{F}$

    ›Reveal solutionSolution

    Using the parallel sum and the energy ratio to find $C_1,C_2$ individually, the series combination works out to $1.8\ \mu$F.

    Concept and Intuition

    At a fixed voltage, the energy stored in a capacitor is directly proportional to its capacitance ($U=\frac12CV^2$), so an energy ratio directly gives a capacitance ratio. Combined with the known parallel sum, both individual capacitances can be found, and then the series formula applied.

    Step-by-Step Solution

    1. Parallel combination: $C_1+C_2=20\ \mu F$.
    2. At $V=1$ V: $U_1=\frac12C_1(1)^2=\frac12C_1$; $U_2=\frac12C_2$.
    3. Given $U_2=9U_1 \Rightarrow C_2=9C_1$.
    4. Substituting: $C_1+9C_1=20 \Rightarrow C_1=2\ \mu F,\ C_2=18\ \mu F$. …

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12. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
    Q.In the given circuit, the potential difference across 5 μF capacitor is [FIGURE] (a circuit with three parallel branches each containing a capacitor - 4 μF, 8 μF, and 4 μF respectively - all connected across a fourth branch containing a 63 V battery in series with a 5 μF capacitor) (A) 48 V (B) 24 V (C) 63 V (D) 21 V

    ›Reveal solutionSolution

    The three capacitors on the top rungs are in parallel, so their equivalent capacitance is $4+8+4 = 16\ \mu\text{F}$. This parallel combination is in series with the $5\ \mu\text{F}$ capacitor and the 63 V battery. Using the voltage divider rule for series capacitors, the voltage across the $5\ \mu\text{F}$ capacitor is $V_5 = \frac{16}{5+16}\times 63 = 48\ \text{V}$. The correct option is (A).

    ---

    Concept and Intuition

    When you see a circuit with capacitors and a battery, the first question is always: How are the capacitors connected?

    Here, the top three capacitors (4, 8, and 4 μF) each connect directly between the same two nodes (the left and right rails). That means they are in parallel — they share the same voltage. The bottom branch has a battery in series with a 5 μF capacitor, and this branch also connects between the same two rails. So the whole circuit is: a parallel group (16 μF) in series with a 5 μF capacitor and a battery.

    The key insight: In a series combination of capacitors, the charge on each is the same, and the voltage divides inversely with capacitance. So we can treat the parallel group as a single capacitor, then use the series voltage divider formula.

    ---

    Step-by-step solution

    1. Identify the parallel group The three capacitors on the top rungs (4 μF, 8 μF, 4 μF) all have one plate connected to the left rail and the other to the right rail. They are in parallel. Equivalent capacitance:

    $C_{\text{parallel}} = 4 + 8 + 4 = 16\ \mu\text{F}$

    2. Redraw the circuit

       Replace the three parallel capacitors with a single 16 μF capacitor. Now the circuit is a simple series loop:

       Battery (63 V) → 5 μF capacitor → 16 μF capacitor → back to battery.

       Both capacitors are in series with each other and with the battery.

    3. Apply the series capacitor voltage divider

       For two capacitors $C_1$ and $C_2$ in series across a total voltage $V_{\text{total}}$, the voltage across $C_1$ is:

    $V_1 = \frac{C_2}{C_1 + C_2} \times V_{\text{total}}$

    This comes from the fact that the charge $Q$ is the same on both: $Q = C_1 V_1 = C_2 V_2$, and $V_1 + V_2 = V_{\text{total}}$.

    Here we want the voltage across the 5 μF capacitor. Let $C_1 = 5\ \mu\text{F}$ and $C_2 = 16\ \mu\text{F}$. Then:

    $$$$

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