Q.A network of four $10\ \mu\text{F}$ capacitors is connected to a $500\ \text{V}$ supply, as shown in Fig. 2.29.

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1. COMEDK 2026Set 2026-M1 markMCQ
   Q.What is the charge on $15 \mu F$ in the circuit given? (A) $260 \mu \mathrm{C}$ (B) $230 \mu \mathrm{C}$ (C) $160 \mu \mathrm{C}$ (D) $130 \mu \mathrm{C}$

   ›Reveal solutionSolution

   The key is to find the voltage across the 15 µF capacitor by first computing the equivalent capacitance of the series branch, then using charge conservation and the fact that parallel branches share the same voltage. The charge on the 15 µF capacitor is 130 µC, which corresponds to option (D).

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   Concept & Intuition

   When capacitors are in series, they each store the same charge (because the charge on one plate comes from the adjacent plate of the next capacitor). When branches are in parallel, they share the same voltage across their endpoints. Here, the 15 µF and 30 µF are in series, forming one branch; that branch is in parallel with the 10 µF capacitor. The battery supplies 13 V across both parallel branches. So we first find the equivalent capacitance of the series pair, then the charge on that equivalent capacitor — which is exactly the charge on each of the series capacitors. That gives us the charge on the 15 µF directly.

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   Step-by-step solution

   1. Identify the series combination The upper branch has a 15 µF and a 30 µF capacitor in series. For capacitors in series, the reciprocal of the equivalent capacitance is the sum of the reciprocals:

   $\frac{1}{C_{\text{eq,series}}} = \frac{1}{15} + \frac{1}{30} = \frac{2}{30} + \frac{1}{30} = \frac{3}{30} = \frac{1}{10}$

   So $C_{\text{eq,series}} = 10\ \mu\text{F}$.

   2. Interpret the parallel structure

      This 10 µF equivalent capacitor (representing the 15 µF + 30 µF series branch) is in parallel with the 10 µF capacitor in the middle branch. Both are connected directly across the 13 V battery. Therefore, the voltage across the series branch is exactly 13 V.

   3. Find the charge on the series branch

      For any capacitor (or equivalent capacitor), $Q = C V$. The charge on the 10 µF equivalent capacitor is:

   $Q_{\text{series branch}} = (10\ \mu\text{F}) \times (13\ \text{V}) = 130\ \mu\text{C}.$

   4. Apply the series property …

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2. KCET 2026Set C21 markMCQ
   Q.In the circuit shown in the figure , the potential difference across the $4\mu F$ capacitor is (A) 3 V (B) 4 V (C) 9 V (D) 12 V

   ›Reveal solutionSolution

   In a series combination, every capacitor carries the same charge $Q$; use $Q = C_{eq}V$ for the whole combination, then $V = Q/C$ for the capacitor in question.

   Step 1 — Reduce the parallel combination

   The $9\mu F$ and $3\mu F$ capacitors are in parallel:

   $C_{parallel} = 9\mu F + 3\mu F = 12\mu F$

   Step 2 — Find the equivalent capacitance of the series combination

   This $12\mu F$ is in series with the $4\mu F$ capacitor:

   $\frac{1}{C_{eq}} = \frac{1}{4} + \frac{1}{12} = \frac{3+1}{12} = \frac{4}{12} \implies C_{eq} = 3\mu F$

   Step 3 — Find the total charge supplied by the source

   $Q = C_{eq}V = 3\mu F \times 12\,V = 36\mu C$ …

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3. COMEDK 2025Set 2025-A1 markMCQ
   Q.By connecting two given capacitors, a technician was able to make two new capacitors having the effective capacitance $12.5 \mu \mathbf{F}$ and $2 \mu \mathbf{F}$. What would be the capacitance of the given capacitors? (A) $8.5 \mu \mathrm{~F}$ and $4 \mu \mathrm{~F}$ (B) $10 \mu \mathrm{~F}$ and $2.5 \mu \mathrm{~F}$ (C) $6.5 \mu \mathrm{~F}$ and $6 \mu \mathrm{~F}$ (D) $10.5 \mu \mathrm{~F}$ and $2 \mu \mathrm{~F}$

   ›Reveal solutionSolution

   The problem involves two unknown capacitors that, when connected in series and in parallel, yield two specific effective capacitances. Solving the system of equations gives the individual capacitances as $10\,\mu\text{F}$ and $2.5\,\mu\text{F}$, matching option (B).

   The key idea is that connecting two capacitors in series and in parallel produces two distinct effective capacitances. The larger effective capacitance comes from the parallel combination (sum of the two), and the smaller from the series combination (reciprocal sum). By setting up and solving these two equations, we find the individual values.

   1. Set up the equations.

      Let the two unknown capacitances be $C_1$ and $C_2$ (in $\mu\text{F}$).

      • In parallel: $C_{\text{parallel}} = C_1 + C_2 = 12.5$
      • In series: $C_{\text{series}} = \frac{C_1 C_2}{C_1 + C_2} = 2$

   2. Substitute the parallel sum into the series equation.

      From the parallel equation, $C_1 + C_2 = 12.5$.

      Plug into the series equation:

   $\frac{C_1 C_2}{12.5} = 2 \quad \Rightarrow \quad C_1 C_2 = 25$

   3. Solve the system. We now have:

   $C_1 + C_2 = 12.5, \quad C_1 C_2 = 25$

   These are the sum and product of the roots of the quadratic $x^2 - (C_1+C_2)x + C_1 C_2 = 0$, i.e.,

   $x^2 - 12.5x + 25 = 0$

   Solve using the quadratic formula:

   $x = \frac{12.5 \pm \sqrt{12.5^2 - 4 \cdot 25}}{2} = \frac{12.5 \pm \sqrt{156.25 - 100}}{2} = \frac{12.5 \pm \sqrt{56.25}}{2}$

   Since $\sqrt{56.25} = 7.5$, we get:

   $x = \frac{12.5 \pm 7.5}{2}$

   So the two values are:

   $x_1 = \frac{20}{2} = 10, \quad x_2 = \frac{5}{2} = 2.5$ …

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4. COMEDK 2025Set 2025-E1 markMCQ
   Q.A network of capacitors is as shown below. If the voltage supply is 100 V , find the energy stored in the $6 \mu \mathrm{~F}$ capacitor. $C_1=3 \mu F, C_2=6 \mu F, C_3=3 \mu F$ and $C_4=4 \mu F$ (A) 1.2 mJ (B) 12 mJ (C) 2.2 mJ (D) 4.2 mJ

   ›Reveal solutionSolution

   $C_3,\,C_2,\,C_1$ form a series string ($=1.2\ \mu\text{F}$) that is in parallel with $C_4$, so the full $100\ \text{V}$ appears across the string. The common charge is $120\ \mu\text{C}$, giving $V_{C_2}=20\ \text{V}$ and stored energy $\tfrac12 C_2V_{C_2}^2=1.2\ \text{mJ}$ — option (A).

   Concept

   For capacitors in series the charge is common and voltages add; for capacitors in parallel the voltage is common and charges add. Energy stored in a capacitor is $U=\tfrac12 CV^2$, using the voltage across that particular capacitor.

   Step-by-step solution

   1. Series string. $C_3=3\,\mu\text{F}$, $C_2=6\,\mu\text{F}$, $C_1=3\,\mu\text{F}$ are in series:

   $\frac{1}{C_s}=\frac{1}{3}+\frac{1}{6}+\frac{1}{3}=\frac{5}{6}\ \Rightarrow\ C_s=1.2\ \mu\text{F}.$

   2. Voltage across the string. This string is connected in parallel with $C_4$ directly across the supply, so it carries the full supply voltage:

   $V_s=100\ \text{V}.$

   3. Common charge. In the series string, …

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5. COMEDK 2025Set 2025-E1 markMCQ
   Q.A capacitor of capacitance $4 \mu \mathrm{~F}$ is charged to a potential of 24 V and then connected in parallel to an uncharged capacitor of capacitance $6 \mu \mathrm{~F}$. The final potential difference across each capacitor will be: (A) 6.9 V (B) 8.2 V (C) 9.6 V (D) 7.4 V

   ›Reveal solutionSolution

   When a charged capacitor is connected in parallel to an uncharged one, charge redistributes until both share the same voltage. The final voltage is found by conserving total charge and using the equivalent parallel capacitance. The answer is 9.6 V.

   Concept & Intuition

   The key idea is charge conservation combined with the fact that in a parallel connection, both capacitors end up at the same potential difference. The total charge initially stored on the charged capacitor cannot go anywhere — it simply spreads across the two capacitors. Since the total capacitance in parallel is the sum of the individual capacitances, the final common voltage is just the initial charge divided by the total capacitance.

   A common mistake is to think the voltage halves or follows some other simple ratio — but the correct ratio depends on the capacitances, not just the number of capacitors.

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   1. Find the initial charge on the charged capacitor. The capacitor of $4 \, \mu\text{F}$ is charged to $24 \, \text{V}$.

   $Q_{\text{initial}} = C_1 V = (4 \times 10^{-6}) \times 24 = 96 \times 10^{-6} \, \text{C} = 96 \, \mu\text{C}$

   2. Understand what happens when they are connected in parallel.

      The uncharged $6 \, \mu\text{F}$ capacitor initially has zero charge. When connected in parallel, charge flows from the charged capacitor to the uncharged one until the voltage across both is equal. The total charge in the system remains $96 \, \mu\text{C}$.

   3. Find the equivalent capacitance of the parallel combination.

      For capacitors in parallel, the total capacitance is the sum:

   $C_{\text{eq}} = C_1 + C_2 = 4 + 6 = 10 \, \mu\text{F}$

   4. Calculate the final common voltage. …

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6. COMEDK 2025Set 2025-M1 markMCQ
   Q.Two capacitors $\mathrm{C}_1$ and $\mathrm{C}_2$ are charged to 100 V and 120 V respectively. It is found that upon connecting them together in parallel, the potential on each one of them is zero. Therefore (A) $\mathrm{C}_1+3 \mathrm{C}_2=0$ (B) $5 \mathrm{C}_1=3 \mathrm{C}_2$ (C) $5 \mathrm{C}_1+6 \mathrm{C}_2=0$ (D) $5 \mathrm{C}_1=6 \mathrm{C}_2$

   ›Reveal solutionSolution

   When two charged capacitors are connected in parallel and the final voltage is zero, the net charge on the combination must be zero. This gives the relation $5C_1 = 6C_2$, so option (D) is correct.

   Concept & Intuition

   The key idea is conservation of charge. When capacitors are connected in parallel, the total charge before connection equals the total charge after connection. If the final voltage across both is zero, then the net charge on the combination must be zero. That means the positive charge on one capacitor exactly cancels the positive charge on the other — but careful: the sign of the charge depends on how they are connected. The problem implies they are connected so that the plates of opposite polarity are joined, leading to cancellation.

   Step-by-step reasoning

   1. Initial charges Capacitor $C_1$ is charged to $100\,\text{V}$, so its charge is

   $Q_1 = C_1 \times 100 = 100C_1.$

   Capacitor $C_2$ is charged to $120\,\text{V}$, so its charge is

   $Q_2 = C_2 \times 120 = 120C_2.$

   2. Connection in parallel

      When connected in parallel, the positive plate of one is joined to the negative plate of the other (this is the only way the final voltage can become zero — otherwise you'd just get a common nonzero voltage). So the charges are effectively opposite in sign relative to the connection.

   3. Net charge after connection

      The total charge on the combined system is

   $Q_{\text{net}} = Q_1 - Q_2 \quad \text{(if we take the polarity such that they oppose)}.$

   After connection, the final voltage is zero, meaning no net charge remains on the combination (since $Q = C_{\text{eq}} V$ and $V=0$ gives $Q=0$). Therefore

   $Q_1 - Q_2 = 0 \quad \Rightarrow \quad Q_1 = Q_2.$ …

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7. COMEDK 2024Set 2024-A1 markMCQ
   Q.The potential difference between the points $\mathrm{P}$ and $\mathrm{Q}$ of the arrangement shown in figure is (A) $-9.3$ V (B) 42 V (C) $-42$ V (D) $-3.9$ V

   ›Reveal solutionSolution

   With the two capacitors in series and a net emf of $14$ V, the common charge is $\approx37.3\,\mu C$; the resulting potential difference between P and Q is $\approx-9.3$ V.

   In the steady state no current flows (capacitors block DC), and the $4\,\mu F$ and $8\,\mu F$ capacitors are in series around the loop.

   Series capacitance:

   $C_{eq}=\frac{4\times8}{4+8}=\frac{32}{12}=\frac{8}{3}\ \mu F.$

   Net driving emf (the two cells oppose): $\varepsilon=28-14=14$ V.

   Common charge on each capacitor:

   $q=C_{eq}\,\varepsilon=\frac{8}{3}\times14=37.3\ \mu C.$ …

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8. COMEDK 2024Set 2024-E1 markMCQ
   Q.Figure below shows a network of resistors, cells, and a capacitor at steady state. What is the current through the resistance 4 $\Omega$ ? (A) 1.0 A (B) 0.2 A (C) Zero (D) 0.5 A

   ›Reveal solutionSolution

   At steady state a capacitor is fully charged and carries no current, so its branch behaves as an open circuit and drops out of the current analysis. The remaining resistor–cell network then carries a single steady current, and applying Kirchhoff's voltage law gives the current through the 4 Ω resistor as 0.2 A. The correct option is (B).

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   Concept & Intuition

   The decisive idea is that a capacitor in DC steady state acts as an open circuit: once it is fully charged, no more charge flows onto its plates, so no current passes through the branch that contains it. That branch can be removed for the purpose of finding currents; the capacitor's only remaining role is to hold a fixed steady voltage across itself. With the capacitor branch open, the rest of the network reduces to a simple current-carrying loop, and Kirchhoff's voltage law (KVL) determines the current.

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   Step-by-step reasoning

   1. Open the capacitor branch. At steady state the capacitor is fully charged, so the branch containing it carries zero current. Remove it from the current analysis; it no longer provides a path for charge to flow.

   2. Reduce the network. With that branch open, the remaining cells and resistors form a single conducting loop. The current is the same through every element in series along this loop, including the 4 Ω resistor.

   3. Apply Kirchhoff's voltage law. Going once around the surviving loop, the sum of the EMFs equals the sum of the potential drops across the resistors:

   $\sum \mathcal{E} = I \sum R$

   The net driving EMF of the loop, divided by its total series resistance, gives

   $$$$

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9. COMEDK 2024Set 2024-E1 markMCQ
   Q.The figure shows a network of five capacitors connected to a 20 V battery. Calculate the charge acquired by each 10 $\mu$F capacitor. (A) $2 \times 10^{-4} \mathrm{C}$ (B) $4 \times 10^{-4} \mathrm{C}$ (C) $6 \times 10^{-4} \mathrm{C}$ (D) $1 \times 10^{-4} \mathrm{C}$

   ›Reveal solutionSolution

   The two 10 µF capacitors share node T–Q, across which 10 V appears (equal series split of the 20 V), giving each $Q=10\,\mu\mathrm{F}\times10\,\mathrm{V}=1\times10^{-4}$ C.

   Label the top wire node T, left mid-node P, right mid-node Q; the 20 V battery is across P–Q.

   • Between T and P: 15 µF (left branch) parallel with 5 µF (diagonal) $= 20\ \mu\mathrm{F}$.
   • Between T and Q: 10 µF (right branch) parallel with 10 µF (diagonal) $= 20\ \mu\mathrm{F}$.

   The path P→T→Q is these two 20 µF groups in series. The 20 µF direct branch (P–Q) simply sits across the battery and does not change the voltage split of the P–T–Q path. …

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10. COMEDK 2024Set 2024-M1 markMCQ
    Q.A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage (U) as $\epsilon=2 U$. A similar capacitor with no dielectric is charged to $U_0=78 \mathrm{~V}$. It is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors. (A) 6V (B) 8V (C) 2V (D) 4V

    ›Reveal solutionSolution

    Charge conservation with $\epsilon = 2U$ gives $2U^2+U-78=0\Rightarrow U=6$ V.

    Let $C_0$ be the base (vacuum) capacitance. The first (plain) capacitor is charged to $U_0=78$ V, storing $Q_0 = C_0 U_0$.

    After connection the two capacitors are in parallel and share the final voltage $U$:

    • Plain capacitor: $Q_1 = C_0 U$
    • Dielectric capacitor: capacitance $=\epsilon C_0 = 2U C_0$, so $Q_2 = (2U C_0)\,U = 2C_0 U^2$

    Charge is conserved: …

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11. KCET 2023Set A-31 markMCQ
    Q.Five capacitors each of value $1\,\mu$F are connected as shown in the figure. The equivalent capacitance between A and B is (A) $1\,\mu$F (B) $2\,\mu$F (C) $5\,\mu$F (D) $3\,\mu$F

    ›Reveal solutionSolution

    A symmetry argument kills the middle capacitor (its two plates are at equal potential), leaving two identical series branches in parallel between $A$ and $B$.

    1. Label the four nodes

    Call the top rail $T$ and the bottom rail $U$. The five capacitors (all $1\,\mu$F) connect:

    $A\!-\!T,\quad A\!-\!U,\quad B\!-\!T,\quad B\!-\!U,\quad T\!-\!U$

    ($A$ is the mid-point of the left branch, $B$ the mid-point of the right branch.) This is a bridge network, and the middle capacitor $T\!-\!U$ is the bridge element.

    2. The symmetry (the balanced-bridge condition)

    Suppose we apply a potential difference between $A$ and $B$. Now interchange the labels $T \leftrightarrow U$: the capacitor list maps to itself ($A\!-\!T \leftrightarrow A\!-\!U$, $B\!-\!T \leftrightarrow B\!-\!U$, and $T\!-\!U$ to itself), and the terminals $A$, $B$ are untouched. Because every capacitor is the same $1\,\mu$F, the circuit is identical under this swap, so $T$ and $U$ must sit at the same potential:

    $V_T = V_U = \frac{V_A + V_B}{2}$

    Equivalently, the bridge is balanced: $\dfrac{C_{AT}}{C_{TB}} = \dfrac{C_{AU}}{C_{UB}} \;(=1)$.

    3. Remove the bridge capacitor …

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12. COMEDK 2023Set 2023-M1 markMCQ
    Q.A capacitor of capacity $2 ~\mu \mathrm{F}$ is charged upto a potential $14 \mathrm{~V}$ and then connected in parallel to an uncharged capacitor of capacity $5 ~\mu \mathrm{F}$. The final potential difference across each capacitor will be (A) 6 V (B) 4 V (C) 8 V (D) 14 V

    ›Reveal solutionSolution

    Conserving charge $28\,\mu C$ across the combined capacitance $7\,\mu F$ gives a common potential of $4$ V.

    Initial charge on the $2\,\mu F$ capacitor:

    $Q=C_1V_1=2\,\mu F\times14\,V=28\,\mu C.$ …

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