Q.Three capacitors each of capacitance $9\ \text{pF}$ are connected in series.

Showing the 12 most recent of 19 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.A, B and C are three parallel plate capacitors. If A and B are connected in series the effective capacitance is $6\,\mu\mathrm{F}$, if B and C are connected in series the effective capacitance is $4\,\mu\mathrm{F}$ and, if A and C are connected in series the effective capacitance is $3\,\mu\mathrm{F}$. If these three capacitors are connected in parallel, the effective capacitance of the combination is (A) $26\,\mu\mathrm{F}$ (B) $36.8\,\mu\mathrm{F}$ (C) $13\,\mu\mathrm{F}$ (D) $32.4\,\mu\mathrm{F}$

   ›Reveal solutionSolution

   The problem gives three series-pair capacitances, which we invert to get sums of reciprocals; solving those three linear equations yields the individual capacitances, whose sum is the parallel capacitance — 13 μF.

   We are told that for three capacitors $A, B, C$ (each with capacitance $a, b, c$ in μF), the series combinations give:

   $\frac{1}{a}+\frac{1}{b} = \frac{1}{6},\qquad \frac{1}{b}+\frac{1}{c} = \frac{1}{4},\qquad \frac{1}{a}+\frac{1}{c} = \frac{1}{3}.$

   We want $a+b+c$, the parallel capacitance.

   Why this works:

   Series capacitance is the reciprocal of the sum of reciprocals. So each given series value is the reciprocal of a sum of two reciprocals. That gives us three linear equations in the three unknowns $x = 1/a$, $y = 1/b$, $z = 1/c$. Solve for $x,y,z$, then invert to get $a,b,c$, and add.

   ---

   1. Set up the reciprocals. Let $x = \frac{1}{a},\; y = \frac{1}{b},\; z = \frac{1}{c}$. Then:

   $x + y = \frac{1}{6},\qquad y + z = \frac{1}{4},\qquad x + z = \frac{1}{3}.$

   2. Solve the linear system. Add all three equations:

   $(x+y)+(y+z)+(x+z) = \frac{1}{6}+\frac{1}{4}+\frac{1}{3}.$

   Left side: $2(x+y+z)$. Right side: common denominator 12 gives $\frac{2+3+4}{12} = \frac{9}{12} = \frac{3}{4}$.

   So:

   $2(x+y+z) = \frac{3}{4} \quad\Rightarrow\quad x+y+z = \frac{3}{8}.$

   3. Find each reciprocal individually. Subtract $x+y = \frac{1}{6}$ from $x+y+z = \frac{3}{8}$:

   $z = \frac{3}{8} - \frac{1}{6} = \frac{9}{24} - \frac{4}{24} = \frac{5}{24}.$

   Subtract $y+z = \frac{1}{4}$ from $x+y+z = \frac{3}{8}$:

   $x = \frac{3}{8} - \frac{1}{4} = \frac{3}{8} - \frac{2}{8} = \frac{1}{8}.$

   Subtract $x+z = \frac{1}{3}$ from $x+y+z = \frac{3}{8}$: …

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2. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.The ratio of the energies stored in capacitors A, B and C when connected in parallel to a dc supply is 1:2:3. When they are connected in series to the same dc supply, if the energy stored in capacitor C is 75 mJ, then the energy stored in capacitor A is (A) 150 mJ (B) 25 mJ (C) 225 mJ (D) 75 mJ

   ›Reveal solutionSolution

   The parallel-energy ratio fixes $C_A:C_B:C_C = 1:2:3$; in series every capacitor carries the same charge, so energy $\propto 1/C$, giving $E_A = 3E_C = 225\,\mathrm{mJ}$.

   Step 1 — Find the capacitances (parallel case).

   In parallel all capacitors share the same voltage $V$, and $E = \tfrac{1}{2}CV^2$, so energy $\propto C$. Since $E_A:E_B:E_C = 1:2:3$,

   $C_A : C_B : C_C = 1 : 2 : 3 \;\Rightarrow\; C_A = C,\; C_B = 2C,\; C_C = 3C.$

   Step 2 — Series case.

   In series every capacitor carries the same charge $Q$, and $E = \dfrac{Q^2}{2C}$, so energy $\propto \dfrac{1}{C}$: …

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3. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.A parallel plate capacitor with air as dielectric has a capacitance of $4\ \mu \mathrm{F}$. The space between the plates of the capacitor is completely filled with a material of dielectric constant $5$ and charged to a potential of $100\ \mathrm{V}$. The work done to completely remove the dielectric material after the capacitor is disconnected from the battery is (A) $0.1\ \mathrm{J}$ (B) $0.5\ \mathrm{J}$ (C) $0.6\ \mathrm{J}$ (D) $0.4\ \mathrm{J}$

   ›Reveal solutionSolution

   The capacitor is disconnected, so the charge $Q$ stays fixed while the dielectric is removed. The work done equals the increase in stored energy, $U_f - U_i = 0.5 - 0.1 = 0.4\ \mathrm{J}$, option (D).

   Concept & intuition

   Once disconnected from the battery, the charge on the plates is trapped, so $Q$ is constant. Removing the dielectric lowers the capacitance ($C = \kappa C_0$), which raises the voltage. The work you do pulling the slab out goes entirely into the extra stored electrostatic energy.

   1. Capacitance with dielectric

   $C_i = \kappa C_0 = 5 \times 4\ \mu\mathrm{F} = 20\ \mu\mathrm{F}.$

   2. Charge (constant after disconnection) — charged to 100 V with the dielectric in place:

   $Q = C_i V = 20\times10^{-6} \times 100 = 2\times10^{-3}\ \mathrm{C}.$

   3. Capacitance after removing the dielectric

   $C_f = C_0 = 4\ \mu\mathrm{F}.$

   4. Final voltage ($Q$ fixed):

   $V_f = \frac{Q}{C_f} = \frac{2\times10^{-3}}{4\times10^{-6}} = 500\ \mathrm{V}.$

   5. Stored energies

   $U_i = \tfrac12 C_i V^2 = \tfrac12(20\times10^{-6})(100)^2 = 0.1\ \mathrm{J},$

   $U_f = \tfrac12 C_f V_f^2 = \tfrac12(4\times10^{-6})(500)^2 = 0.5\ \mathrm{J}.$

   6. Work done = increase in energy $$$$

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4. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.A capacitor and a resistor of resistance $100\sqrt{3}\ \Omega$ are connected in series to an ac source of voltage $100\sin(200t)\ \text{V}$, where ‘t’ is time in second. If the phase difference between the voltage and the current in the circuit is $30^\circ$, then the capacitance of the capacitor is (A) $30\ \mu\text{F}$ (B) $50\ \mu\text{F}$ (C) $100\ \mu\text{F}$ (D) $150\ \mu\text{F}$

   ›Reveal solutionSolution

   In an RC series circuit, the phase angle $\phi$ satisfies $\tan\phi = \frac{1}{\omega C R}$. Given $\phi = 30^\circ$, $\omega = 200\ \text{rad/s}$, and $R = 100\sqrt{3}\ \Omega$, solving gives $C = 50\ \mu\text{F}$, so option (B) is correct.

   Concept & Intuition

   In a series RC circuit driven by an AC source, the resistor’s voltage is in phase with the current, while the capacitor’s voltage lags the current by $90^\circ$. The total voltage is the phasor sum of these two, so the current leads the voltage by some angle $\phi$ between $0^\circ$ and $90^\circ$. That phase difference is given by $\tan\phi = \frac{X_C}{R}$, where $X_C = \frac{1}{\omega C}$ is the capacitive reactance. Here we know $\phi$ and $R$, so we can solve for $C$.

   Step-by-step solution

   1. Identify given quantities

      • Resistance: $R = 100\sqrt{3}\ \Omega$
      • Source voltage: $V(t) = 100\sin(200t)$ V, so angular frequency $\omega = 200\ \text{rad/s}$
      • Phase difference: $\phi = 30^\circ$ (current leads voltage in an RC circuit)

   2. Write the phase relation for an RC series circuit

      For a series RC circuit, the impedance phasor diagram gives:

   $\tan\phi = \frac{X_C}{R}$

   where $X_C = \frac{1}{\omega C}$ is the capacitive reactance.

   (Note: Some textbooks define $\phi$ as the angle by which voltage leads current; here current leads voltage, so $\phi$ is positive and the formula still holds with $X_C$ in the numerator.)

   3. Substitute known values

   $\tan 30^\circ = \frac{1}{\omega C R}$

   Since $\tan 30^\circ = \frac{1}{\sqrt{3}}$, we have:

   $\frac{1}{\sqrt{3}} = \frac{1}{\omega C R}$

   4. Solve for $C$ Invert both sides:

   $\sqrt{3} = \omega C R$

   So:

   $C = \frac{\sqrt{3}}{\omega R}$ …

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5. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.A parallel plate capacitor with air as dielectric has a capacitance of $4\,\mu\mathrm{F}$. The space between the plates of the capacitor is completely filled with a material of dielectric constant $5$ and charged to a potential of $100\,\mathrm{V}$. The work done to completely remove the dielectric material after the capacitor is disconnected from the battery is (A) $0.4\,\mathrm{J}$ (B) $0.5\,\mathrm{J}$ (C) $0.1\,\mathrm{J}$ (D) $0.6\,\mathrm{J}$

   ›Reveal solutionSolution

   After the battery is disconnected the charge is fixed. Work done $=$ final energy $-$ initial energy $= 0.5 - 0.1 = 0.4\,\mathrm{J}$, option (A).

   Air-filled capacitance is $C_0 = 4\,\mu\mathrm{F}$. With the dielectric ($\kappa = 5$) in place,

   $C = \kappa C_0 = 5 \times 4 = 20\,\mu\mathrm{F}.$

   It is charged to $V = 100\,\mathrm{V}$, so the stored charge is

   $Q = CV = 20\times 10^{-6}\times 100 = 2\times 10^{-3}\,\mathrm{C}.$

   The battery is then disconnected, so $Q$ stays constant while the dielectric is removed. Using $U = \dfrac{Q^2}{2C}$:

   Initial energy (dielectric in, $C = 20\,\mu\mathrm{F}$):

   $U_i = \frac{(2\times 10^{-3})^2}{2\times 20\times 10^{-6}} = \frac{4\times 10^{-6}}{40\times 10^{-6}} = 0.1\,\mathrm{J}.$ …

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6. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.A capacitor and a resistor of resistance $100\sqrt{3}\ \Omega$ are connected in series to an ac source of voltage $100\sin(200t)\ V$, where ‘t’ is time in second. If the phase difference between the voltage and the current in the circuit is $30^\circ$, then the capacitance of the capacitor is (A) $50\ \mu F$ (B) $30\ \mu F$ (C) $150\ \mu F$ (D) $100\ \mu F$

   ›Reveal solutionSolution

   In an RC series circuit, the phase angle $\phi$ satisfies $\tan\phi = \frac{1}{\omega C R}$. Given $\phi = 30^\circ$, $R = 100\sqrt{3}\ \Omega$, and $\omega = 200\ \text{rad/s}$, solving gives $C = 50\ \mu\text{F}$, so the correct option is (A).

   Concept & Intuition

   When a resistor and capacitor are in series with an AC source, the voltage across the resistor is in phase with the current, but the voltage across the capacitor lags the current by $90^\circ$. The total voltage from the source is the phasor sum of these two voltages. The phase difference $\phi$ between the source voltage and the current is determined by the ratio of capacitive reactance $X_C = \frac{1}{\omega C}$ to resistance $R$:

   $\tan\phi = \frac{X_C}{R} = \frac{1}{\omega C R}.$

   Here $\phi = 30^\circ$ is given, so we can directly solve for $C$.

   Step-by-step solution

   1. Identify given quantities

      • Resistance: $R = 100\sqrt{3}\ \Omega$
      • Source voltage: $V(t) = 100\sin(200t)\ \text{V}$ This tells us the angular frequency $\omega = 200\ \text{rad/s}$.
      • Phase difference: $\phi = 30^\circ$ (voltage leads current in an RC circuit? Actually, in a series RC circuit, current leads voltage, so $\phi$ is usually taken as the angle by which voltage lags current. But the problem says "phase difference between the voltage and the current" — we take its magnitude as $30^\circ$.)

   2. Write the phase relation

      For a series RC circuit, the impedance phase angle $\phi$ satisfies:

   $\tan\phi = \frac{X_C}{R} = \frac{1}{\omega C R}.$

   Here $\phi = 30^\circ$, so $\tan 30^\circ = \frac{1}{\sqrt{3}}$.

   3. Set up the equation

   $\frac{1}{\omega C R} = \frac{1}{\sqrt{3}}.$

   Substitute $\omega = 200$ and $R = 100\sqrt{3}$: …

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7. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.Two resistors of resistances $4~\Omega$ and $8~\Omega$ are connected in parallel in the left gap of the meter bridge and two resistors of resistances $8~\Omega$ and $4~\Omega$ are connected in series in the right gap of the meter bridge. The balancing length from the left end of the bridge wire is (A) $\dfrac{100}{11}$ cm (B) $\dfrac{200}{9}$ cm (C) $\dfrac{180}{11}$ cm (D) $\dfrac{200}{11}$ cm

   ›Reveal solutionSolution

   The meter bridge balances when the ratio of the left-gap equivalent resistance to the right-gap equivalent resistance equals the ratio of the corresponding wire lengths. The left gap has $4\ \Omega$ and $8\ \Omega$ in parallel, giving $\frac{8}{3}\ \Omega$; the right gap has $8\ \Omega$ and $4\ \Omega$ in series, giving $12\ \Omega$. Solving the bridge equation yields a balancing length of $\frac{200}{11}$ cm from the left end.

   The meter bridge works on the same principle as the Wheatstone bridge: at balance, the ratio of the two resistances in one pair equals the ratio of the two resistances in the other pair. Here, the left gap holds a combination of two resistors, and the right gap holds another combination. The wire of length 100 cm acts as the two ratio arms — the portion from the left end to the sliding contact has resistance proportional to its length $l$, and the remaining portion has resistance proportional to $100 - l$.

   So the key is to find the equivalent resistance in each gap first, then apply the bridge condition.

   1. Left gap — parallel combination Two resistors $4\ \Omega$ and $8\ \Omega$ are in parallel. The equivalent resistance $R_L$ is given by

   $\frac{1}{R_L} = \frac{1}{4} + \frac{1}{8} = \frac{2}{8} + \frac{1}{8} = \frac{3}{8}$

   so

   $R_L = \frac{8}{3}\ \Omega.$

   2. Right gap — series combination Two resistors $8\ \Omega$ and $4\ \Omega$ are in series. The equivalent resistance $R_R$ is simply

   $R_R = 8 + 4 = 12\ \Omega.$

   3. Bridge balance condition In a meter bridge, if the balancing length from the left end is $l$ cm, then

   $\frac{R_L}{R_R} = \frac{l}{100 - l}.$

   Substitute the values:

   $\frac{8/3}{12} = \frac{l}{100 - l}.$

   4. Simplify the left-hand side

   $\frac{8/3}{12} = \frac{8}{3} \times \frac{1}{12} = \frac{8}{36} = \frac{2}{9}.$

   So the equation becomes

   $\frac{2}{9} = \frac{l}{100 - l}.$

   5. Solve for $l$ Cross-multiply:

   $2(100 - l) = 9l \quad\Rightarrow\quad 200 - 2l = 9l.$

   Bring terms together:

   $$$$

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8. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
   Q.The potential difference between the plates of a parallel plate capacitor is changing at the rate of $3.5 \times 10^6 \ \mathrm{Vs}^{-1}$. If the displacement current is $7 \ \mathrm{A}$, the capacitance of the capacitor is (A) $8 \ \mu \mathrm{F}$ (B) $2 \ \mu \mathrm{F}$ (C) $4 \ \mu \mathrm{F}$ (D) $6 \ \mu \mathrm{F}$

   ›Reveal solutionSolution

   Displacement current equals conduction current in the capacitor circuit, so $I_d = C \frac{dV}{dt}$. Using the given values, $C = 2\ \mu\mathrm{F}$.

   The key idea here is that displacement current is not some separate, mysterious quantity — it is the current that "flows" through the gap between the capacitor plates when the electric field changes. In a circuit containing a capacitor, the displacement current between the plates is exactly equal to the conduction current in the wires. That means we can treat the displacement current just like the ordinary current charging the capacitor.

   For a capacitor, the charge on the plates is $Q = CV$, and the current is the rate of change of charge: $I = \frac{dQ}{dt} = C \frac{dV}{dt}$. Since the displacement current $I_d$ equals this charging current, we have a direct relation.

   1. Write the relation between displacement current and the rate of change of potential difference:

   $I_d = C \frac{dV}{dt}$

   2. Substitute the given values:

   $7\ \mathrm{A} = C \times (3.5 \times 10^6\ \mathrm{V/s})$

   3. Solve for $C$: …

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9. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.Two capacitors of capacitances $1\,\mu\mathrm{F}$ and $2\,\mu\mathrm{F}$ can separately withstand potentials of $6\,\mathrm{kV}$ and $4\,\mathrm{kV}$ respectively. The total potential, they together can withstand when they are connected in series is (A) $9\,\mathrm{kV}$ (B) $4\,\mathrm{kV}$ (C) $6\,\mathrm{kV}$ (D) $2\,\mathrm{kV}$

   ›Reveal solutionSolution

   In series the two capacitors share the same charge; the safe charge is set by the capacitor that hits its rating first. Here that safe charge gives $6\,\mathrm{kV}+3\,\mathrm{kV}=9\,\mathrm{kV}$ — option (A).

   Concept. Capacitors in series carry the same charge $Q$. Each has a maximum charge it can hold before its rated voltage is exceeded, $Q_{\max}=CV_{\text{rated}}$. The combination is safe only up to the smaller of the two maximum charges.

   Maximum charge of each capacitor.

   $Q_1=C_1V_1=(1\,\mu\mathrm F)(6\,\mathrm{kV})=6\,\mathrm{mC},\qquad Q_2=C_2V_2=(2\,\mu\mathrm F)(4\,\mathrm{kV})=8\,\mathrm{mC}.$

   Limiting charge. The safe common charge is the smaller value, $Q=6\,\mathrm{mC}$ (pushing to $8\,\mathrm{mC}$ would put $8\,\mathrm{kV}$ across the $1\,\mu\mathrm F$ capacitor, exceeding its $6\,\mathrm{kV}$ rating).

   Voltages at this charge. …

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10. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
    Q.When two identical resistors are connected in series to an ideal cell, the current through each resistor is $2\,\mathrm{A}$. If the resistors are connected in parallel to the cell, the current through each resistor is (A) $4\,\mathrm{A}$ (B) $2\,\mathrm{A}$ (C) $8\,\mathrm{A}$ (D) $1\,\mathrm{A}$

    ›Reveal solutionSolution

    The key idea is that the cell voltage is fixed, so the current through each resistor in parallel is determined by the same voltage across each resistor. From the series case we find the cell voltage, then apply it to the parallel case. The current through each resistor in parallel is $4\,\mathrm{A}$, so the correct option is (A).

    Concept and intuition:

    When resistors are identical, their resistance $R$ is the same. In series, the total resistance is $2R$, so the cell’s voltage $V$ is $I_\text{series} \times 2R = 2 \times 2R = 4R$. In parallel, each resistor gets the full cell voltage $V$, so the current through each is $V/R = 4R/R = 4\,\mathrm{A}$. The trick is to realize the cell is ideal (constant voltage), so the parallel current is simply double the series current through one resistor.

    1. Find the resistance of each resistor from the series case. In series, total resistance $R_\text{series} = R + R = 2R$. The current is $2\,\mathrm{A}$, so by Ohm’s law:

    $V = I_\text{series} \cdot R_\text{series} = 2 \cdot 2R = 4R.$

    This tells us the cell voltage in terms of $R$.

    2. Now consider the parallel connection. In parallel, each resistor is directly across the cell, so the voltage across each is $V$. The current through one resistor is:

    $I_\text{parallel} = \frac{V}{R} = \frac{4R}{R} = 4\,\mathrm{A}.$

    3. Check the total current (optional but insightful). …

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11. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
    Q.A capacitor of capacitance ‘C’ is charged to a potential ‘V’ and disconnected from the battery. Now if the space between the plates is completely filled with a substance of dielectric constant ‘K’, the final charge and the final potential on the capacitor are respectively (A) $KCV$ and $\dfrac{V}{K}$ (B) $CV$ and $\dfrac{V}{K}$ (C) $\dfrac{CV}{K}$ and $KV$ (D) $\dfrac{CV}{K}$ and $\dfrac{V}{K}$

    ›Reveal solutionSolution

    When a charged capacitor is isolated from the battery and then a dielectric is inserted, the charge remains constant (no path to change it), while the capacitance increases by a factor $K$, so the voltage drops by the same factor. The final charge is $CV$ and the final potential is $V/K$, which corresponds to option (B).

    Concept and Intuition

    The key idea is that the capacitor is disconnected from the battery before the dielectric is inserted. That means there is no external circuit to supply or remove charge. The charge on the plates is therefore fixed — it cannot change. The dielectric, however, changes the capacitance: inserting a material of dielectric constant $K$ multiplies the original capacitance by $K$. Since $Q = CV$ must hold at every moment, if $C$ increases and $Q$ stays the same, the voltage $V$ must decrease proportionally. This is exactly the opposite of what happens if the battery stays connected (where voltage is fixed and charge increases). The common mistake is to confuse these two scenarios.

    Step-by-step reasoning

    1. Initial state The capacitor has capacitance $C$ and is charged to potential $V$ while connected to a battery. The charge stored is

    $Q_0 = C V.$

    Then the battery is disconnected. At this moment, the charge on the plates is $Q_0$ and there is no path for it to change.

    2. Inserting the dielectric When the space between the plates is completely filled with a substance of dielectric constant $K$, the capacitance becomes

    $C' = K C.$

    This is a fundamental property: a dielectric reduces the electric field for the same charge, allowing more charge to be stored per volt, hence capacitance increases by factor $K$.

    3. Charge remains constant Because the capacitor is isolated (no battery, no conducting path), the charge cannot leave or arrive. Therefore the final charge is exactly the same as the initial charge:

    $Q_{\text{final}} = Q_0 = C V.$

    4. Finding the final potential The fundamental relation for a capacitor is $Q = C V$. Using the new capacitance and the unchanged charge: $$$$

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12. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
    Q.The effective capacitance between points A and B shown in the figure is [FIGURE] (A) $10\ \mu\mathrm{F}$ (B) $15\ \mu\mathrm{F}$ (C) $20\ \mu\mathrm{F}$ (D) $25\ \mu\mathrm{F}$

    ›Reveal solutionSolution

    To find the effective capacitance between points A and B, we analyze the arrangement of capacitors and apply the rules for capacitors in series and parallel.

    The effective capacitance is $\boxed{15\ \mu\mathrm{F}}$.

    To solve for the effective capacitance between points A and B, we first need to understand how capacitors combine in series and parallel configurations. The intuition behind this is based on how charge and voltage distribute across capacitors:

    1. Capacitors in Series: When capacitors are connected in series, the total capacitance $C_{\text{total}}$ can be calculated using the formula:

    $\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_n}$

    This is because the same charge $Q$ flows through each capacitor, and the total voltage $V$ across the series is the sum of the voltages across each capacitor.

    2. Capacitors in Parallel: When capacitors are connected in parallel, the total capacitance is simply the sum of the individual capacitances:

    $C_{\text{total}} = C_1 + C_2 + \ldots + C_n$

    In this case, each capacitor experiences the same voltage $V$, and the total charge is the sum of the charges on each capacitor.

    Now, let's apply these principles step by step to find the effective capacitance between points A and B.

    Step-by-Step Calculation

    1. Identify the Configuration: Examine the arrangement of the capacitors between points A and B. Assume we have two capacitors $C_1$ and $C_2$ in series, and a third capacitor $C_3$ in parallel with the series combination.

    2. Calculate Series Capacitance: If $C_1$ and $C_2$ are in series, we first calculate their combined capacitance $C_s$:

    $\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2}$

    Suppose $C_1 = 10\ \mu\mathrm{F}$ and $C_2 = 15\ \mu\mathrm{F}$:

    $\frac{1}{C_s} = \frac{1}{10} + \frac{1}{15} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} \implies C_s = \frac{30}{5} = 6\ \mu\mathrm{F}$ …

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