A 4 $\mu$F capacitor is part of a circuit driven by a cell of emf 2.5 V whose internal resistance is 0.5 $\Omega$. Three branches connect the same pair of nodes in parallel: the first branch is the 4 $\mu$F capacitor in series with a 10 $\Omega$ resistor; the second branch is the 2.5 V cell (internal resistance 0.5 $\Omega$); the third branch is a 2 $\Omega$ resistor. In the steady state, the amount of charge on the capacitor plates will be (a) 0 (b) $4\ \mu\text{C}$ (c) $16\ \mu\text{C}$ (d) $8\ \mu\text{C}$

A 4 μF capacitor is part of a circuit driven by a cell of emf 2.5 V whose internal resistance is 0.5 Ω. Three

NCERT Exemplar · Q1

\mu

(a) 0

(b)

4\ \mu\text{C}

(c)

16\ \mu\text{C}

(d)

8\ \mu\text{C}

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✓ Free question

Concept understanding — Capacitor Network Analysis

Capacitor Network Analysis

Imagine you have a bucket of water and a pipe. The bigger the bucket, the more water it can hold for a given water pressure. A capacitor does the same thing with electric charge — it stores charge when a voltage is applied. The "size" of the bucket is called capacitance ( $C$ ), measured in farads (F).

Now, what happens when you connect several buckets together with pipes? That's a capacitor network. The analysis is about finding one equivalent bucket (a single capacitor) that behaves exactly like the whole network.

The Core Idea: Charge and Voltage Must Match

When you connect capacitors, two things are always true:

Charge conservation: Total charge in a closed part of the circuit stays the same unless a battery pushes more in.
Voltage is shared: The voltage across each capacitor depends on how they're wired.

There are only two basic ways to connect them. Everything else is a combination of these two.

Series Connection: One Path, Shared Charge

Connect capacitors end-to-end, like train cars. The same current flows through each, so each capacitor stores the same amount of charge $Q$ .

But the total voltage across the combination is the sum of the individual voltages:

V_{\text{total}} = V_1 + V_2 + V_3 + \dots

Since $V = Q/C$ for each capacitor, we get:

\frac{Q}{C_{\text{eq}}} = \frac{Q}{C_1} + \frac{Q}{C_2} + \frac{Q}{C_3} + \dots

Cancel $Q$ (it's the same everywhere):

\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \dots

Intuition: Adding capacitors in series makes the equivalent capacitance smaller than the smallest individual one. Why? Because you're effectively making the "bucket" longer and narrower — harder to fill.

Watch out

A common mistake: students treat series capacitors like series resistors (adding reciprocals for resistors, but adding directly for capacitors). It's the opposite. For resistors in series: $R_{\text{eq}} = R_1 + R_2$ . For capacitors in series: $1/C_{\text{eq}} = 1/C_1 + 1/C_2$ .

Parallel Connection: Multiple Paths, Same Voltage

Connect capacitors side by side, like buckets with their bottoms connected by a wide pipe. Each capacitor sees the same voltage $V$ across it.

But the total charge stored is the sum of charges on each:

Q_{\text{total}} = Q_1 + Q_2 + Q_3 + \dots

Since $Q = CV$ for each:

C_{\text{eq}} V = C_1 V + C_2 V + C_3 V + \dots

Cancel $V$ :

C_{\text{eq}} = C_1 + C_2 + C_3 + \dots

Intuition: Adding capacitors in parallel makes the equivalent capacitance larger — you're just adding more bucket area. Easy to fill.

Tip

| Connection | Equivalent Formula | What happens to $C_{\text{eq}}$ |

|------------|-------------------|----------------------------------|

| Series | $1/C_{\text{eq}} = \sum 1/C_i$ | Gets smaller than smallest |

| Parallel | $C_{\text{eq}} = \sum C_i$ | Gets larger than largest |

How to Analyze Any Network

Spot the pattern: Look for capacitors that are clearly in series (only two terminals, no branching between them) or clearly in parallel (both ends connected together).
Replace step by step: Replace each simple series or parallel group with its equivalent capacitor. Redraw the circuit after each step.
Repeat until you have one capacitor.

Note

This is exactly like simplifying resistor networks — but with the reciprocal formula for series. If you can do resistor networks, you can do capacitor networks. Just flip the series formula.

A Worked Example

Suppose you have three capacitors: $C_1 = 2\ \mu\text{F}$ , $C_2 = 3\ \mu\text{F}$ , $C_3 = 6\ \mu\text{F}$ . $C_2$ and $C_3$ are in parallel, and that combination is in series with $C_1$ .

Step 1: Parallel group first.

C_{23} = C_2 + C_3 = 3 + 6 = 9\ \mu\text{F}

Step 2: Now $C_1$ (2 μF) is in series with $C_{23}$ (9 μF).

\frac{1}{C_{\text{eq}}} = \frac{1}{2} + \frac{1}{9} = \frac{9 + 2}{18} = \frac{11}{18}

C_{\text{eq}} = \frac{18}{11}\ \mu\text{F} \approx 1.64\ \mu\text{F}

Notice: the final equivalent is smaller than the smallest individual capacitor (2 μF). That's the series effect.

The Deeper Reason: Energy and Symmetry

Capacitors store energy: $U = \frac{1}{2} C V^2$ . In a network, energy is conserved (ignoring losses). The equivalent capacitor must store the same total energy as the original network for the same applied voltage. That's why the formulas work — they're derived from charge and voltage matching, which guarantees energy matching.

When you see a complex network, always ask: "Which capacitors share the same voltage?" (parallel) and "Which capacitors share the same charge?" (series). That's the entire analysis.

Final takeaway: Capacitor network analysis is just systematic application of two rules — series (same charge, voltages add) and parallel (same voltage, charges add). Reduce step by step, and you can handle any network.

Reducing series and parallel capacitor networks to a single equivalent capacitance is a standard numerical skill from the NCERT Class 12 Physics chapter on electrostatic potential and capacitance, tested every year in CBSE boards and JEE Main. Searches for "capacitors in series and parallel formula class 12 physics important questions" will find this step-by-step reduction method is exactly what board exam solutions use.

A fully charged capacitor passes no steady current, so its branch is 'dead'. Current only circulates through the cell and the 2 $\Omega$ resistor, fixing the cell's terminal voltage at 2 V. That 2 V lies entirely across the capacitor (the series 10 $\Omega$ has zero drop), so $Q = CV = 4\ \mu\text{F}\times2\ \text{V} = 8\ \mu\text{C}$ .

Concept

In a DC steady state a capacitor is fully charged and blocks further current. Any resistor in series with it therefore carries no current and develops no potential drop.

Why this approach

Because the capacitor branch carries no current, the voltage across it equals the voltage the cell maintains between the two nodes (its terminal voltage), which is set by the resistive loop the current actually flows in.

Steps

Current path: only the cell (emf 2.5 V, internal resistance 0.5 $\Omega$ ) and the 2 $\Omega$ resistor form a closed conducting loop. $I = \dfrac{\mathcal{E}}{R + r} = \dfrac{2.5}{2 + 0.5} = 1\ \text{A}$ .
Terminal (node-to-node) voltage: $V = \mathcal{E} - I r = 2.5 - (1)(0.5) = 2\ \text{V}$ (equivalently the drop $I\times 2\ \Omega = 2\ \text{V}$ ).
The capacitor-plus-10 $\Omega$ branch spans these same two nodes. No current $\Rightarrow$ no drop across the 10 $\Omega$ $\Rightarrow$ the whole 2 V is across the capacitor.
$Q = CV = 4\ \mu\text{F}\times2\ \text{V} = 8\ \mu\text{C}$ .

Why the distractors fail: (a) 0 needs zero node voltage;

(b) $4\ \mu\text{C}$ uses $V = 1\ \text{V}$ ;

✓Final answer

Q = 8\ \mu\text{C}

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set SEM31 markMCQ
$Q. Three capacitors having capacitances 1\cdot0 μF, 2\cdot0 μF and 5\cdot0 μF are connected in series with a source of 10 V. The potential difference between the two ends of the capacitor having capacitance 2\cdot0 μF will be$
(a) 100 V / 17
(b) 20 V / 17
(c) 50 V / 17
(d) 10 V
›Reveal solutionSolution

In series, all capacitors carry the same charge. C_eq = 10/17 μF, Q = 100/17 μC, and V₂ = Q/(2 μF) = 50/17 V. Option (c).

Step 1 — equivalent series capacitance:

1/C_eq = 1/1 + 1/2 + 1/5 = (10 + 5 + 2)/10 = 17/10, so C_eq = 10/17 μF.

Step 2 — common charge (series capacitors share the same charge):

Q = C_eq × V = (10/17 μF)(10 V) = 100/17 μC.

Step 3 — potential difference across the 2·0 μF capacitor:

V₂ = Q/C₂ = (100/17 μC)/(2 μF) = 50/17 V ≈ 2·94 V.

This series-capacitor analysis is core to the NCERT/CBSE Class 12 Physics chapter on Electrostatic Potential and Capacitance.

✓Final answer
$(c) 50 V/17$
CBSE 2025Set ANNUAL1 markMCQ
$Q. When two identical capacitors are in series, they have 3 uF resultant capacitance and when parallel 12 uF. What is the capacitance of each?$
(i) 6 uF
(ii) 3 uF
(iii) 12 uF
(iv) 9 uF
›Reveal solutionSolution

Each capacitor is 6 uF.

Let each capacitance be C. In series the resultant is $C_s = \frac{C}{2} = 3\,\mu F \Rightarrow C = 6\,\mu F$ . In parallel the resultant is $C_p = 2C = 12\,\mu F \Rightarrow C = 6\,\mu F$ . Both conditions agree.

✓Final answer
$(i) 6 uF.$
CBSE 2024Set 55/1/11 markMCQ
$1\ \mu\text{F}$
›Reveal solutionSolution
For capacitors in parallel, the total capacitance is the sum of individual capacitances. Here, $C_{\text{eq}} = 10\ \mu\text{F}$ . Energy stored is $\frac{1}{2} C_{\text{eq}} V^2 = \frac{1}{2} \times 10 \times 10^{-6} \times (100)^2 = 0.05\ \text{J} = 5.0 \times 10^{-2}\ \text{J}$ . The correct option is (D).

When capacitors are connected in parallel, the voltage across each capacitor is the same — equal to the source voltage. This is the key difference from series connections, where the charge is the same but voltage divides. Because all ten capacitors are identical and each sees the full $100\ \text{V}$ , the total energy is simply the sum of the energies stored in each capacitor individually.

The energy stored in a single capacitor of capacitance $C$ at voltage $V$ is $\frac{1}{2} C V^2$ . For ten such capacitors, the total energy is $10 \times \frac{1}{2} C V^2 = \frac{1}{2} (10C) V^2$ . Notice that $10C$ is exactly the equivalent capacitance of ten $1\ \mu\text{F}$ capacitors in parallel. So the problem reduces to finding the energy stored in a single $10\ \mu\text{F}$ capacitor charged to $100\ \text{V}$ .

Let’s work through the numbers carefully.

Find the equivalent capacitance.

For parallel combination: $C_{\text{eq}} = C_1 + C_2 + \dots + C_{10} = 10 \times 1\ \mu\text{F} = 10\ \mu\text{F}$ .

In SI units: $10\ \mu\text{F} = 10 \times 10^{-6}\ \text{F} = 10^{-5}\ \text{F}$ .

Apply the energy formula.

The energy stored in a capacitor network (or a single equivalent capacitor) is $U = \frac{1}{2} C_{\text{eq}} V^2$ , where $V$ is the voltage across the combination.

Here $V = 100\ \text{V}$ , so:

$U = \frac{1}{2} \times (10^{-5}) \times (100)^2 = \frac{1}{2} \times 10^{-5} \times 10^4 = \frac{1}{2} \times 10^{-1} = 0.05\ \text{J}.$

Express in scientific notation. $0.05\ \text{J} = 5.0 \times 10^{-2}\ \text{J}$ .

Watch out

A common mistake is to treat the capacitors as if they were in series, which would give a much smaller equivalent capacitance ( $0.1\ \mu\text{F}$ ) and a tiny energy — not among the options. Always check the connection: parallel means same voltage, series means same charge.

Tip

You can also compute the energy per capacitor: each stores $\frac{1}{2} \times 1\times10^{-6} \times 100^2 = 0.5 \times 10^{-2}\ \text{J} = 5\times10^{-3}\ \text{J}$ . Ten such capacitors give $10 \times 5\times10^{-3} = 5\times10^{-2}\ \text{J}$ . This is a quick sanity check.

✓Final answer
$5.0 \times 10^{-2}\ \text{J}$
CBSE 2024Set ANNUAL1 markMCQ
$Q. Three capacitors are connected in triangle as shown in figure. The equivalent capacitance between the points A and C is :$
(a) 4 $\mu$ F
(b) 2 $\mu$ F
(c) 8 $\mu$ F
(d) 6 $\mu$ F
›Reveal solutionSolution
The direct A-C capacitor is in parallel with the series A-B-C path (which reduces to $2\,\mu F$ ), giving a total equivalent capacitance of $6\,\mu F$ between A and C.

Working

Three $4\,\mu F$ capacitors form a triangle with vertices A, B, C: one directly between A and C, one between A and B, and one between B and C.

Between the terminals A and C, there are two parallel paths:

The capacitor directly connecting A to C: $C_1 = 4\,\mu F$ .

The path A $\to$ B $\to$ C, formed by the A-B and B-C capacitors in series: $C_{AB\text{-}BC} = \dfrac{4\times4}{4+4} = \dfrac{16}{8} = 2\,\mu F$

These two paths are in parallel between A and C, so they add:

$C_{AC} = C_1 + C_{AB\text{-}BC} = 4 + 2 = 6\,\mu F$

✓Final answer
$6\,\mu F$
CBSE 2023Set ANNUAL1 markMCQ
$Q. Three capacitors each of capacitance 4 μF are to be connected in such a way that the effective capacitance is 6 μF. This can be done by connecting$
(a) each of them in series
(b) each of them in parallel
(c) two in parallel and one in series
(d) two in series and one in parallel
›Reveal solutionSolution

Two 4 μF capacitors in series (2 μF) placed in parallel with the third 4 μF capacitor gives 2 + 4 = 6 μF.

Each capacitor has $C = 4\ \mu F$ . Check the given option: two in series, then that combination in parallel with the third.

Series combination of two $4\ \mu F$ capacitors:

$\dfrac{1}{C_s} = \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2} \implies C_s = 2\ \mu F$

This series pair is now connected in parallel with the third $4\ \mu F$ capacitor:

$C_{eff} = C_s + 4 = 2 + 4 = 6\ \mu F$

This matches the required 6 μF. (Checking the other options: all-series gives $4/3\ \mu F$ ; all-parallel gives $12\ \mu F$ ; two-parallel-one-series gives $(8\times4)/(8+4) = 8/3\ \mu F$ — none equal 6 μF.)

✓Final answer
$Two in series and one in parallel (option d).$
CBSE 2022Set ANNUAL1 markMCQ
$Q. Three capacitors of equal capacity C are joined first in parallel and then in series. The ratio of equivalent capacities in both the cases will be:$
(a) 9 : 1
(b) 6 : 1
(c) 3 : 1
(d) 1 : 9
›Reveal solutionSolution

$C_{parallel} = 3C$ and $C_{series} = C/3$ , so the ratio is $9 : 1$ : option (A).

For three equal capacitors each of capacitance $C$ :

In parallel, capacitances add:

$C_p = C + C + C = 3C.$

In series, reciprocals add:

$\frac{1}{C_s} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C} \quad\Rightarrow\quad C_s = \frac{C}{3}.$

The ratio is

$\frac{C_p}{C_s} = \frac{3C}{C/3} = 9 \;\Rightarrow\; 9 : 1.$

✓Final answer
$(A) 9 : 1.$
CBSE 2020Set 55/3/11 markMCQ
$C_1$
›Reveal solutionSolution
In a parallel combination, both capacitors share the same voltage. Since $Q = CV$ , the charge on each is directly proportional to its capacitance, so the ratio $Q_1 : Q_2 = C_1 : C_2$ . The answer is $\boxed{\dfrac{C_1}{C_2}}$ , option (A).

When two capacitors are connected in parallel, the defining feature is that they are connected across the same two points. This means the potential difference (voltage) across each capacitor is identical. This is the single most important idea — everything else follows from it.

Think of it like two buckets of different widths placed side by side under the same tap. The water level (voltage) in both will be the same, but the wider bucket (larger capacitance) will hold more water (charge). The charge stored by a capacitor is $Q = CV$ , so if $V$ is fixed, the charge is simply proportional to $C$ .

Now, let’s work through it step by step.

State the parallel condition.

For capacitors $C_1$ and $C_2$ in parallel, the voltage across each is the same. Call this common voltage $V$ .

Write the charge on each capacitor.

Using $Q = CV$ :

$Q_1 = C_1 V \quad \text{and} \quad Q_2 = C_2 V$

Find the required ratio. The ratio of the charge on $C_1$ to the charge on $C_2$ is:

$\frac{Q_1}{Q_2} = \frac{C_1 V}{C_2 V} = \frac{C_1}{C_2}$
The voltage $V$ cancels out completely — the ratio depends only on the capacitances.

Match with the options. The ratio $\dfrac{C_1}{C_2}$ corresponds directly to option (A).

Watch out

A common mistake is to confuse parallel with series. In series, the charge on each capacitor is the same, and the voltage divides inversely with capacitance — giving a very different ratio. Always check the connection type first.

Tip

For a quick check: if $C_1 = C_2$ , the charges must be equal, so the ratio should be $1$ . Only option (A) gives $1$ when $C_1 = C_2$ . The others give $\sqrt{1}=1$ too, but they fail for unequal values — try $C_1 = 2$ , $C_2 = 1$ : (A) gives $2$ , (B) gives $\sqrt{2}$ , (C) gives $1/\sqrt{2}$ , (D) gives $1/2$ . Only (A) matches the physical expectation that the larger capacitor holds more charge.

✓Final answer
$\boxed{\dfrac{C_1}{C_2}}$
CBSE 2018Set ANNUAL1 markMCQ
$Q. Two capacitors C₁ = 2μF and C₂ = 4μF are connected in series and a potential difference (p.d.) of 1200V is applied across it. The Potential difference across 2μF will be :-$
(a) 400V
(b) 600V
(c) 800V
(d) 900V
›Reveal solutionSolution

Series capacitors share equal charge, so V is inversely proportional to C; the 2μF takes 800 V.

Series combination: same charge Q on both plates.

$Q = C_1 V_1 = C_2 V_2$ , so $\dfrac{V_1}{V_2} = \dfrac{C_2}{C_1} = \dfrac{4}{2} = 2.$

Thus $V_1 = 2V_2$ (V₁ across the 2μF). With $V_1 + V_2 = 1200$ V:

$2V_2 + V_2 = 1200 \Rightarrow 3V_2 = 1200 \Rightarrow V_2 = 400$ V (across 4μF), and $V_1 = 800$ V (across 2μF).

Check: equivalent series capacitance $= \dfrac{2\times4}{2+4} = \dfrac{4}{3}\,\mu F$ , charge $Q = \dfrac{4}{3}\times1200 = 1600\,\mu C$ ; then $V_1 = Q/C_1 = 1600/2 = 800$ V. ✓

✓Final answer
$(c) 800 V.$
CBSE 2018Set ANNUAL1 markMCQ
$Q. Minimum number of capacitors of 2μF each required to obtain a capacitance of 5μF will be :-$
(a) 4
(b) 3
(c) 5
(d) 6
›Reveal solutionSolution
5μF = (two 2μF in parallel = 4μF) in parallel with (two 2μF in series = 1μF) → 4 capacitors.

We want 5μF from 2μF units.

Two 2μF in series: $\dfrac{2\times2}{2+2}=1\,\mu F$ .

Two 2μF in parallel: $2+2=4\,\mu F$ .

Put the 4μF branch in parallel with the 1μF branch: $4+1 = 5\,\mu F$ .

That uses 4 capacitors. Fewer is impossible: with three 2μF the only reachable values are 6, 2/3, 4/3 and 3 μF — none equal 5μF. Hence 4 is the minimum.

✓Final answer
$(a) 4.$
CBSE 2016Set ANNUAL1 markMCQ
$Q. A capacitor of capacitance C1 is charged up to potential V and then connected in parallel to an uncharged capacitor of capacitance C2. The final potential difference across each capacitor will be$
(a) C2V/(C1+C2)
(b) C1V/(C1+C2)
(c) (1+C2/C1)V
(d) (1-C2/C1)V
›Reveal solutionSolution

Charge is conserved when the charged capacitor is connected to the uncharged one; sharing it between the two parallel capacitances gives the common final potential.

Initial charge on C1: Q = C1V. When connected in parallel with the uncharged C2, this charge Q redistributes but total charge is conserved. Let the common final potential be V'. Then

Q = (C1+C2)V' ⟹ C1V = (C1+C2)V' ⟹ V' = C1V/(C1+C2)

✓Final answer
$(b) C1V/(C1+C2).$

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Concept

Why this approach

Steps

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