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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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.

KEAM 2026Set eng-2026-04184 marksMCQ
$pF$
›Reveal solutionSolution

The two 100 pF caps in parallel give 200 pF; the two 200 pF caps in series give 100 pF; connecting these two blocks across each other places them in parallel: $200+100=300\,pF$ .

Parallel combination of two $100\,pF$ :
$C_1=100+100=200\,pF$
Series combination of two $200\,pF$ :
$C_2=\frac{200\times200}{200+200}=\frac{40000}{400}=100\,pF$
"Connected across" one another means $C_1$ and $C_2$ share the same two terminals — a parallel connection:
$C_{eff}=C_1+C_2=200+100=300\,pF$
✓Final answer
$The correct option is (E) .$
KEAM 2025Set eng-2025-04234 marksMCQ
$Q. In a circuit, the capacitance C is connected. The effective capacitance of the circuit can be reduced by (A) introducing a metal plate between the plates of the capacitor (B) introducing a dielectric slab between the plates (C) reducing the potential difference between the plates (D) connecting another capacitor in series with it (E) connecting another capacitor in parallel with it$
›Reveal solutionSolution
The effective capacitance is reduced by connecting another capacitor in series with it.

Concept and Intuition

Capacitors in series give a combined capacitance smaller than the smallest individual one, since $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$ . All the other listed actions either increase capacitance or do not change it.

Step-by-Step Solution

Series: $\frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{C'} \Rightarrow C_{eq} < C$ .

A metal plate or dielectric increases capacitance; parallel connection increases it; potential difference does not change capacitance.

Only series reduces it.

Common Mistakes

Thinking a parallel capacitor reduces capacitance; parallel actually adds capacitances.

✓Final answer
$The correct option is (D) — connecting another capacitor in series with it.$

ANSWER: D
KEAM 2025Set eng-2025-04264 marksMCQ
$\mu$
›Reveal solutionSolution

For $n$ equal capacitors, series gives $C/n$ and parallel gives $nC$ . Multiplying the two results eliminates $n$ : $C^2 = 0.4\times10 = 4$ , so $C = 2\ \mu\text{F}$ .

Series equivalent:
$\dfrac{C}{n} = 0.4\ \mu\text{F}$
Parallel equivalent:
$nC = 10\ \mu\text{F}$
Multiply the two equations:
$\dfrac{C}{n}\times nC = C^2 = 0.4 \times 10 = 4$ $C = 2\ \mu\text{F}$
(Then $n = C/0.4 = 5$ , consistent with $nC = 10$ .)

✓Final answer
$The correct option is (A) .$
KEAM 2025Set pha-2025-0424F4 marksMCQ
$12\ \mu F$
›Reveal solutionSolution

Series equivalent $=12/3=4\,\mu F$ ; charge drawn $Q=C_{eq}V=4\times12=48\,\mu C$ .

For $n$ equal capacitors $C$ in series, the equivalent capacitance is
$C_{eq}=\frac{C}{n}=\frac{12\,\mu F}{3}=4\,\mu F.$
The charge drawn from the battery (which equals the charge on each series capacitor) is
$Q=C_{eq}V=4\,\mu F\times 12\,V=48\,\mu C.$
✓Final answer
$The correct option is (C) .$
KEAM 2024Set eng-2024-06064 marksMCQ
$Q. Three capacitances 1 µF, 4 µF and 5 µF are connected in parallel with a supply voltage. If the total charge flowing through the capacitors is 50 µC, then the supply voltage is (A) 2 V (B) 10 V (C) 6 V (D) 3 V (E) 5 V$
›Reveal solutionSolution

Capacitances in parallel add; the supply voltage is total charge divided by total capacitance.

For capacitors in parallel the equivalent capacitance is the sum:
$C=1+4+5=10\ \mu\text{F}.$
Using $Q=CV$ with total charge $Q=50\ \mu$ C,
$V=\frac{Q}{C}=\frac{50\ \mu\text{C}}{10\ \mu\text{F}}=5\text{ V}.$
✓Final answer
$The correct option is (E) .$
KEAM 2023Set eng-2023-P1-A14 marksMCQ
$1\,\mu$
›Reveal solutionSolution
With $N = 10^4$ capacitors, the series equivalent is $10^{-10}$ F.

Concept and Intuition

First find N from the parallel configuration: total parallel capacitance is $Q/V$ , and it equals $N$ times the individual $1\ \mu$ F. For identical capacitors in series, the equivalent capacitance is the individual value divided by N.

Step-by-Step Solution

Parallel: $C_{par} = \dfrac{Q}{V} = \dfrac{1}{100} = 10^{-2}$ F.

$N \times 1\ \mu\mathrm{F} = 10^{-2}$ F $\Rightarrow N = 10^{4}$ .

Series: $C_{ser} = \dfrac{1\ \mu\mathrm{F}}{N} = \dfrac{10^{-6}}{10^{4}} = 10^{-10}$ F.

Common Mistakes

Multiplying instead of dividing by N for the series combination.

✓Final answer
$10^{-10}$

ANSWER: C
KEAM 2022Set eng-2022-P1-A14 marksMCQ
$A$
›Reveal solutionSolution
Solving the symmetric ladder (with the shorted rightmost branch removed) gives about 1 uF between A and B.

Concept and Intuition

The far-right wire ties the two rails together, so the 3 uF capacitor drawn across that shorted junction has both plates at the same node and carries no independent charge. What remains is a symmetric network: A and B feed 3 uF top and bottom rails whose internal junctions are bridged by 2 uF capacitors, terminating at a common right node.

Step-by-Step Solution

Short the right ends: the rightmost 3 uF vertical branch is bypassed and drops out.

Set A = V, B = 0; by top-bottom symmetry the right node sits at V/2.

Write charge-balance (Kirchhoff) equations at the internal top nodes P1, P2, using V_Q = V - V_P.

Solving gives V_P1 approx 0.665 V, so charge from A = 3(V - V_P1) approx 1.0 V.

C_eq = Q/V approx 1.0 uF.

Common Mistakes

Forgetting that the closing wire short-circuits the rightmost vertical capacitor.

✓Final answer
$The correct option is (B) — 1 uF.$

ANSWER: B
KEAM 2021Set eng-2021-P1-A14 marksMCQ
$C_1$
›Reveal solutionSolution
The air capacitor drops 10 V and the dielectric one drops 2 V.

Concept and Intuition

Series capacitors carry the same charge $Q$ , so $V=Q/C$ : the smaller capacitor takes the larger voltage. The dielectric raises $C_2$ five-fold, so it takes the smaller share.

Step-by-Step Solution

$C_1=C$ (air), $C_2=5C$ (dielectric constant 5).

Same $Q$ : $V_1/V_2=C_2/C_1=5$ .

$V_1+V_2=12$ V with $V_1=5V_2\Rightarrow 6V_2=12\Rightarrow V_2=2$ V, $V_1=10$ V.

Common Mistakes

Assigning the larger voltage to the higher-capacitance (dielectric) capacitor.

✓Final answer
$The correct option is (B) — 10 V and 2 V.$

ANSWER: B

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Concept

Why this approach

Steps

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