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}

Uttar Pradesh UpmspMCQ· 1mImportance★★★★★

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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}

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Concept

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