Q.In a circuit a 9 V cell is connected in the top line first to a capacitor $C_1 = 6\ \mu\text{F}$, then to a key $K_1$, then to a key $K_2$. A capacitor $C_2 = 3\ \mu\text{F}$ joins the point between $K_1$ and $K_2$ down to the common return wire, and a capacitor $C_3 = 3\ \mu\text{F}$ joins the point just beyond $K_2$ down to the return wire, which goes back to the cell's negative terminal (take the unit capacitance $C = 1\ \mu\text{F}$, so $C_1 = 6\ \mu\text{F}$ and $C_2 = C_3 = 3\ \mu\text{F}$). Initially $K_1$ is closed and $K_2$ is open — find the charge on each capacitor. Then $K_1$ is opened and $K_2$ is closed (the order is important) — find the new charge on each capacitor.

Concept

Capacitors in series carry equal charge; capacitors in parallel share a common voltage. An isolated capacitor keeps its charge, and a group of connected isolated capacitors conserves total charge.

Phase 1 — $K_1$ closed, $K_2$ open

1. Charging path: cell $\to C_1 \to (K_1) \to C_2 \to$ back to cell. So $C_1 = 6\ \mu\text{F}$ and $C_2 = 3\ \mu\text{F}$ are in series across 9 V.
2. $C_{series} = \dfrac{C_1 C_2}{C_1 + C_2} = \dfrac{6\times 3}{6 + 3} = 2\ \mu\text{F}$.
3. $Q = C_{series}\times E = 2\ \mu\text{F}\times 9\ \text{V} = 18\ \mu\text{C}$. Series $\Rightarrow$ each of $C_1$ and $C_2$ carries $18\ \mu\text{C}$ (with $V_1 = 18/6 = 3\ \text{V}$ and $V_2 = 18/3 = 6\ \text{V}$, summing to 9 V).
4. $K_2$ open $\Rightarrow C_3$ is disconnected $\Rightarrow Q_3 = 0$.

Phase 2 — $K_1$ opened, then $K_2$ closed

5. Opening $K_1$ isolates $C_1$'s inner plate, so its $18\ \mu\text{C}$ is trapped and unchanged: $Q_1 = 18\ \mu\text{C}$. …

Method: Analysing Capacitor Networks That Change with Switches

Use this for circuits where opening/closing keys changes which capacitors are connected to the cell and to each other — a very common exam pattern.

Steps

Step 1: Redraw (or re-imagine) the circuit separately for each switch configuration

Don't try to reason about the whole multi-phase problem at once. For each distinct state of the keys, trace which components are actually joined into a single conductive path, and identify whether that path includes the EMF source.

Step 2: Classify each connected group as series, parallel, or isolated

• Series (same charge flows through each): capacitors connected end-to-end with no other path branching off between them.
• Parallel (same voltage across each): capacitors whose corresponding terminals are tied directly together.
• Isolated: a capacitor whose plate has no closed path back to the rest of the circuit — it cannot exchange charge with anything, so whatever charge it was holding stays exactly as is.

Step 3: Solve the phase that's connected to the cell using $Q = CV$

Find the equivalent capacitance of the group actually connected to the EMF (series: $\frac{1}{C_{eq}} = \sum \frac{1}{C_i}$; parallel: $C_{eq} = \sum C_i$), then $Q_{eq} = C_{eq} \times \mathcal{E}$. For a series group, every capacitor in it carries this same $Q_{eq}$.

Step 4: When the switches change, track trapped charge vs. newly available charge …