Series: 1/C = 1/C₁ + 1/C₂. Parallel: C = C₁ + C₂.

Capacitors in Series and Parallel — Equivalent Capacitance

Question

Three capacitors of capacitances C₁ = 2 μF, C₂ = 3 μF, and C₃ = 6 μF are connected first in series, then in parallel across a 12 V battery. Find the equivalent capacitance and total charge stored in each case.

Solution — Step by Step

Case 1: Series Connection

For capacitors in series, the reciprocals add:

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

\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3 + 2 + 1}{6} = \frac{6}{6} = 1

So $C_{eq} = 1\ \mu F$ .

In series, the same charge $Q$ sits on every capacitor — this is because each capacitor’s inner plate gets charge only from its neighbour, not from the battery directly.

Q = C_{eq} \times V = 1 \times 12 = 12\ \mu C

Every capacitor in the series combination carries 12 μC.

Case 2: Parallel Connection

For capacitors in parallel, capacitances simply add up (all plates share the same potential difference):

C_{eq} = C_1 + C_2 + C_3 = 2 + 3 + 6 = 11\ \mu F

Q_{total} = C_{eq} \times V = 11 \times 12 = 132\ \mu C

Each capacitor sees the full 12 V, so their individual charges are:

$Q_1 = 2 \times 12 = 24\ \mu C$
$Q_2 = 3 \times 12 = 36\ \mu C$
$Q_3 = 6 \times 12 = 72\ \mu C$

Cross-check: $24 + 36 + 72 = 132\ \mu C$ ✓

Final Answers:

Series: $C_{eq} = 1\ \mu F$ , $Q = 12\ \mu C$ on each
Parallel: $C_{eq} = 11\ \mu F$ , $Q_{total} = 132\ \mu C$

Why This Works

In series, capacitors share the same charge but split the voltage. Think of it like resistors in parallel — the effective capacitance always comes out smaller than the smallest individual capacitor. Here, 1 μF is less than even the 2 μF capacitor. Whenever you see the equivalent is smaller than any individual value, your answer is probably right.

In parallel, capacitors share the same voltage but split the charge. The battery sees a single larger plate area, so capacitance adds directly. The effective capacitance is always larger than the largest individual capacitor.

This duality — series for charge sharing, parallel for voltage sharing — is the core idea that appears repeatedly in both CBSE boards and JEE Main.

Alternative Method

For the series part, notice that 2, 3, and 6 have LCM = 6. Finding LCM first makes the arithmetic faster:

\frac{1}{C_{eq}} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1 \implies C_{eq} = 1\ \mu F

In competitive exams, spotting the LCM shortcut saves 30–40 seconds per problem.

You can also verify using the charge-voltage method for series: since Q is the same on all capacitors, the voltages are $V_1 = Q/C_1 = 6\ V$ , $V_2 = 4\ V$ , $V_3 = 2\ V$ . Sum = 12 V — matches the battery. This cross-check takes 10 seconds and catches arithmetic errors before they cost you marks.

Common Mistake

Swapping the formulas. A very common slip in CBSE 2024 and JEE Main papers: students write $C_{eq} = C_1 + C_2 + C_3$ for series (the resistor-parallel formula) and get 11 μF — which is actually the parallel answer.

Memory trick: capacitors in series behave opposite to resistors in series. For resistors in series, $R$ adds. For capacitors in series, $1/C$ adds. Keep this contrast in mind and you will never mix them up again.