Three capacitors 2μF, 3μF, 6μF in series — find equivalent capacitance and charge

Question

Three capacitors of capacitances $2\,\mu\text{F}$, $3\,\mu\text{F}$, and $6\,\mu\text{F}$ are connected in series across a $12\,\text{V}$ battery. Find the equivalent capacitance and the charge on each capacitor.

(CBSE 2023, 3 marks)

Solution — Step by Step

Why This Works

In a series connection, the same charge must flow through each capacitor (there’s nowhere else for it to go — the plates are isolated). The voltage divides across the capacitors inversely proportional to their capacitance — the smallest capacitor gets the largest voltage share.

The reciprocal formula for series capacitors is analogous to resistors in parallel. The equivalent capacitance is always less than the smallest individual capacitance. Here, $C_{eq} = 1\,\mu\text{F} < 2\,\mu\text{F}$.

Alternative Method — Work with ratios

Since $1/C_1 : 1/C_2 : 1/C_3 = 3 : 2 : 1$ (voltage divides in this ratio):

$V_1 = \frac{3}{6} \times 12 = 6\,\text{V}$, $V_2 = 4\,\text{V}$, $V_3 = 2\,\text{V}$.

Then $Q = C_1 V_1 = 2 \times 6 = 12\,\mu\text{C}$.

Common Mistake