NEET · Physics Electrostatics Capacitors

Find the shared node — then it's just parallel, parallel, series

Four capacitors that look tangled collapse the moment you spot the central node M. Everything left of M adds, everything right of M adds, and the two halves combine in series.

Question

Four capacitors are connected as shown. Find the effective capacitance (in μF) between points A and B.

Fig 1 · Two capacitors meet A at node M on the left; two meet B on the right.

(a) 28/9
(b) 4
(c) 5
(d) 18

Given

A → M (upper)2 μF
A → M (lower)2 μF
M → B (upper)12 μF
M → B (lower)2 μF

Asked

The effective capacitance $C_{AB}$ between terminals A and B.

Concept

Label the central node M. Both of the left capacitors run from A to M — same two endpoints, so they're in parallel. Both right capacitors run from M to B — also parallel.

That turns the whole network into just two effective capacitors in a line:

$$A \;\xrightarrow{\;C_{left}\;}\; M \;\xrightarrow{\;C_{right}\;}\; B$$

and a single node between two capacitors means they're in series.

Method & Steps

Left of M — parallel. $$C_{left}=2+2=4\,\mu\text{F}$$
Right of M — parallel. $$C_{right}=12+2=14\,\mu\text{F}$$
Across M — series. $$C_{AB}=\frac{C_{left}\,C_{right}}{C_{left}+C_{right}}=\frac{4\times 14}{4+14}=\frac{56}{18}=\frac{28}{9}\,\mu\text{F}$$

Reduced: 4 μF and 14 μF in series → 28/9 μF

Easy Trick

Same two endpoints = parallel (add). One shared node between two = series (product/sum). Find M first and the problem almost solves itself: $4$ on the left, $14$ on the right.

For two in series use $\dfrac{\text{product}}{\text{sum}}=\dfrac{4\times14}{18}=\dfrac{56}{18}=\dfrac{28}{9}\approx 3.1$. The traps: 5 appears if you wrongly hang the lower-right 2 μF straight across A–B, and 4 if you lump all three small caps on one side.

Solution

Left pair → 4 μF, right pair → 14 μF, combined in series:

$$C_{AB}=\frac{4\times 14}{4+14}=\frac{56}{18}=\frac{28}{9}\,\mu\text{F}\approx 3.11\,\mu\text{F}$$

a 289 μF ≈ 3.11 μF