NEET · Physics Electrostatics Capacitors

Two capacitors on the same rails — that's the whole trick

A four-capacitor network between P and Q. It looks tangled, but it's a single series chain once you notice the 2 μF and 3 μF are wired across the same two nodes.

Question

In the circuit diagram shown, find the resultant capacitance between points P and Q.

Fig 3 · 12 μF and 20 μF feed a 2 μF ∥ 3 μF middle section

(a) 47 μF
(b) 3 μF
(c) 60 μF
(d) 10 μF

Given

From P12 μF
Middle branch 12 μF
Middle branch 23 μF
From Q20 μF

Asked

The single equivalent capacitance $C_{PQ}$ measured between terminals P and Q.

Concept

Trace the nodes, not the picture. The 12 μF plate and the tops of the 2 μF and 3 μF all sit on one wire — node M (top rail). The 20 μF plate and the bottoms of 2 μF and 3 μF all sit on another wire — node N (bottom rail).

So the 2 μF and 3 μF join the same two nodes (M and N) → they're in parallel:

$$C_{mid}=2+3=5\,\mu\text{F}$$

The path is then one straight line: $P \to 12\,\mu F \to M \to 5\,\mu F \to N \to 20\,\mu F \to Q$. Three capacitors in series.

Method & Steps

Collapse the middle (parallel). $$C_{mid}=2+3=5\,\mu\text{F}$$
Combine the chain (series), LCM 60. $$\frac{1}{C_{PQ}}=\frac{1}{12}+\frac{1}{5}+\frac{1}{20}=\frac{5+12+3}{60}=\frac{20}{60}=\frac{1}{3}$$
Invert. $$C_{PQ}=3\,\mu\text{F}$$

Easy Trick

Same two endpoints = parallel. The 2 μF and 3 μF start and finish on the same wires, so just add: $2+3=5$. Then three in series with LCM 60: numerators $5,12,3$ sum to $20$, giving $C_{PQ}=\tfrac{60}{20}=3$.

The $60$ in option (c) is the LCM trap; $47$ in (a) is wrong-adding everything. A series result must fall below the smallest capacitor (below 5) — only 3 μF qualifies.

Solution

Middle pair in parallel gives 5 μF; then 12, 5 and 20 μF in series:

$$\frac{1}{C_{PQ}}=\frac{1}{12}+\frac{1}{5}+\frac{1}{20}=\frac{1}{3}\;\Rightarrow\;C_{PQ}=3\,\mu\text{F}$$

b 3 μF between P and Q