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 path once you notice the 2 μF and 3 μF are wired across the same two nodes.
In the circuit shown, find the resultant capacitance between points P and Q.
- From P12 μF
- Middle branch 12 μF
- Middle branch 23 μF
- From Q20 μF
The single equivalent capacitance \(C_{PQ}\) measured between terminals P and Q.
Trace the nodes, not the picture. The 12 μF plate, the tops of the 2 μF and 3 μF, all sit on one wire — call it 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 connect the same two nodes (M and N) → they are in parallel:
$$C_{mid}=2+3=5\,\mu\text{F}$$The current path is now 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.
- 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. $$C_{PQ}=3\,\mu\text{F}$$
Same two endpoints = parallel. Any time two capacitors start and finish on the same pair of wires, just add them. Here \(2+3=5\) instantly.
For the series step, take LCM 60 as the denominator: \(\tfrac{1}{12},\tfrac15,\tfrac{1}{20}\) become \(\tfrac{5}{60},\tfrac{12}{60},\tfrac{3}{60}\). Numerators add to \(20\), so \(C_{PQ}=\tfrac{60}{20}=3\).
The \(60\) in option (c) is the LCM trap, and \(47\) in (a) is what you'd get by wrongly adding everything. A series result must be below the smallest capacitor (here below 5) — only 3 μF qualifies.
Middle pair in parallel gives 5 μF; then 12, 5 and 20 μF in series:
$$\frac{1}{C_{PQ}}=\frac{1}{12}+\frac{1}{5}+\frac{1}{20}=\frac{1}{3}\;\Rightarrow\;C_{PQ}=3\,\mu\text{F}$$