Collapse the box first, then it's three in a line
The rectangle in the middle is just two parallel routes between a left node and a right node. Reduce it to one value and the whole circuit is a simple series chain.
Find the equivalent capacitance between A and B (all values in farad).
- At A12 F
- Top edge8 F then 4 F
- Bottom edge4 F
- At B16 F
The equivalent capacitance \(C_{AB}\) between terminals A and B.
Name the two vertical sides L and R. Each side is a plain wire, so the whole left edge is one node (L) and the whole right edge is another (R).
Between L and R there are exactly two routes: the top (8 F in series with 4 F) and the bottom (a single 4 F). Same endpoints → those two routes are in parallel.
Once the box is one capacitor, the circuit is just \(A \to 12 \to L \to (\text{box}) \to R \to 16 \to B\): three capacitors in series.
- Top route (series). $$\frac{8\times 4}{8+4}=\frac{32}{12}=\frac{8}{3}\,\text{F}$$
- Box = top ∥ bottom (parallel). $$\frac{8}{3}+4=\frac{8}{3}+\frac{12}{3}=\frac{20}{3}\,\text{F}$$
- Whole chain (series), LCD 240. $$\frac{1}{C_{AB}}=\frac{1}{12}+\frac{3}{20}+\frac{1}{16}=\frac{20+36+15}{240}=\frac{71}{240}$$
- Invert. $$C_{AB}=\frac{240}{71}\,\text{F}$$
Inner box before outer chain. Two parallel routes collapse to \(\tfrac83+4=\tfrac{20}{3}\). Then three in series — for the reciprocals use LCD 240: \(\tfrac{1}{12},\tfrac{3}{20},\tfrac{1}{16}\) become \(\tfrac{20}{240},\tfrac{36}{240},\tfrac{15}{240}\), summing to \(\tfrac{71}{240}\).
A series result must sit below the smallest member (below \(\tfrac{20}{3}\approx 6.7\)); \(\tfrac{240}{71}\approx 3.4\) fits. The tiny \(\tfrac{1}{31}\) and the other fractions come from skipping the box reduction.
Box reduces to \(\tfrac{20}{3}\) F; then 12, \(\tfrac{20}{3}\) and 16 F in series:
$$\frac{1}{C_{AB}}=\frac{1}{12}+\frac{3}{20}+\frac{1}{16}=\frac{71}{240}\;\Rightarrow\;C_{AB}=\frac{240}{71}\,\text{F}$$