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Complex capacitor networks appear regularly in JEE. A systematic approach saves time and prevents errors.

Step 1 — Identify topology: Determine which capacitors are in series (same charge flows through them) and which are in parallel (same voltage across them). Redraw the circuit if needed to clarify connections.

Step 2 — Check for symmetry: If the network has a symmetry axis, points on the axis are at equal potential. Any capacitor connecting two equipotential points carries zero charge and can be removed (Wheatstone bridge condition). This dramatically simplifies many problems.

Step 3 — Reduce series-parallel groups: Combine series groups (1/$C_{eq}$ = sum of 1/$C_i$) and parallel groups ($C_{eq}$ = sum of $C_i$) iteratively until a single equivalent capacitance remains.

Step 4 — For irreducible networks: Use Kirchhoff's laws adapted for capacitors. At each node, charge is conserved (sum of charges = total charge supplied). Around each loop, the sum of potential drops = applied voltage. This gives simultaneous equations to solve.

Step 5 — Star-delta transformation: For networks with three capacitors in a star (Y) or delta configuration, convert one to the other. Star to delta: $C_{ab}$ = $\frac{C_a*C_b + C_b*C_c + C_a*C_c}{C_c}$ (where c is the opposite arm).

Common pitfalls: (1) Not recognizing that capacitors in series have the same charge, not the same voltage. (2) Forgetting that voltage across a capacitor is V = $\frac{Q}{C}$, which means the smaller capacitor has the larger voltage in series. (3) Confusing series/parallel combinations with resistor rules (they are swapped).