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Asked in CBSE AIPMT 1999 · Capacitor networks
Nodes: A (joined to X), B, C, D (joined to Y). C₁: A–B; C₃: A–C; C₅: C–B; C₂: B–D; C₄: C–D.
This is a bridge with C₅ across the middle.
Balance test: (C₁)/(C₂)=6/6=(C₃)/(C₄), so B and C are at the same potential and C₅ holds no charge.
Drop C₅: arm A–B–D is 6 and 6 in series =3 μF; arm A–C–D is also 3 μF.
The arms are in parallel: C_XY=3+3=6 μF.
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