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Asked in GUJCET 2008 · Capacitor networks
Given: capacitors of 4 μF in X–M, M–Y, X–N and N–Y, and a 4 μF capacitor between M and N.
Idea: this is a bridge. It is balanced because (C_XM)/(C_MY) = (C_XN)/(C_NY) (both ratios are 1), so M and N are at the same potential.
The middle capacitor then has no potential difference across it and holds no charge, so it can be removed.
Each branch: 4 and 4 μF in series give (4×4)/(4+4) = 2 μF.
The two branches in parallel: C_XY = 2 + 2 = 4 μF.
So the equivalent capacitance is 4 μF.
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