Figure shows a network of seven capacitors. If charge on 5 μF capacitor is 10 μC, find the potential difference between points A and C (in V).

Charge across 5 μF capacitor 10 μC.Potential difference across 5 μF capacitor is qC=10×10−65×10−6=2voltAs 3, 4, and 5 μF capacitors are in parallel, so potential difference across each of the them equals 2 V.Charge on 3μF capacitor =CV=3×10−6×2=6μCCharge on 4μF capacitor 4×10−6×2=8μCTotal charge flowing in upper branch of circuit is10μC+6μC+8μC=24μCPotential difference across C2 is 24μC4μC=6VTotal potential difference across AB is = 6 + 2 = 8 V.Equivalent capacitance of lower branch of circuit is 6×36+3=2μFSo, charge flowing is =2×10−6×8=16μCTherefore, potential difference between A and C is16μC3μF=163=5.33V