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In the circuit diagram shown all the capacitors are in mF the equivalent capacitance between points A and B is
The capacitors with capacitance areC1 = 6 C2 = 8 C3 = 10 C4 = 3 C5 = 4 and C6 = 7The capacitors C1 , C2 and C4 , C5 forms Wheatstone bridge such that 6/8 = 3/4So C3 is eliminatedNow C2 and C5 are in seriesSo C25 = 8×4/8+4 = 8/3 Now C1 and C4 are in seriesSo C14 = 6×3/6+3 = 2 C14 and C25 are parallelCeq1 = 8/3 + 2 = 14/3 Ceq1 and C6 are in series Ceq = (14/3 × 7) ÷ (14/3 + 7) = (14 × 7)/3 ÷ 35 / 3×7 = 14×7×7 / 35 = 19.6 mF
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