Three parallel connected conductors A, B and C have a total charge of 48µC. The ratio of their capacitances are 1 : 3 : 2. The charges on them individually are

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8µC,18µC, 22µC

16µC,16µC,16µC

8µC, 24µC,16µC

24µC,12µC,12µC

answer is C.

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Detailed Solution

$\begin{array}{l} Q_{1} = \frac{C_{1} Q}{C_{1} + C_{2} + C_{3}} = \frac{1 \times 10^{- 6} \times 48 \times 10^{- 6}}{(1 + 3 + 2) \times 10^{- 6}} = 8 μC \\ Q_{2} = \frac{C_{2} Q}{C_{1} + C_{2} + C_{3}} = \frac{3 \times 10^{- 6} \times 48 \times 10^{- 6}}{(1 + 3 + 2) \times 10^{- 6}} = 24 μC \\ Q_{3} = \frac{C_{2} Q}{C_{1} + C_{2} + C_{3}} = \frac{2 \times 10^{- 6} \times 48 \times 10^{- 6}}{(1 + 3 + 2) \times 10^{- 6}} = 16 μC \end{array}$

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