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Question Number 124932 by aurpeyz last updated on 07/Dec/20

Answered by som(math1967) last updated on 07/Dec/20

$$\mathrm{Equivalent}\mathrm{capacitance}$$$$\frac{1}{\mathrm{C}}=\frac{1}{4+2}+\frac{1}{3}=\frac{3}{6}$$$$\mathrm{C}=2\mu \mathrm{F}$$$$\mathrm{charge}=12\times 2=24\times {10}^{-6}\mathrm{coulomb}$$$$\therefore \mathrm{charge}\mathrm{on}3\mu \mathrm{F}=24\times {10}^{-6}\mathrm{coulombans}$$$$\mathrm{p}.\mathrm{d}\mathrm{across}3\mu \mathrm{F}=\frac{24}{3}=8\mathrm{v}$$$$\therefore \mathrm{p}.\mathrm{d}\mathrm{across}4\mu \mathrm{F}\mathrm{and}2\mu \mathrm{F}$$$$12-8=4\mathrm{v}$$$$\mathrm{charge}\mathrm{on}2\mu \mathrm{F}=4\times 2=8\times {10}^{-6}\mathrm{coulomb}\mathrm{ans}$$$$\mathrm{charge}\mathrm{on}4\mu \mathrm{F}=4\times 4=16\times {10}^{-6}\mathrm{coulomb}\mathrm{ans}$$

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