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Question Number 55970 by necx1 last updated on 07/Mar/19

Answered by tanmay.chaudhury50@gmail.com last updated on 07/Mar/19

$$10\mu Fand10\mu Fareparallel$$$$so10+10=20\mu F$$$$15\mu Fand5\mu Fareparallel$$$$so15+5=20\mu F$$$$nowthis20\mu Fand20\mu Fareseries$$$$c_{eq}=\frac{1}{\frac{1}{20}+\frac{1}{20}}=\frac{20}{2}=10\mu F$$$$nowthis10\mu Fand20\mu Fareparallel$$$$so10+20=30\mu F$$$$Qcharge=30\times {10}^{-6}\times 10=300\times {10}^{-6}Columb...$$$$C_{15}:C_5:C_{20}=15:5:20=3:1:4$$$$soithinkchargeon5\mu F$$$$is=\frac{1}{3+1+4}\times 300\times {10}^{-6}=37.5\times {10}^{-6}C$$$$\mathit{P}lscommentandrectifyifiamwrong...$$$$$$

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