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Question-8674




Question Number 8674 by tawakalitu last updated on 20/Oct/16
Answered by sandy_suhendra last updated on 21/Oct/16
when series :  C_1 =0.10 μF  C_2 =0.20 μF  (1/C_(series) ) = (1/(0.10))+(1/(0.20))=(3/(0.20)) ⇒ C_(series) =((0.20)/3) μF  the charge ⇒Q_(series) =Q_1 =Q_2 =C_(series) .V=((0.20)/3)×100=6.67 μC    when parallel :  Q_1 +Q_2 =C_(parallel) .V_(parallel)   6.67+6.67=(0.10+0.20).V_(parallel)   V_(paeallel) =((13.34)/(0.30))=44.47 volt
$$\mathrm{when}\:\mathrm{series}\:: \\ $$$$\mathrm{C}_{\mathrm{1}} =\mathrm{0}.\mathrm{10}\:\mu\mathrm{F} \\ $$$$\mathrm{C}_{\mathrm{2}} =\mathrm{0}.\mathrm{20}\:\mu\mathrm{F} \\ $$$$\frac{\mathrm{1}}{\mathrm{C}_{\mathrm{series}} }\:=\:\frac{\mathrm{1}}{\mathrm{0}.\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{0}.\mathrm{20}}=\frac{\mathrm{3}}{\mathrm{0}.\mathrm{20}}\:\Rightarrow\:\mathrm{C}_{\mathrm{series}} =\frac{\mathrm{0}.\mathrm{20}}{\mathrm{3}}\:\mu\mathrm{F} \\ $$$$\mathrm{the}\:\mathrm{charge}\:\Rightarrow\mathrm{Q}_{\mathrm{series}} =\mathrm{Q}_{\mathrm{1}} =\mathrm{Q}_{\mathrm{2}} =\mathrm{C}_{\mathrm{series}} .\mathrm{V}=\frac{\mathrm{0}.\mathrm{20}}{\mathrm{3}}×\mathrm{100}=\mathrm{6}.\mathrm{67}\:\mu\mathrm{C} \\ $$$$ \\ $$$$\mathrm{when}\:\mathrm{parallel}\:: \\ $$$$\mathrm{Q}_{\mathrm{1}} +\mathrm{Q}_{\mathrm{2}} =\mathrm{C}_{\mathrm{parallel}} .\mathrm{V}_{\mathrm{parallel}} \\ $$$$\mathrm{6}.\mathrm{67}+\mathrm{6}.\mathrm{67}=\left(\mathrm{0}.\mathrm{10}+\mathrm{0}.\mathrm{20}\right).\mathrm{V}_{\mathrm{parallel}} \\ $$$$\mathrm{V}_{\mathrm{paeallel}} =\frac{\mathrm{13}.\mathrm{34}}{\mathrm{0}.\mathrm{30}}=\mathrm{44}.\mathrm{47}\:\mathrm{volt} \\ $$
Commented by tawakalitu last updated on 21/Oct/16
Thank you sir. i appreciate.
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{appreciate}. \\ $$

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