Find charge on positive plate of 8 μF (assuming before connecting to battery all capacitors were uncharged)
A
36 μC
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B
12 μC
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C
24 μC
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D
6 μC
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Solution
The correct option is C 24 μC Let s represent series and p represent parallel then to calculate effective capacitance, circuit can be represented by {12μF}(s){8μF(p)40μF} which is equivalent to {12μF}(s){12μF} =6μF ⇒qsupplied=Ceff.ΔV=6×6=36μC Now on 8μF&4μF charge distributes propotional to their capacitance. Then charge on 8 μF capacitor is given by q=412×36=24μC