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In the network shown, if the potential difference between the plates of 3μF capacitor is 5 volt, the charge stored in 2μF capacitor is 

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a
22.5 μC
b
10 μC
c
7.5 μC
d
15 μC

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detailed solution

Correct option is D

Charge on 3μF capacitor =Q1=3×5μC=15μC∴ Charge on 3μF capacitor =15μCCe=Equivalent capacitance of 3μF and 6μF capacitors =3×63+6μF=2μF∴ VA−VB=QCe=152volt=7.5 volt∴ Charge on 2μF capacitor =2×(VA−VB)=2×7.5 μC=15 μC


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Two capacitors C1 and C2 having capacitance 2μF and 4μF respectively are connected as shown in the figure. Initially C1 has charge 4μC and C2 is uncharged. After long time closing the switch S :


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