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A capacitor C1=1.0 μF is charged up to a voltage V=60 V by connecting it to battery B through switch (1). Now C1 is disconnected from batter

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\begin{array}{l} Q=1 \mu F \times 60 \\ =60 \mu C \\ 1(x-0)+2(x-0) \\ \Rightarrow 3 x=60 \\ \Rightarrow x=20 \\ \therefore Q 2 \mu F=20 \times 2=40 \mu C \end{array}

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Satyajeet Kumar

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