Two capacitors of capacitances $3\,\mu \,F$ and $6\,\mu F$ are charged to a potential of $12 \,V$ each. They are now connected to each other, with the positive plate of each joined to the negative plate of the other. The potential difference across each will be......$volt$
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(b) $V = \frac{{{C_1}{V_1} - {C_2}{V_2}}}{{{C_1} + {C_2}}} = \frac{{6 \times 12 - 3 \times 12}}{{3 + 6}} = 4\,volt$
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