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CGP EDU Academic Team
Published on: September 12, 2026
Two capacitors of capacitances $3 \mu F$ and $6 \mu \mathrm{F}$ 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
Text Solution
Verified by ExpertsThe correct answer is:
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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