Question

Two capacitors of capacitances $3~μ F$ and $6~μ F$ 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:

• A. 4 V
• B. 12 V
• C. zero
• D. 3 V

Hint:

When opposite plates are connected, charges subtract.

Answer 1

The Correct Option is A

Step 1: Initial charges: $Q₁ = 36 μ C$, $Q₂ = 72 μ C$. Net charge = $72 - 36 = 36 μ C$.

Step 2: Equivalent capacitance = $3 + 6 = 9 μ F$. $V = \frac369 = 4$ V.