Q. Two capacitors of capacitance $3\mu\ F$ 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
A) $6\ volt$
B) $4\ volt$
C) $3\ volt$
D) Zero
Explanation
(b) $V = \frac{C_{1}V_{1} - C_{2}V_{2}}{C_{1} + C_{2}} = \frac{6 \times 12 - 3 \times 12}{3 + 6} = 4\ volt$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.