Question

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 V
• B. 4 V
• C. 3 V
• D. Zero

Answer 1

Answer: (B)

Here $q_1 = C_1V = 3 \times 10^{-6} \times12 $ = 36 $\mu$C and
$q_2 = C_2V= 6 \times 10^{-6} \times 12 $ = 72 $\mu$C
Net charge = 72 - 36 = 36 $\mu$C
This charge gets distributed among two capacitors.
$\therefore $ $V_1 = \frac{12}{3} $ = 4V and $V_2 = \frac{24}{6}$ = 4 V