Question

Two capacitors of capacitance $3\mu F$ and $6\mu F$ are charged to a potential of $12V$ each. They are now connected to each other with the positive plate of joined to the negative plate o the other.The potential difference across each will be:

• A. $3v$
• B. $Zero$
• C. $6v$
• D. $4v$

Answer 1

The correct option is D $4v$

$C_1=3\mu F$

$C_2=6\mu F$

$V=12V$ each

$\therefore Q=CV{Q}_1=36\mu F{Q}_2=72\mu F$

After positive and negative plates of both capacitors joined together. then,

$Q_{net}=36\mu C(72-36\mu F)$

and $C_{net}=(3+6)\mu F$

$\therefore$ or $V=\frac{Q}{C}$

$V=\frac{36\mu C}{9\mu F}$

or $V=4V$