Question

A $20F$ capacitor is charged to $5V$ and isolated. It is then connected in parallel with an uncharged $30F$ capacitor. The decrease in the energy of the system will be :

• A. $25J$
• B. $100J$
• C. $125J$
• D. $150J$

Answer 1

The correct option is D $150J$

Initial energy $=\frac{1}{2}C{V}^2$
$=\frac{1}{2}\times 20\times 25$ $=250J$
$Q=CV$
$Q=20\times 5$
$Q=100C$
Now another capacitor is connected in parallel, total charge is constant and voltage across the capacitors is same.
$100=(C_1+C_2)V=50V\Rightarrow V=2V$
Now energy$=\frac{1}{2}\times (C_1+C_2)(2{)}^2$
$=100$
$=100$
decrease in energy $=250-100=150J$