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. 25 J
• B. 200 J
• C. 125 J
• D. 150 J

Answer 1

Answer: (D)

Charge stored on $20F$ capacitor $=20\times 5=100C$
When $20F$ capacitor is connected to $30F$ capacitor,
$V=\frac{q}{C_1+C_2}=\frac{100}{20+30}=2V$
Energy of the system,
$E_f=\frac{1}{2}\times (30+20)\times (2{)}^2$
$=100J$
Energy before isolation,
$E_i=\frac{1}{2}\times 20\times (5{)}^2=250J$
Therefore, decrease in energy $=250-100=150J$