Question

A $20\, F$ capacitor is charged to $5\, V$ and isolated. It is then connected in parallel with an uncharged $30\, F$ 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 $20\, F$ capacitor $=20 \times 5=100\, C$
When $20\, F$ capacitor is connected to $30\, F$ capacitor,
$V=\frac{q}{C_{1}+C_{2}}=\frac{100}{20+30}=2\, V$
Energy of the system,
$E_{f}=\frac{1}{2} \times(30+20) \times(2)^{2}$
$=100\, J$
Energy before isolation,
$E_{i}=\frac{1}{2} \times 20 \times(5)^{2}=250\, J$
Therefore, decrease in energy $=250-100=150\, J$