Question

A capacitor of capacitance value $1\mu F$ is charged to $30V$ and the battery is then disconnected. If the remaining circuit is connected across a $2\mu F$ capacitor, the energy lost by the system is

• A. $300\mu J$
• B. $450\mu J$
• C. $225\mu J$
• D. $245.0J$
• E. $100\mu J$

Answer 1

Answer: (A)

Energy $E_1=\frac{1}{2}{C}_1{V}_1^2=\frac{1}{2}\times 1\times {10}^{-6}\times {(30)}^2$
$=450\times {10}^{-6}J$ Common potential
$V=\frac{q_1+q_2}{C_1+C_2}$
$=\frac{1\times 30+0}{1+2}=10volt$
$E_2=\frac{1}{2}(C_1+C_2)V^2$
$=\frac{1}{2}(1+2)\times {10}^{-6}\times {(10)}^2$
$=1.5\times 100\times {10}^{-6}$
$=150\times {10}^{-6}J$
Loss of energy $=E_2-E_1=300\mu J$