Question

A $4\mu F$ capacitor, charged to $50V$, is connected to another $2\mu F$ capacitor, charged to $100V$. Final energy of the combination is:

• A. $13.33\times {10}^{-3}J$
• B. $20\times {10}^{-3}J$
• C. $5\times {10}^{-3}J$
• D. $10\times {10}^{-3}J$

Answer 1

Answer: (A)

After connecting capacitors, the potential of two capacitor will be same.
And total charge will be conserved.
$\Rightarrow$ $Q_f=Q_i$
$\Rightarrow$ $C_1{V}_1+C_2{V}_2=(C_1+C_2)V$
$\Rightarrow$ $4\times 50+2\times 100=(4+2)V$
$\Rightarrow$ $V=\frac{400}{6}=\frac{200}{3}$
$\therefore$ final energy $=\frac{1}{2}(C_1+C_2)V^2=\frac{1}{2}(4+2)\times {10}^{-6}\times {\left(\frac{200}{3}\right)}^2$$=13.33\times {10}^{-3}J$