Question

A $4\mu F$ capacitor is charged to $50V$ and another capacitor of $2\mu F$ is charged to $100V$. The two condensers are connected such that their charged plates are connected together. The total energy of the system before and after joining will be: (in multiple of ${10}^{-2}J$)

• A. $3.0$ and $2.67$
• B. $2.67$ and $3.0$
• C. $1.5$ and $1.33$
• D. $1.33$ and $1.5$

Answer 1

The correct option is C $1.5$ and $1.33$

Total energy before joining is $U_b=\frac{1}{2}{C}_1{V}_1^2+\frac{1}{2}{C}_2{V}_2^2$

$U_b=\frac{1}{2}\times 4\times {10}^{-6}\times (50{)}^2+\frac{1}{2}\times 2\times {10}^{-6}\times (100{)}^2=0.5\times {10}^{-2}+{10}^{-2}=1.5\times {10}^{-2}J$

After joining the common potential is

$V_c=\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}=\frac{(4\times 50)+(2\times 100)}{4+2}=66.67V$

Total energy after joining is $U_a=\frac{1}{2}(C_1+C_2)V_c^2$

$U_a=\frac{1}{2}(4+2)\times {10}^{-6}\times (66.67{)}^2=1.33\times {10}^{-2}J$