Question

A $2\mu F$ capacitor $C_1$ is charged to a voltage $100V$ and a $4\mu F$ capacitor $C_2$ is charged to a voltage $50V$. The capacitors are then connected in parallel. What is the loss of energy due to parallel connection?

• A. $1.7\times {10}^{-4}J$
• B. $1.7\times {10}^{-5}J$
• C. $1.7\times {10}^{-2}J$
• D. $1.7\times {10}^{-3}J$

Answer 1

The correct option is D $1.7\times {10}^{-3}J$

The initial energy is $U_i=\frac{1}{2}{C}_1{V}_1^2+\frac{1}{2}{C}_2{V}_2^2$

After connecting in parallel ,the common potenial will be

$V_c=\frac{Q_1+Q_2}{C_1+C_2}$

$=\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}$

$=\frac{(2\times 100)+(4\times 50)}{2+6}$

$=\frac{400}{6}$

$=\frac{200}{3}V$

Final energy is $U_f=\frac{1}{2}(C_1+C_2)V_c^2$

Energy loss $=U_i-U_f$

$=\frac{1}{2}{C}_1{V}_1^2+\frac{1}{2}{C}_2{V}_2^2-\frac{1}{2}(C_1+C_2)V_c^2$

$=[\left(\frac{1}{2}\right)\times 2\times (100{)}^2+\left(\frac{1}{2}\right)\times 4\times (50{)}^2-\left(\frac{1}{2}\right)\times (2+4)\times {\left(\frac{200}{3}\right)}^2]\times {10}^{-6}$

$=1.7\times {10}^{-3}$