Question

A capacitor of capacitance $4\mu \text{F}$ is charged upto $80\text{V}$ and another capacitor of capacitance $6\mu \text{F}$ is changed upto $30\text{V}$. When they are connected together the energy lost by the $4\mu \text{F}$ capacitor is

• A. $2.5\text{mJ}$
• B. $3.2\text{mJ}$
• C. $4.6\text{mJ}$
• D. $7.8\text{mJ}$

Answer 1

The correct option is D $7.8\text{mJ}$

$V_{\text{common}}=\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}=\frac{(4\times 80)+(6\times 30)}{(4+6)}$

$V_c=\frac{500}{10}=50\text{V}$

Initial and final energy of $4\mu \text{F}$ capacitor is,

$U_i=\frac{1}{2}{C}_1{V}_1^2$

$U_f=\frac{1}{2}{C}_1{V}_c^2$

Energy loss is, $\Delta U=U_i-U_f$

$=\frac{1}{2}{C}_1[V_1^2-V_c^2]$

$=\frac{1}{2}\times 4\times {10}^{-6}\left[\right(80{)}^2-\left(50{)}^2\right]$

$=2\times {10}^{-6}\times 3900$

$=7.8\times {10}^{-3}\text{J}=7.8\text{mJ}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (d) is the correct answer.