Question

A capacitor of capacitance $6\mu F$ is charged to a potential of $150V$. Its potential falls to 90V. When another capacitor is connected to it. Find the capacitance of the second capacitor and the amount of energy lost due to the connection.

Answer 1

Given,

$\begin{array}{c}C=6\mu F\\ V=150Vother90V\end{array}$

$Q$ is remain conserved .

$\begin{array}{c}{C}_1{V}_1=\left(C_1+C_2\right)V_2\\ 6\mu \times 150=\left(6\mu +c\right)90\\ \left(6\mu +C_2\right)=\frac{6\mu \times 150}{90}=10\mu \end{array}$

$C_2=4\mu F$

Energy loss $=\left(E_f-E_i\right)$

$\begin{array}{c}=\frac{1}{2}{C}_f{V}^2-\frac{1}{2}{C}_i{V}^2\\ =\frac{u}{2}\left(10\mu \times {\left(90\right)}^2\right)-\left(6\mu \times {\left(150\right)}^2\right)\\ =\frac{54000}{2}\times {10}^{-6}\\ =27\times {10}^{-3}\\ Loss=0.027J\end{array}$