Question

Two identical capacitors each of capacitance $5\mu F$ are charged to potential $2KV$ and $1KV$ respectively. There negative ends are connected together, when the positive ends are also connected together, the loss of the energy of the system is.

Answer 1

Step 1: Given data

$C_{1}and{C}_2$= $5\mu F$
$V_1=2KV$

$V_2=1KV$

Step 2: Write the formula

Loss of energy: $\mathit{U}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\frac{{\mathbf{C}}_{\mathbf{1}\mathbf{}}{\mathbf{C}}_{\mathbf{2}}}{(C_1+C_2)}{\mathbf{(}{\mathbf{V}}_{\mathbf{1}}\mathbf{-}{\mathbf{V}}_{\mathbf{2}}\mathbf{)}}^{\mathbf{2}}$

Where,

$U$ is the energy loss.

$C_{1}and{C}_2$ are the two identical capacitors.

$V_{1}and{V}_2$ are the potential.

Step 3: Write the required conversions.

$1\mu F=1\times {10}^{-6}F$

$1KV=1000V$

Step 4: Find the energy loss of the system.

Replacing the values in the formula.

$\begin{array}{rcl}\mathit{U}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{2}}\frac{{\mathbf{C}}_{\mathbf{1}\mathbf{}}{\mathbf{C}}_{\mathbf{2}}}{(C_1+C_2)}{\mathbf{(}{\mathbf{V}}_{\mathbf{1}}\mathbf{-}{\mathbf{V}}_{\mathbf{2}}\mathbf{)}}^{\mathbf{2}}\\ U& =& \frac{1}{2}\frac{5\times {10}^{-6}\times 5\times {10}^{-6}{\left(2000-1000\right)}^2}{\left(5+5\right)\times {10}^{-6}}\\ U& =& \frac{5\times 5}{2\times 10}\\ U& =& 1.25J\end{array}$

Therefore, the loss of energy in the system is $\mathbf{1}\mathbf{.}\mathbf{25}\mathit{J}$.