A body of capacity $4 \mu \mathrm{F}$ is charged to $80 \mathrm{~V}$ and another body of capacity $6 \mu…

A body of capacity $4 \mu \mathrm{F}$ is charged to $80 \mathrm{~V}$ and another body of capacity $6 \mu \mathrm{F}$ is charged to $30 \mathrm{~V}$. When they are connected the energy lost by $4 \mu \mathrm{F}$ capacitor is
  1. $9.8 \mathrm{~mJ}$
  2. $4.6 \mathrm{~mJ}$
  3. $3.2 \mathrm{~mJ}$
  4. $2.5 \mathrm{~mJ}$

Solution

$\mathrm{C}_2=4 \mu \mathrm{F}, V_1=80$ volt, $C_2=6 \mu \mathrm{F}, V_2=30$ volt Energy loss, $\begin{aligned} \Delta U & =\frac{1}{2} \frac{C_1 C_2}{C_2+C_2}\left(V_1-V_2\right)^2 \\ & =\frac{1}{2} \times \frac{4 \times 10^{-6} \times 6 \times 10^{-4}}{4 \times 10^{-6}+6 \times 10^{-4}}(80-30)^2 \\ & =\frac{12 \times 10^{-6}}{10} \times 2500=1.2 \times 25 \times 10^{-4} \\ & =30 \times 10^{-4}=3 \times 10^{-3} \mathrm{~J}\end{aligned}$ Initial energy of $4 \mu \mathrm{F}$ capacitor. $\begin{aligned} E_{\mathrm{n}} & =\frac{1}{2} C_1 V_1^2 \\ & =\frac{1}{2} \times 4 \times 10^{-4} \times(80)^2 \\ & =2 \times 10^{-4} \times 6400 \\ & =12.8 \times 10^{-3} \mathrm{~J}\end{aligned}$ Energy lost by $C_1=12.8 \times 10^{-1}-3 \times 10^{-1}$ $\begin{aligned} & =9.8 \times 10^{-3} \mathrm{~J} \\ & =9.8 \mathrm{~mJ}\end{aligned}$ .

Asked in: MHT CET Full Test 13

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